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Let $X$ be a normed space, $Y$ a Banach space, and $A_n:X\to Y$ be bounded linear operators with $C={\sup }_n\Vert A_n\Vert <\infty$. Convergence of $A_{n}d$ for every $d$ in a dense subset $D$ implies convergence of $A_{n}x$ for every $x\in X$ to a bounded linear operator $A$, with $\Vert A\Vert \le {\mathrm{liminf}}_n\Vert A_n\Vert$. Indeed, $\Vert A_{n}x-A_{m}x\Vert \le 2C\Vert x-d\Vert +\Vert A_{n}d-A_{m}d\Vert$ makes $(A_{n}x)$ a Cauchy sequence; completeness gives its limit, and linearity and the bound pass to that limit. If $X$ is a Banach space, pointwise convergence conversely implies the uniform bound by the Uniform boundedness principle. If the limit bounded linear operator is specified in advance, completeness of $Y$ is unnecessary.

For a bounded operator $K$ between Hilbert spaces, compactness is equivalent to $f_n\rightharpoonup f$ implying $K{f}_n\to Kf$ in norm. The Uniform boundedness principle bounds a weakly convergent sequence. Relative compactness of its images and the unique possible weak limit then give norm convergence. Conversely, every bounded domain sequence has a weakly convergent subsequence, so the image of the unit ball is relatively compact.

A bounded linear operator between Banach spaces is completely continuous when it maps weakly convergent sequences to norm-convergent sequences. Equivalently, it maps weakly compact sets to norm-compact sets: the Eberlein-Šmulian theorem gives the forward direction, while a weakly convergent sequence together with its limit is weakly compact and gives the reverse direction. The inclusion $C(\mathbb{T})\to L^2(\mathbb{T})$ for finite Lebesgue measure is completely continuous by pointwise convergence, the Uniform boundedness principle and the dominated convergence theorem, but is not a compact operator because of the orthogonal Fourier modes.

For operators on a complex Hilbert space, pointwise boundedness of $⟨Tx,x⟩$ over a family implies uniform boundedness of their operator norms. The complex polarization identity recovers each $⟨Tx,y⟩$ from four diagonal values. Apply the Uniform boundedness principle first to the functionals in $y$ for fixed $x$, then to the operators in $x$. On a real Hilbert space the claim fails: multiples of a nonzero skew-symmetric operator have zero diagonal forms but unbounded norms. For example, rotations through a quarter turn on ${\mathbb{R}}^2$ multiplied by integers suffice.

For each integer $m\ge 1$, the set of continuous functions satisfying $|S_N(f,0)|\le m$ for every $N$ is closed in the uniform norm. If it contained a ball of radius $r$, subtracting two members of that ball would bound every partial-sum functional norm by $4m/r$. The Dirichlet kernel harmonic lower bound excludes this, so these sets are nowhere dense. By the Baire category theorem, their complement is a dense $G_{\delta }$ set. Every function in that complement has unbounded Fourier partial sums at zero. Translation gives the same conclusion at any prescribed point. This is a direct Uniform boundedness principle argument on the real Banach space of continuous functions on the circle.

Every algebraically irreducible representation of a Banach algebra that is normed is continuous. The continuous orbit vectors form an invariant subspace. If all nonzero orbit maps were discontinuous, apply the countable annihilation sequence in an irreducible Banach module and the gliding-hump continuity principle to right multiplication on the algebra and evaluation of the operator-valued representation. Off-diagonal maps vanish, whereas diagonal maps are discontinuous, a contradiction. Once every orbit map is continuous, the Uniform boundedness principle on the Banach algebra gives the common operator-norm bound. Completing the original normed representation space supplies the Banach operator space needed in the argument.

For orthonormal Legendre polynomials on $[-1,1]$, the polynomial $q_n$ represents evaluation at zero on polynomials of degree at most $n$: $⟨r,q_n⟩=r(0)$. This follows by expanding $r$ in the orthonormal basis, or from the Riesz representation theorem. With $M=\lfloor n/2\rfloor$ and $c_m=\left(\genfrac{}{}{0ex}{}{2m}{m}\right)/4^m$, orthonormality givesThe last identity telescopes using $c_m/c_{m-1}=(2m-1)/(2m)$. The norms are unbounded: $c_m\ge 1/(2m+1)$ makes the sum dominate a divergent harmonic series. Thus the evaluation identities hold eventually for every fixed polynomial but cannot extend to convergence for every $L^2$ vector, by the Uniform boundedness principle. Point evaluation is also not defined on $L^2$ equivalence classes.

