For $n\ge 1$ put $C_n=\overline{T(n{\overline{B}}_X)}$, with closure taken in $Y$. ThenIf every $C_n$ has empty interior, then each $C_n$ is a nowhere dense set, because it is closed. Each $T(n{\overline{B}}_X)$ is then a nowhere dense set too. Consequently $T(X)$ is a meagre set, that is, a set of the first category in $Y$.

Otherwise some $C_n$ has nonempty interior. Scaling gives nonempty interior of $\overline{T({\overline{B}}_X)}$. The geometric correction for approximate surjectivity proved in the preceding part gives an open ball about zero inside $T({\overline{B}}_X)$. Since $T(X)$ is a vector subspace, it then equals $Y$: scale any $y\in Y$ into this ball and scale its preimage back. Hence $T(X)$ is a closed set.

Thus, in fact, the stronger operator range category dichotomy holds:In particular $T(X)$ is closed or of the first category, as required. The only substantive theorem used here is the preceding geometric correction for approximate surjectivity: for a bounded linear operator from a Banach space to a normed space, nonempty interior of the closure of the image of the unit ball implies nonempty interior of the image itself. Its series proof was given there, so this proof does not require the Baire category theorem or completeness of $Y$.

Use the closed unit ball ${\overline{B}}_X$ first, and put $C=\overline{T({\overline{B}}_X)}$. This is a convex set symmetric about zero. If $B_Y(y_0,r)\subseteq C$, then $-y_0\in C$, so taking midpoints givesWrite $\delta =r/2$. By scaling,Fix $y$ with $\Vert y\Vert <\delta /2$. Starting with $r_0=y$, recursively choose $x_j\in X$ such thatAt each step the previous residual lies in the open ball on the left of the scaled inclusion, so the definition of closure permits this approximation. The series ${\sum }_j{x}_j$ converges to some $x$ because $X$ is a Banach space, and the triangle inequality gives $\Vert x\Vert \le {\sum }_j{2}^{-j}=1$. Continuity of $T$ gives $Tx={\lim }_N{\sum }_{j=1}^{N}T{x}_j=y$, since $r_N\to 0$. ThusScaling by $1/2$ now gives the slightly smaller open ballThis proves the assertion whether the original unit ball is taken open or closed. Only completeness of $X$ was used; the normed space $Y$ need not be complete. The argument is geometric correction for approximate surjectivity.

The open mapping theorem states that a surjective bounded linear operator $T:X\to Y$ between Banach spaces is an open map: $T(U)$ is an open set in $Y$ whenever $U$ is an open set in $X$. Equivalently, there is $\delta >0$ such thatwhere both displayed balls are open. In particular, a bijective bounded linear operator between Banach spaces has a bounded inverse, the bounded inverse theorem.

Composition with $t\mapsto t^{1+{\alpha }_n}$ preserves continuity, so $A_n$ maps the space of continuous functions on a compact space $C([0,1])$ into itself and is linear. Since $t\mapsto t^{1+{\alpha }_n}$ maps $[0,1]$ onto itself,For $0<t\le 1$, the mean value theorem applied to $r\mapsto t^{1+r}$ givesThe same bound holds at $t=0$. Each $x\in C([0,1])$ has uniform continuity by the Heine-Cantor theorem. With its modulus of continuity ${\omega }_x(\delta )={\sup }_{|s-t|\le \delta }|x(s)-x(t)|$, we obtainAlternatively, the bound proves convergence on every polynomial; the Weierstrass approximation theorem makes these a dense subset of $C([0,1])$, and the Banach-Steinhaus convergence theorem with $\Vert A_n\Vert =1$ gives the same conclusion.

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 orthogonal complement $x^{\perp }$ is infinite-dimensional: otherwise $X=\mathrm{span}\{x\}\oplus x^{\perp }$ would be finite-dimensional. Inductively choose an orthonormal sequence $(e_n)$ in $x^{\perp }$, each new vector being orthogonal to the preceding finite linear span. The Bessel inequality gives ${\sum }_n|⟨e_n,z⟩{|}^2\le \Vert z{\Vert }^2$ for every $z\in X$, so $⟨e_n,z⟩\to 0$. Thus $(e_n)$ is a weakly null orthonormal sequence.

SetBy orthogonality, $\Vert x_n{\Vert }^2=\Vert x{\Vert }^2+(1-\Vert x{\Vert }^2)=1$, while $⟨x_n-x,z⟩\to 0$ for every $z$. Hence $x_n\rightharpoonup x$. When $\Vert x\Vert =1$, this formula simply gives the constant sequence $x_n=x$. In particular, the weak closure of the unit sphere contains the whole closed unit ball.

First suppose the convex set $K$ is closed in the norm topology, and $x_n\in K$ satisfies $x_n\rightharpoonup x$. Apply the weak Banach–Saks theorem in a Hilbert space to the weakly null sequence $x_n-x$. Along the resulting subsequence,in norm. Each $z_m$ is a convex combination of points in $K$, so $z_m\in K$. Since $K$ is a closed set in the norm topology, $x\in K$.

