With $\hat{f}(k)=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }f(s)e^{-iks}ds$, the classical Fourier partial sum is $s_N^f(t)={\sum }_{|k|\le N}\hat{f}(k)e^{ikt}$. Its Cesàro mean isFor the uniform Fejér theorem, extend the continuous endpoint-matching $f$ periodically. The given Fejér kernel is nonnegative, even and has total integral $\pi$. ThusGiven $\epsilon >0$, uniform continuity supplies $\delta >0$ so the bracket has modulus less than $\epsilon$ for $|u|<\delta$, uniformly in $t$. That part contributes at most $\epsilon$. On the remainder, $F_N(u)\le [2(N+1){\sin }^2(\delta /2){]}^{-1}$, and the bracket is at most $2\Vert f{\Vert }_{\infty }$. HenceLetting $\epsilon$ decrease to zero proves uniform convergence of the Fejér sums.

The later assertions attached to the printed factorial-frequency example are inconsistent. The following calculations resolve them while keeping the Fourier definition just given. Put $a_j=e^{i\sqrt{j+1}}-e^{i\sqrt{j}}$ and consider its formal trigonometric series. For $t=\pi p/q$, there is $J_0$ such that $\cos (j!t)=1$ for every $j\ge J_0$, because $2q$ divides $j!$. At $N_J=(J+1)!-1$, its ordinary partial sum therefore telescopes toThe exponential has subsequences tending to $1$ and $-1$: take $J+1$ to be integers nearest $(2\pi m{)}^2$ and $((2m+1)\pi {)}^2$. Thus the ordinary sums diverge at every $t\in \pi \mathbb{Q}$.

But their genuine Fourier Cesàro means also diverge there. Indeed,Since $|a_j|\le 2$ and ${\sum }_{j=0}^{J}j!\le 2J!$ for $J\ge 3$, its absolute value is at most $4/(J+1)\to 0$, uniformly in $t$. Thereforewhich has the same incompatible limit points. This is an instance of Cesàro sums at factorial cutoffs track ordinary sums. The printed Cesàro-convergence claim is false. Averaging the sparse series by its index $j$ would be a different operation from Fourier Cesàro summation by frequency.

There is a further issue: no $L^1$ function has exactly this Fourier series. Here is a direct proof using strongly lacunary polynomial L1-L2 bound. For a trigonometric polynomial $p$ supported on $\{\pm j!:j\ge 3\}$, consecutive positive frequencies have ratio at least four. Every zero sum of four signed frequencies pairs into cancelling equal frequencies: a largest unpaired frequency would exceed the sum of the other three. Expanding the fourth moment with normalized measure $dt/(2\pi )$ and bounding the three possible pairings givesThe Holder inequality gives $\Vert p{\Vert }_2\le \Vert p{\Vert }_1^{1/3}\Vert p{\Vert }_4^{2/3}$, hence $\Vert p{\Vert }_2\le \sqrt{3}\Vert p{\Vert }_1$.

If the asserted $f\in L^1$ existed, positivity and unit mass of the summation kernel would bound $\Vert {\sigma }_N^f{\Vert }_1\le \Vert f{\Vert }_1$. Subtract the finitely many modes from $j<3$; their norms stay uniformly bounded, and the remainder is a polynomial of the type above. Its $L^2$ norms would therefore be uniformly bounded. Orthogonality then forces ${\sum }_{j\ge 3}|a_j{|}^2<\infty$, by fixing any finite coefficient block and letting $N\to \infty$. However,so that sum diverges. This contradiction disproves the stated existence premise.

Even considered only formally, these coefficients rule out an $L^2$ function by Bessel's inequality, and hence rule out a bounded function on the finite interval. Thus a bounded continuous representative cannot satisfy the asserted series either. The asserted $L^1$ function does not exist.

The norm-preserving form of the Hahn-Banach theorem states: if $M$ is a linear subspace of a real or complex normed vector space $X$ and $f:M\to \mathbb{K}$ is a bounded linear functional, there exists $F\in X^{\ast }$ with $F{|}_M=f$ and $\Vert F\Vert =\Vert f\Vert$.