Use the convention that the inner product is linear in its first argument. If $v\ne 0$, nonnegativity of $\Vert u-⟨u,v⟩v/\Vert v{\Vert }^2{\Vert }^2$ givesThe case $v=0$ is immediate. This Cauchy-Schwarz inequality shows that $f_v$ is bounded linear with norm at most $\Vert v\Vert$. For $v\ne 0$, evaluation at $v/\Vert v\Vert$ gives equality; for $v=0$ both norms vanish. The Banach dual is the space of all bounded linear functionals with the operator norm.

The Riesz representation theorem asserts that every $f\in V^{\prime }$ is uniquely $f(u)=⟨u,v⟩$ for some $v\in V$, with $\Vert f\Vert =\Vert v\Vert$. If $f=0$, take $v=0$. Otherwise put $W=\ker f$, a closed subspace, and use the projection theorem to obtain $w\in W^{\perp }$ with $f(w)=1$. Since $u-f(u)w\in W$, $⟨u,w⟩=f(u)\Vert w{\Vert }^2$. Thus $v=w/\Vert w{\Vert }^2$ represents $f$. If two vectors represent it, their difference has inner product zero with every vector, in particular itself, giving uniqueness. The norm statement was already proved.

The space $P_n$ is finite-dimensional and hence complete; evaluation at zero is bounded there by equivalence of finite-dimensional norms. Riesz therefore supplies exactly one $q_n\in P_n$. In any orthonormal polynomial basis,For the real Legendre functions the bars may be omitted.

The printed even-degree normalization has an error. Orthogonality forces a degree-two polynomial to be a scalar multiple of $3t^2-1$, whose squared integral on $[-1,1]$ is $8/5$. Hence any orthonormal choice has $|p_2(0){|}^2=5/8$, whereas the printed formula gives $3/8$. The factor under its square root must be $4m+1$ instead of $2m+1$. Signs of the individual polynomials can be chosen so that the values at zero are positive; these signs do not affect $q_n$ or its norm.

Put $M=\lfloor n/2\rfloor$ and $c_m=\left(\genfrac{}{}{0ex}{}{2m}{m}\right)/4^m$. With the consistent orthonormal normalization the norm isThe finite sum telescopes because $c_m=(2m-1)c_{m-1}/(2m)$, so $(2m+1{)}^2{c}_m^2-(2m-1{)}^2{c}_{m-1}^2=(4m+1)c_m^2$. If one substitutes the erroneous printed values formally instead, one obtains $\frac{1}{2}{\sum }_{m=0}^M(2m+1)c_m^2$; that is not the norm of the actual orthonormal Legendre evaluation representer.

For a polynomial $r$ of degree $d$, the defining identity gives $⟨r,q_n⟩=r(0)$ for every $n\ge d$, so the requested limit is eventually constant. It does not extend to all of $L^2$. First, evaluation at a point is not well-defined on equivalence classes. More strongly, the functional norms $\Vert q_n\Vert$ are unbounded. Indeed the largest binomial coefficient in row $2m$ is at least the average, so $c_m\ge 1/(2m+1)$, and the displayed sum dominates a divergent constant multiple of the harmonic series. The Uniform boundedness principle then provides an $L^2$ function for which ${\sup }_n|⟨r,q_n⟩|=\infty$. Thus no choice of point representative can restore the assertion for all $L^2$ functions.

The Baire category theorem states that a nonempty complete metric space cannot be a countable union of closed nowhere dense sets. The Uniform boundedness principle says that if $V$ is a Banach space, $W$ a normed vector space, and $\mathcal{T}\subseteq B(V,W)$ is pointwise bounded, meaning ${\sup }_{T\in \mathcal{T}}\Vert Tx\Vert <\infty$ for every $x\in V$, then ${\sup }_{T\in \mathcal{T}}\Vert T\Vert <\infty$.

To prove it, let $E_n=\{x:{\sup }_{T\in \mathcal{T}}\Vert Tx\Vert \le n\}$. Each set is closed by continuity of the individual operators, and their union is $V$. The Baire category theorem gives a ball $B(x_0,r)$ inside some $E_N$. For $\Vert h\Vert <r$, both $x_0+h$ and $x_0$ are in $E_N$, so $\Vert Th\Vert \le \Vert T(x_0+h)\Vert +\Vert T{x}_0\Vert \le 2N$ for every $T$. Scaling gives $\Vert T\Vert \le 2N/r$ uniformly. The zero-space and empty-family cases are immediate.

For the first family on a Hilbert space, fix $x$. The functionals $y\mapsto ⟨y,Tx⟩$ have operator norm $\Vert Tx\Vert$ by the Cauchy-Schwarz inequality and equality on the direction of $Tx$. Their values are pointwise bounded by the hypothesis, after taking complex conjugates if necessary. Apply the Uniform boundedness principle to obtain ${\sup }_{T\in \mathcal{E}}\Vert Tx\Vert <\infty$. Apply it a second time to the operator family itself to conclude ${\sup }_{T\in \mathcal{E}}\Vert T\Vert <\infty$.

For the diagonal hypothesis, the answer depends on the scalar field. Over $\mathbb{C}$, let $q_T(x)=⟨Tx,x⟩$, using the convention that the inner product is linear in its first variable. The complex polarization identity givesThe assumed bounds at these four fixed vectors make every matrix coefficient uniformly bounded, so the preceding argument applies. Thus the complex-space answer is yes, even without assuming self-adjointness.

Over $\mathbb{R}$, take $X={\mathbb{R}}^2$ andEvery diagonal value $⟨nJx,x⟩$ is zero, whereas $\Vert nJ\Vert =n$. Thus the real-space answer is no in general. Real diagonal values do not detect the skew-symmetric part of an operator. This is diagonal boundedness of operator families over real and complex Hilbert spaces.

(iii) implies (ii): the preliminary bounds show ${\sup }_k\Vert P_{k}x\Vert <\infty$ for every $x$. Applying the Uniform boundedness principle to the bounded maps $P_k$ gives ${\sup }_k\Vert P_k\Vert <\infty$.

(iii) implies (iv): the normalization $\Vert x_n\Vert =1$ givesHence every coordinate functional is bounded.

The Baire category theorem says that a nonempty complete metric space is not a countable union of closed sets with empty interior; equivalently, every countable intersection of open dense sets is dense. To prove the latter, start with any nonempty open set $V$ and open dense sets $G_n$. Choose closed balls of positive radii withEach choice is possible by density and openness. Their centers form a Cauchy sequence; its limit exists by completeness and belongs to every closed ball, hence to $V\cap \bigcap G_n$. This proves density and the theorem.

The Uniform boundedness principle says that a pointwise bounded family $\mathcal{T}$ of bounded linear maps from a Banach space $X$ to a normed space $Z$ has uniformly bounded operator norms. Put $E_m=\{x:{\sup }_{T\in \mathcal{T}}\Vert Tx\Vert \le m\}$. These sets are closed and cover $X$. Baire gives an open ball $B(x_0,r)\subset E_m$. For $\Vert h\Vert <r$, the triangle inequality applied to $T(x_0+h)-T{x}_0$ gives $\Vert Th\Vert \le 2m$. Taking $h=(r/2)u$ for unit $u$ proves ${\sup }_T\Vert T\Vert \le 4m/r$.

For the particular expansion, uniqueness makes all coefficient maps and partial-sum maps linear. Since $P_{k}x\to x$, the norms $\Vert P_{k}x\Vert$ form a bounded sequence with limit $\Vert x\Vert$. Also, with $P_0=0$,This proves both preliminary bounds. The supremum therefore defines a finite norm $\Vert \cdot {\Vert }^{\prime }$, with $\Vert x\Vert \le \Vert x{\Vert }^{\prime }$.

To prove (i) implies (ii), the identity map from $(X,\Vert \cdot {\Vert }^{\prime })$ to $(X,\Vert \cdot \Vert )$ is a bounded linear bijection. If (i) holds both spaces are Banach, so the bounded inverse theorem makes its inverse bounded: $\Vert x{\Vert }^{\prime }\le C\Vert x\Vert$. This is precisely (ii).

Assume first that $\Vert P_n\Vert \le M$ for all $n$. Given $x\in X$ and $ϵ>0$, choose a finite linear combination $y$ of the $x_i$ with $\Vert x-y\Vert <ϵ/(M+1)$. For all sufficiently large $n$, $P_{n}y=y$. ThereforeThus $P_{n}x\to x$, and the formula in part (iii) yields the norm-convergent expansionThis proves the stated existence of coefficients.

Conversely, suppose every $x$ has some norm-convergent expansion $x={\sum }_i{\alpha }_i{x}_i$. Applying the continuous functional $x_j^{\ast }$ to its partial sums and passing to the limit gives ${\alpha }_j=x_j^{\ast }(x)$. Hence those partial sums are precisely $P_{n}x$, so $P_{n}x\to x$ for every $x$. In particular ${\sup }_n\Vert P_{n}x\Vert <\infty$ for each fixed $x$. The Uniform boundedness principle, using completeness of $X$, now givesThis proves the reverse implication. The coefficients are automatically unique, so the equivalence is the uniform partial-sum criterion for a Schauder basis for the given minimal total sequence. Its projection bound is the basis constant.