For a real separable $X$, put $C=\Vert f\Vert$. The case $C=0$ is immediate. To extend over one vector $x_0\notin M$, choose a real number $c$ betweenThe lower endpoint is no greater than the upper: for any $m,n\in M$,The endpoints are finite because testing zero bounds them between finite numbers. Define $F(m+t{x}_0)=f(m)+tc$. The upper inequality ensures $F(z)\le C\Vert z\Vert$ for vectors with positive $x_0$ coefficient, and the lower inequality gives it for negative coefficient; it already holds on $M$. Applying the same bound to $-z$ gives $|F(z)|\le C\Vert z\Vert$. Thus the extension preserves the norm.

Now add a countable dense sequence of vectors one at a time. The consistent extensions define a bounded functional of norm $C$ on the union of the increasing domains. That union is dense in $X$, so extension by continuity gives the desired functional on all of $X$. This proves the real separable case without a maximality argument.

For the density application, suppose the closed linear span $M$ were proper. The Hahn-Banach theorem supplies a nonzero continuous linear functional annihilating $M$. By duality of sequence spaces, it has the form $f(z)={\sum }_{j\ge 1}{a}_j{z}_j$ with $a\in {\ell }_1$, not identically zero. Let $m$ be the first index with $a_m\ne 0$. Writing $t_n=(n+1{)}^{-1}$, annihilation of $x_n$ givesThe tail has absolute value at most $t_n{\sum }_{j>m}|a_j|\to 0$. Letting $n\to \infty$ contradicts $a_m\ne 0$. ThereforeThis is geometric sequences with accumulating ratios span c0.

With the classical real coefficientsthe partial sum is $s_n^f=a_0/2+{\sum }_{j=1}^n(a_j\cos jx+b_j\sin jx)$. It is the orthogonal projection onto the first trigonometric modes. Completeness of the trigonometric orthonormal system therefore gives $\Vert s_n^f-f{\Vert }_2\to 0$.

For $f\in C^1$ periodic, integration by parts has zero endpoint terms and gives $a_j(f^{\prime })=j{b}_j(f)$, $b_j(f^{\prime })=-j{a}_j(f)$, and $a_0(f^{\prime })=0$. ThusFor periodic $C^2$ functions, another integration by parts gives $a_j(f)=-a_j(f^{\prime \prime })/j^2$ and $b_j(f)=-b_j(f^{\prime \prime })/j^2$. Riemann-Lebesgue makes the numerator sequences tend to zero, in particular bounded. The Weierstrass M-test and $\sum j^{-2}<\infty$ prove absolute and uniform convergence. The resulting continuous sum agrees with $f$ in $L^2$ by the first result. Two continuous functions equal almost everywhere are equal everywhere, so the pointwise sum is $f$ for every real $x$.

For the polynomial example the periodic extension is $C^2$: the function, its first derivative and its second derivative match at the endpoints. It is even, so all sine coefficients vanish. Direct integration givesFor example, integration by parts gives ${\int }_0^{\pi }{x}^2\cos (jx)dx=2\pi (-1{)}^j/j^2$ and ${\int }_0^{\pi }{x}^4\cos (jx)dx=4{\pi }^3(-1{)}^j/j^2-24\pi (-1{)}^j/j^4$, from which the cancellation of the $j^{-2}$ terms follows. ThereforeAt $x=0$, the left side vanishes, yielding

Write $L_n(t)={\sum }_{k=0}^n{a}_k{t}^k$. Matching coefficients in the differential equation givesTaking $a_0=1$ yieldsThe top coefficient is $(-1{)}^n/n!\ne 0$, so this is a real polynomial of degree exactly $n$.

Put $D_n=L_n^{\prime }-L_n$. Integration by parts, with vanishing endpoint terms because of the factor $t$ at zero and exponential decay at infinity, givesThe last step uses $D_n+t{D}_n^{\prime }=t{L}_n^{\prime \prime }+(1-t)L_n^{\prime }-L_n=-(n+1)L_n$. The left side is symmetric in $m,n$. Interchanging them therefore gives $(n-m)\int L_n{L}_m{e}^{-t}=0$, proving orthogonality for distinct indices. These integrals are exactly $⟨{\phi }_n,{\phi }_m⟩$, and each ${\phi }_n$ is square-integrable and nonzero. Thus they form an orthogonal sequence in $L^2(0,\infty )$.

Multiplying the equations by $z^n$ and summing givesConversely equality of power-series coefficients recovers every polynomial equation. For the proposed expression,Substitution makes the PDE left side zero. Its coefficient of $z^n$ has value one at $t=0$, so it corresponds to the normalization $L_n(0)=1$ chosen above.