Define the continuous linear functionals $f_n(z)=⟨x_n,z⟩$. For each fixed $z\in X$, weak convergence makes $f_n(z)$ a convergent, hence bounded, scalar sequence. The Uniform boundedness principle states that a pointwise bounded family of bounded linear operators from a Banach space to a normed space has uniformly bounded operator norms. Since a Hilbert space is a Banach space, it applies to $(f_n)$.

The Cauchy-Schwarz inequality gives $\Vert f_n\Vert \le \Vert x_n\Vert$, and evaluation at $z=x_n/\Vert x_n\Vert$ gives equality when $x_n\ne 0$; equality also holds at zero. Consequently

The Banach-Steinhaus convergence theorem can be stated as follows. Let $X$ be a normed space, $Y$ a Banach space, and $A_n\in \mathcal{B}(X,Y)$. Ifand $A_{n}d$ converges in $Y$ for every $d$ in a dense subset $D\subseteq X$, then $A_{n}x$ converges for every $x\in X$. The pointwise limit $A$ is a bounded linear operator andIf $X$ is also a Banach space, the uniform operator norm bound is necessary: pointwise convergence implies pointwise boundedness, so the Banach-Steinhaus theorem, namely the Uniform boundedness principle, supplies it. The limit is convergence in the strong operator topology; it need not be convergence in operator norm.

To see the sufficient direction, write $C={\sup }_n\Vert A_n\Vert$. For $d\in D$ the triangle inequality givesFirst approximate $x$ by $d$, then let $m,n\to \infty$. Completeness of $Y$ gives the limit. Linearity passes to limits, and $\Vert Ax\Vert ={\lim }_n\Vert A_{n}x\Vert \le ({\mathrm{liminf}}_n\Vert A_n\Vert )\Vert x\Vert$ gives the displayed bound. If a candidate bounded linear operator $A:X\to Y$ is already given and $A_{n}d\to Ad$ on $D$, the same approximation proves convergence to $Ax$ even when $Y$ is merely a normed space.

The Weierstrass approximation theorem asserts that every real continuous function on a compact interval is a uniform limit of polynomials. An affine change of variable reduces the interval to $[0,1]$. There are two complementary routes: construct positive approximation operators, or construct an algebra dense enough to approximate absolute values and hence pointwise maxima and minima.

For the first route, the Korovkin theorem reduces convergence of positive linear operators on continuous functions $L_n$ to the three tests $1,t,t^2$. Here is its proof. Given $f\in C[0,1]$ and $\epsilon >0$, uniform continuity supplies a $\delta >0$ such that $|f(t)-f(x)|<\epsilon$ for $|t-x|<\delta$. For all $t,x$ one then hasApply positivity to the two real inequalities obtained from this bound. At $x$ this givesThe expression in brackets is nonnegative and tends uniformly to zero by the three test assumptions. Also $L_{n}1\to 1$ uniformly. Taking the upper limit of the supremum norm and then letting $\epsilon \downarrow 0$ proves convergence for every $f$. This is the quadratic barrier estimate for positive approximation operators.

The Bernstein polynomial supplies the constructive corollaryThe binomial theorem makes its nonnegative weights sum to one. Differentiating the binomial theorem, or using the first two binomial distribution moments, givesThe Korovkin theorem proves $B_{n}f\to f$ uniformly, and each $B_{n}f$ is a polynomial, proving the Weierstrass approximation theorem. The same criterion gives the Schoenberg spline operator convergence in Solution 3. For periodic functions its counterpart uses $1,\cos t,\sin t$: the nonnegative barrier $1-\cos (t-x)$ is bounded away from zero outside any fixed small arc. This periodic Korovkin test set gives convergence of the normalized Fejér sums and density of trigonometric polynomials.

The Lebesgue convergence criterion for approximation operators addresses stability. Suppose bounded linear operators $L_n:C[a,b]\to C[a,b]$ converge to the identity on every polynomial. Then they converge on every continuous function if and only if ${\sup }_n\Vert L_n\Vert <\infty$. Sufficiency is the extension of approximation operators from a dense subspace: for a polynomial $p$ with $\Vert f-p\Vert <\eta$ and a common bound $M$,First let $n\to \infty$, then $\eta \downarrow 0$. Necessity follows from the Uniform boundedness principle, since a convergent $L_{n}f$ is bounded for every $f$. Applied to approximation projectors, the Lebesgue inequality for approximation projectors isIndeed $P_{n}v=v$ makes $f-P_{n}f=(f-v)-P_n(f-v)$; take the infimum. Thus density of the ranges and a uniform bound on the Lebesgue constants of interpolation imply convergence. Density alone does not control a sequence of interpolation projectors.

There is also a kernel version of this stability criterion, useful for trigonometric polynomial approximation. If integrable periodic kernels $K_n$ have uniformly bounded $L^1$ norms, their integrals tend to one, and ${\int }_{|t|\ge \delta }|K_n(t)|dt\to 0$ for every $\delta >0$, then convolution with $K_n$ converges uniformly to every continuous periodic function. Subtract $f(x)\int K_n$ and split the error into a small arc, bounded by $\omega (f,\delta ){\sup }_n\Vert K_n{\Vert }_1$, and its complement, bounded by $2\Vert f{\Vert }_{\infty }{\int }_{|t|\ge \delta }|K_n|$. The remaining mass error tends to zero. This approximate identity argument shows directly why nonnegative unit-mass Fejér kernels work even though ordinary Fourier partial sums need not converge uniformly for every continuous target.

Lebesgue's constructive approach to the Weierstrass approximation theorem starts instead with the absolute value. On $[-1,1]$ define the polynomial recurrence for the absolute valueInduction gives $0\le p_r(x)\le |x|$, monotonicity, andThe pointwise limit satisfies $p(x{)}^2=x^2$ and is nonnegative, hence equals $|x|$. Dini's theorem makes the convergence uniform. More explicitly, the continuous decreasing errors have, for any positive tolerance, open sublevel sets covering the compact interval; a finite subcover supplies one index valid everywhere. Every continuous piecewise linear function has the formThe $c_j$ are its changes in slope. Scaling and translating the absolute-value approximation shows that $L$ is uniformly approximable by polynomials. Solution 4, or polygonal interpolation and uniform continuity, makes these piecewise linear functions dense in $C[a,b]$. Approximating $f$ first by $L$ and then by a polynomial is another complete proof of the Weierstrass approximation theorem.

The Stone-Weierstrass theorem generalizes from intervals to a compact Hausdorff space $K$: a real subalgebra $A\subseteq C(K,\mathbb{R})$ containing the constants and separating points is uniformly dense. Its proof uses the same absolute-value mechanism. The uniform closure of a function algebra is an algebra. If $a$ belongs to this closure, approximation of $|t|$ by polynomials on a bounded interval containing its range proves that $|a|$ also belongs to the closure. Therefore the closure containsPoint separation and constants allow exact interpolation of any two real values: choose $h\in A$ with $h(s)\ne h(t)$ and solve for $\alpha ,\beta$ in $\alpha h+\beta$. The lattice approximation from two-point approximation proved in Solution 4 now makes the closure all of $C(K,\mathbb{R})$.

The complex version requires a unital, point-separating subalgebra closed under complex conjugation. Its real-valued members separate points using real or imaginary parts, so the real result approximates both parts of every complex target. Conjugation closure is essential: polynomials in $z$ alone on the closed unit disk cannot approximate $\overline{z}$ uniformly, because ${\int }_{|z|=1}p(z)dz=0$ whereas ${\int }_{|z|=1}\overline{z}dz=2\pi i$.

Among the corollaries are density of multivariate polynomials on compact subsets of real Euclidean space, since coordinates separate points, and density of trigonometric polynomials on the circle, since sine and cosine separate points there. On a compact interval, polynomials with rational coefficients are a countable dense family, giving separability of $C[a,b]$. The real and complex approximation results above concern uniform approximation of continuous targets; neither asserts convergence of arbitrary interpolation schemes or arbitrary Fourier series.

Assume instead that $T$ sends weakly compact sets to norm-compact sets. If $x_n\rightharpoonup 0$, then $K=\{0,x_1,x_2,\dots \}$ is weakly compact: an open cover has a member containing zero, which contains all but finitely many terms of the convergent sequence. Hence $T(K)$ is norm compact. A bounded linear operator is weak-to-weak continuous, because each $g\in F^{\ast }$ pulls back to $g\circ T\in E^{\ast }$. Thus $T{x}_n\rightharpoonup 0$. If $\Vert T{x}_n\Vert$ did not tend to zero, a subsequence with norms bounded below by some $\epsilon >0$ would have a norm-convergent further subsequence with limit $y$. That further subsequence also converges weakly to both $y$ and zero; uniqueness of limits in the Hausdorff weak topology forces $y=0$, contradicting the lower bound. Therefore $\boxed{\Vert T{x}_n\Vert \to 0}$, proving the equivalence and the characterization of completely continuous operators.