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 spectrum of a bounded operator is $\sigma (T)=\{\lambda :\lambda I-T\text{ has no bounded two-sided inverse}\}$. Its point spectrum is the set of eigenvalues, where $\lambda I-T$ is not injective.

Here $T$ is linear and takes continuous functions to continuously differentiable functions. Also $\Vert Tx{\Vert }_{\infty }\le 2\Vert x{\Vert }_{\infty }$; the constant function one attains this bound at $t=1$, so $\Vert T\Vert =2$. Its range is contained in $C^1([0,1])$, a proper subset of $C([0,1])$, so zero is in its spectrum. It is not an eigenvalue: $Tx=0$ gives $x(0)=0$ at zero and $x=0$ upon differentiation.

For a nonzero eigenvalue, $\lambda x=Tx$ implies $x^{\prime }=x/\lambda$ and $(\lambda -1)x(0)=0$. A nonzero solution is therefore possible exactly at $\lambda =1$, with eigenfunctions $x(t)=c{e}^t$. Hence $p\sigma (T)=\{1\}$.

For the proposed resolvent putThe proposed $x$ is $y/\lambda +y(0)e^{t/\lambda }/[\lambda (\lambda -1)]+F/{\lambda }^2$. Consequently $x(0)=y(0)/(\lambda -1)$ andAdding $x(0)$ and subtracting from $\lambda x(t)$ gives exactly $y(t)$. Thus $\lambda I-T$ is onto for $\lambda \ne 0,1$. It is also injective by the eigenvalue calculation. The displayed formula defines a bounded map in the supremum norm, since its exponential kernels are bounded on the compact interval. It is therefore the bounded inverse, and

The complex Hahn-Banach theorem says that every bounded complex-linear functional on a linear subspace of a normed complex space extends to the whole space with the same norm.

For the separation assertion choose $x_0\notin M$ and $d=\mathrm{dist}(x_0,M)>0$. On $M+\mathbb{C}{x}_0$ define $g_0(m+\lambda x_0)=\lambda d$. The inequality $\Vert m+\lambda x_0\Vert \ge |\lambda |d$ makes its norm at most one; approximating the distance $d$ shows the norm is exactly one. Hahn-Banach gives an extension $g$ of norm one annihilating $M$.

Each geometric sequence satisfies $\Vert x_k{\Vert }_1=(1-|{\beta }_k|{)}^{-1}\le (1-\delta {)}^{-1}$. Let $M$ be their closed linear span. If $M\ne {\ell }^1$, the separation just proved gives a nonzero bounded functional annihilating all $x_k$. Represent it by $b\in {\ell }^{\infty }$ and formThe series converges locally uniformly and defines a holomorphic function. It vanishes at every ${\beta }_k$. These distinct points lie in the compact disk $|z|\le \delta <1$, so have an accumulation point inside the domain. The identity theorem gives $F=0$, hence every Taylor coefficient $b_j=0$, contradicting the functional's nonzero norm. Therefore

The dual space $X^{\prime }$ consists of bounded linear maps $g:X\to \mathbb{C}$, with operator norm $\Vert g\Vert ={\sup }_{\Vert x\Vert \le 1}|g(x)|$. It is Banach even if $X$ is not: an operator-norm Cauchy sequence has pointwise limits, these define a bounded linear functional, and the uniform Cauchy estimate on the unit ball gives operator-norm convergence.

Restriction $R(g)=g{|}_Z$ is linear and contractive. If $Z$ is dense, for every $x\in X$ choose $z_n\in Z$ with $z_n\to x$. The bound $|g(z_n)|\le \Vert Rg\Vert \Vert z_n\Vert$ passes to the limit, proving $|g(x)|\le \Vert Rg\Vert \Vert x\Vert$. Hence $\Vert g\Vert =\Vert Rg\Vert$ and restriction is injective. Conversely, for $h\in Z^{\prime }$ define $g(x)=\lim h(z_n)$. The estimate $|h(z_n)-h(z_k)|\le \Vert h\Vert \Vert z_n-z_k\Vert$ gives existence and independence of the approximating sequence. It also gives linearity and $\Vert g\Vert \le \Vert h\Vert$. Thus restriction is an onto linear isometry of dual spaces.

For the sequence subspace in this part, the norm is the unweighted ${\ell }^1$ norm; only its membership condition uses the weight $j$. It contains all finitely supported sequences, hence is dense in ${\ell }^1$. Applying restriction to $X={\ell }^1$ and using its known dual yields