Now consider the inclusion of the space of continuous functions on a compact space $C(\mathbb{T})$ into L2 space for normalized Lebesgue measure. It is a bounded linear operator, since $\Vert f{\Vert }_2\le \Vert f{\Vert }_{\infty }$. For a weakly null sequence $(f_n)$ in $C(\mathbb{T})$, point evaluations are continuous linear functionals, so $f_n(t)\to 0$ for every $t$. The Uniform boundedness principle, applied to the canonical embeddings $f_n\in C(\mathbb{T}{)}^{\ast \ast }$, gives a uniform bound $\Vert f_n{\Vert }_{\infty }\le M$: each dual functional is bounded on the weakly convergent sequence. The dominated convergence theorem now yieldsThus $j_2$ satisfies both equivalent conditions.

It is nevertheless not a compact operator. Over complex scalars the functions $f_n(t)=e^{int}$ lie in the unit ball of $C(\mathbb{T})$ and form an orthonormal sequence in $L^2(\mu )$, soThere is no norm-convergent subsequence. Over real scalars use $\cos (nt)$, whose distinct images have distance $1$. Complete continuity controls weakly compact sets, rather than all bounded sets.

First prove the Baire category theorem in its complete-metric form. Let $X$ be a nonempty complete metric space, let $G_n$ be open dense sets, and let $V$ be any nonempty open set. Choose a closed ball $\overline{B}(x_1,r_1)\subset V\cap G_1$ with $0<r_1<1/2$. Inductively, density and openness permit the choiceFor $m>n$, $x_m$ belongs to the $n$th closed ball, so $d(x_m,x_n)\le r_n$. The centers form a Cauchy sequence and converge to some $x\in X$ by completeness. Each closed ball contains all later centers and therefore contains their limit. Hence $x\in V\cap {\bigcap }_n{G}_n$. Since this holds for every $V$, every countable intersection of open dense sets is dense. Equivalently, a nonempty complete metric space cannot be a countable union of closed sets with empty interior; this is the form needed below.

For the Banach-Steinhaus theorem, let $X$ be a Banach space, $Y$ a normed vector space, and $\mathcal{T}$ a family of bounded linear operators from $X$ to $Y$ such that ${\sup }_{T\in \mathcal{T}}\Vert Tx\Vert <\infty$ for every $x$. DefineEach $E_m$ is closed, as an intersection of inverse images of closed balls under continuous maps, and pointwise boundedness gives $X={\bigcup }_m{E}_m$. The Baire category theorem gives some $E_m$ with nonempty interior. Choose $B(x_0,r)\subset E_m$. If $\Vert h\Vert \le r/2$, both $x_0+h$ and $x_0$ lie in $E_m$, so $\Vert Th\Vert \le 2m$ for every $T$. Applying this to $h=(r/2)u$ with $\Vert u\Vert \le 1$ provesThis proves the Uniform boundedness principle; completeness is required for the domain, but not for $Y$.

Now fix the prescribed point ${\theta }_0$ on the unit circle, parametrized by angles modulo $2\pi$. On the real Banach space $C(\mathbb{T};\mathbb{R})$ with the supremum norm, consider the continuous linear functionals $T_{N}f=S_{N}f({\theta }_0)$, where $S_N$ is the symmetric Fourier partial sum. Summing the finite geometric series gives its Dirichlet kernel representationThe continuous extension at $t=0$ is $2N+1$. The kernel bound gives $\Vert T_N\Vert \le {\Lambda }_N=(2\pi {)}^{-1}\int |D_N|$. Equality holds: the continuous periodic functions defined byhave supremum norm at most one, and dominated convergence gives $T_N{f}_{\delta }\to {\Lambda }_N$ as $\delta \downarrow 0$.

These norms are unbounded. Indeed, evenness and $\sin (t/2)\le t/2$ for $0<t\le \pi$ giveFor the second inequality, integrate over $[(j-1)\pi ,j\pi ]$, where $1/u\ge 1/(j\pi )$ and the integral of $|\sin u|$ is two. The harmonic series diverges, proving the Dirichlet kernel harmonic lower bound. If every continuous $f$ had bounded partial sums at ${\theta }_0$, the Banach-Steinhaus theorem would bound all the operator norms, a contradiction. Consequently there exists a continuous real function $f$ with ${\sup }_N|S_{N}f({\theta }_0)|=\infty$, and its Fourier series diverges at the prescribed point. The function depends on the chosen point; the argument requires no claim of simultaneous divergence everywhere.

A representation of a Banach algebra is a complex-linear multiplicative map $\pi :A\to {\mathrm{End}}_{\mathbb{C}}(X)$. A nonzero irreducible action of a unital algebra automatically has $\pi (1)=I$: the image of the idempotent $\pi (1)$ is a nonzero invariant subspace and hence is all of $X$. An algebraically irreducible representation of a Banach algebra is a nonzero action if it has no nonzero proper invariant submodule, whether closed or not. A normed representation of a Banach algebra has $\pi (a)$ bounded for each $a$ on a preassigned normed vector space $X$; continuity of $\pi$ as an operator-norm-valued map is a further conclusion. This must not be confused with topological irreducibility, which excludes only closed invariant subspaces. Unitizing the algebra extends a nonunital action by $\pi (a+\lambda 1)=\pi (a)+\lambda I$, so it is enough to discuss the unital case.

For $\xi \ne 0$, irreducibility gives $A\xi =X$. The annihilator $L_{\xi }=\{a:\pi (a)\xi =0\}$ is a maximal left ideal: an intermediate left ideal gives an intermediate submodule of $X$. It is closed by the Neumann-series argument for maximal left ideals. Consequently $A/L_{\xi }$ transports a complete quotient norm to $X$, and left multiplication is contractive in this norm. Every algebraically irreducible representation of a Banach algebra can therefore be realized continuously on a Banach space. This alone does not prove continuity for its preassigned norm.

The kernel of $\pi$ is the intersection of the closed left ideals $L_{\xi }$ and is a closed primitive ideal. The Jacobson radical is the intersection of all primitive ideals, equivalently the intersection of maximal left ideals. Thus these quotient representations detect semisimplicity. Equivalent algebra representations are related by a bijective module homomorphism.

The algebra of module endomorphisms is exactly $\mathbb{C}I$. Indeed, if a module endomorphism $D$ sends a fixed cyclic vector $\xi$ to $a\xi$, then $D(b\xi )=ba\xi$. It is induced by bounded right multiplication by $a$ on $A/L_{\xi }$, so is bounded for the quotient norm on an irreducible Banach module. Every nonzero module endomorphism is injective and surjective by irreducibility, and its inverse is bounded by the same argument. This commutant of an operator algebra is a closed division subalgebra of the bounded operators on this Banach module; the Gelfand-Mazur theorem makes it $\mathbb{C}I$.

The Jacobson density theorem now says that for finitely many linearly independent ${\xi }_1,\dots ,{\xi }_n$ and arbitrary ${\eta }_1,\dots ,{\eta }_n$, one element of $A$ sends all ${\xi }_j$ to ${\eta }_j$. Here is the induction behind it. The one-vector assertion is $A{\xi }_1=X$. Assuming the first $n-1$ coordinates can be prescribed, set $L=\{a:a{\xi }_j=0(j<n)\}$. Its image $L{\xi }_n$ is an invariant subspace. If it is $X$, it adjusts the last image without changing the first ones. If it is zero, the rule $(a{\xi }_1,\dots ,a{\xi }_{n-1})\mapsto a{\xi }_n$ is a well-defined module map $X^{n-1}\to X$. Each coordinate map is scalar by the module endomorphism result, forcing ${\xi }_n$ to be a linear combination of the earlier vectors, a contradiction. This proves the assertion.

We can now give Johnson's continuity argument with its crucial construction. If $X$ is finite dimensional, the closed kernel has finite codimension and $A/\ker \pi$ is finite dimensional, so $\pi$ is continuous. Suppose instead that $X$ is infinite dimensional. LetThis is an invariant submodule, because the orbit map of $\pi (b)\xi$ is the orbit map of $\xi$ composed with the bounded map $a\mapsto ab$. Thus either $Y=X$ or $Y=0$.

To rule out $Y=0$, choose linearly independent unit vectors ${\xi }_0,{\xi }_1,\dots$. There are elements $a_n\in A$ such that, with $c_n=a_n\cdots a_1$,For completeness, construct them in the canonical quotient-norm Banach module. At stage $n$ the current tail vectors are independent. In the closed left annihilator of its first vector, density permits arbitrary images of any finite list of the remaining vectors. The condition that finitely many such images are independent is open in this Banach subspace and dense: perturb any element by $t$ times an element prescribing independent images; a suitable determinant is a nonzero polynomial in $t$, with only finitely many exceptional values. The Baire category theorem intersects these open dense conditions for all initial finite tail lists. The resulting $a_n$ kills the first current vector and retains independence of the entire remaining tail. This is the countable annihilation sequence in an irreducible Banach module.

The gliding-hump continuity principle supplies the other ingredient: if $E,F$ are Banach spaces, $S:E\to F$ is linear, $T_j:E\to E$ and $U_n:F\to F_n$ are bounded, and $U_{n}S{T}_1\cdots T_m$ is continuous whenever $m>n$, then $U_{n}S{T}_1\cdots T_n$ is continuous eventually. The product form of this principle is recorded in Proposition 1.3 of Thomas's paper. Here is the contradiction mechanism. Rescale so $\Vert T_j\Vert ,\Vert U_n\Vert \le 1$ and put $P_m=T_1\cdots T_m$. If infinitely many diagonal maps are discontinuous, choose increasing $n_j$ and successively tiny $x_j$ soAlso choose $\Vert x_j\Vert \le 2^{-j}/(1+{\max }_{i<j}\Vert U_{n_i}S{P}_{n_i+1}\Vert )$. The convergent sum $x={\sum }_j{P}_{n_j}{x}_j$ has, after its $j$th term, a tail factoring through $P_{n_j+1}$. Applying that continuous composite bounds the tail contribution by one. The display forces $\Vert U_{n_j}Sx\Vert >j$, contradicting $\Vert U_{n_j}Sx\Vert \le \Vert Sx\Vert$. Only finite decompositions and the continuous tail composites are used; no unjustified interchange of a discontinuous map with an infinite sum occurs.

Apply this principle with $E=A$, $F=\mathcal{B}(\overline{X})$, where $\overline{X}$ is the completion in the original norm, and $S(a)$ the bounded extension of $\pi (a)$. Set $T_j(a)=a{a}_j$ and $U_n(R)=R{\xi }_n$. ThenFor $m>n$ this is zero, hence continuous. For $m=n$ it is the orbit map of a nonzero vector, hence discontinuous if $Y=0$. This contradicts the principle. Therefore $Y=X$.

Finally, apply the Uniform boundedness principle on the Banach space $A$ to the continuous maps $a\mapsto \pi (a)\xi$ indexed by $\Vert \xi \Vert \le 1$. They are pointwise bounded because each individual $\pi (a)$ is bounded in the original norm. It givesThis proves Johnson's continuity theorem for irreducible normed representations, including when the original normed vector space is incomplete. The quotient norm, countable density construction, and gliding hump explain the substantive steps rather than assuming the desired continuity in advance.

For $x,y\in B$, write $L_x(y)=xy$ and $R_y(x)=xy$. Separate continuity and linearity make both maps bounded linear operators. For each fixed $y$,The Uniform boundedness principle therefore gives $C<\infty$ such that $\Vert L_x\Vert \le C$ whenever $\Vert x\Vert \le 1$. Scaling yieldsFor completeness, the relevant Uniform boundedness principle follows from the Baire category theorem as follows. For a pointwise bounded family $\mathcal{T}$ of bounded linear operators on a Banach space, the closed sets $E_m=\{y:{\sup }_{T\in \mathcal{T}}\Vert Ty\Vert \le m\}$ cover the space. One $E_m$ contains an open ball $B(y_0,r)$. Subtracting the bounds for $y_0$ and $y_0+h$, with $\Vert h\Vert <r$, gives ${\sup }_T\Vert Th\Vert \le 2m$. Applying this to $h=(r/2)u$, $\Vert u\Vert \le 1$, bounds every operator norm by $4m/r$.

Define the new norm byIt is a norm because $L_{x}e=x$, so $L_x=0$ implies $x=0$. The triangle inequality and scalar homogeneity follow from those of the operator norm. Moreover,so it is equivalent to the original norm and remains complete. By associativity, $L_{xy}=L_x{L}_y$, and the submultiplicativity of the operator norm provesThis is renorming a separately continuous Banach algebra. It even gives $\Vert e{\Vert }_{\ast }=1$. If the algebra is the zero space, the conclusion is immediate separately.

The real space $C(\mathbb{T})$ is a Banach space in the uniform norm: a uniformly Cauchy sequence has a uniform pointwise limit by completeness of $\mathbb{R}$, and that uniform limit is continuous. Each functional $L_N$ from the preceding part is bounded, but their norms are unbounded.

For integers $m\ge 1$, defineEach $E_m$ is closed, being an intersection of inverse images of closed intervals under continuous functionals. It has empty interior. Otherwise some ball $B(f_0,r)$ would be contained in it. For every $h$ of norm at most one, both $f_0$ and $f_0+(r/2)h$ would belong to that ball, soTaking the supremum over $h$ would give $\Vert L_N\Vert \le 4m/r$ for all $N$, contradicting the preceding part.

Thus the $E_m$ are nowhere dense. The Baire category theorem says their union cannot cover $C(\mathbb{T})$. A function outside that union satisfiesso its Fourier series diverges at zero. In fact these functions form a dense $G_{\delta }$ subset of $C(\mathbb{T})$. This proves the relevant Uniform boundedness principle argument directly and establishes generic unbounded Fourier sums at a fixed point.