Conversely, suppose $K$ is a sequentially weakly closed set. If $x\in \overline{K}$ in the norm topology, choose $x_n\in K$ with $\Vert x_n-x\Vert <1/n$. The Cauchy-Schwarz inequality gives $|⟨x_n-x,z⟩|\le \Vert x_n-x\Vert \Vert z\Vert \to 0$ for every $z$, hence weak convergence to $x$. The assumed closure property gives $x\in K$. The empty set satisfies both properties too. Thus

By weakly convergent sequence is bounded, $M={\sup }_n\Vert x_n{\Vert }^2$ is finite. Choose any $n_1$. Once $n_1<\cdots <n_k$ have been chosen, put $s_k={\sum }_{j=1}^k{x}_{n_j}$. Since $s_k$ is fixed, weak convergence to zero gives $⟨s_k,x_n⟩\to 0$. Choose $n_{k+1}>n_k$ with $|⟨s_k,x_{n_{k+1}}⟩|\le 1$. This gives the required one-sided inequality as well.

The useful estimate is for the squared norm. Indeed, $\Vert s_1{\Vert }^2\le M$, and the inner product expansion givesBy mathematical induction, $\Vert s_k{\Vert }^2\le Mk+2(k-1)\le (M+2)k$. HenceThis proves norm convergence of the averages, the weak Banach–Saks theorem in a Hilbert space. The printed inequality omits the square on its left side. Our stronger estimate also proves that printed bound, since $\sqrt{(M+2)k}\le (M+2)k$ for $k\ge 1$, but the printed bound alone only bounds the averages and cannot imply their norm convergence to zero.

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

In a real Hilbert space, weak convergence in a Hilbert space meansBy the Riesz representation theorem, every continuous linear functional on $X$ has the form $u\mapsto ⟨u,z⟩$. Thus this is exactly weak convergence, defined by convergence under every continuous linear functional.

From $T{e}_0=e_0\ne 0$, one is an eigenvalue. Zero is also an eigenvalue: the nonzero continuous function $q(x)=x(1-x)$ has both endpoint values zero, so $Tq=0$.

For $\lambda \notin \{0,1\}$, the linear projection identity $T^2=T$ gives an explicit bounded inverse:Multiplying on either side by $\lambda I-T$ yields $(I-T)+T=I$. There are no other points in the spectrum, and thus the spectrum of a nontrivial bounded projection here is

Let $M=\Vert T\Vert$, as calculated in part (ii). Its endpoint values ensure $1\le M<\infty$. Since $T^j=T$ for all $j\ge 1$,This verifies the spectral radius formula explicitly:

Let $M=\Vert |e_0|+|e_1|{\Vert }_{\infty }$. Since both linear functionals have operator norm one, for every $f\in X$ the triangle inequality givesTaking the essential supremum proves $\Vert T\Vert \le M$.

For the reverse bound, continuity and the extreme value theorem give $x_{\ast }\in [0,1]$ with $|e_0(x_{\ast })|+|e_1(x_{\ast })|=M$. Choose complex $a_0,a_1$ of modulus one so that $a_j{e}_j(x_{\ast })=|e_j(x_{\ast })|$, taking $a_j=1$ if $e_j(x_{\ast })=0$. The continuous functionhas supremum norm one by the triangle inequality and its endpoint values. Since ${\ell }_j(f_{\ast })=a_j$,The output is continuous, so its L-infinity norm equals its supremum norm, including when $x_{\ast }$ is an endpoint. Therefore $\Vert T\Vert \ge M$, and the exact operator norm isThis is the norm of a two-endpoint interpolation projection.

Because the linear functionals ${\ell }_0,{\ell }_1$ agree with endpoint point evaluations on continuous functions,The endpoint values also make $e_0,e_1$ linearly independent. By its definition, the image of a linear map $T$ lies in $E=⟨e_0,e_1⟩$. Conversely $T{e}_j=e_j$, so $T$ acts as the identity operator on $E$. ThusFor every $f$, $Tf\in E$, whence $T^{2}f=Tf$. It is a linear projection, and an annihilating polynomial is

Write $\psi ={\psi }_{x_0}$; its value at $x_0$, if that point is in $(0,1)$, may be assigned arbitrarily because Lp spaces identify functions equal almost everywhere. Its L-infinity norm is one.

The supplied estimate shows that $\psi$ cannot agree almost everywhere with any continuous function $q$: choosing $g=-q$ would give $0\ge |q(x_0)|+1$. Henceis a direct sum, so every element has a unique expression $g+c\psi$.

For any complex $a$ with $|a|\le 1$, define the linear functionalIf $c\ne 0$, the supplied estimate, applied to $g/c$, and absolute homogeneity of a norm giveFor $c=0$, the same bound follows from point evaluation. Thus $\Vert L_a\Vert \le 1$, and $L_a(1)=1$ makes its operator norm exactly one. The complex Hahn-Banach theorem extends $L_a$ to a norm-one linear functional on all of $X$, with value $a$ at $\psi$.

Conversely every norm-one extension satisfies $|{\ell }_{x_0}(\psi )|\le \Vert \psi {\Vert }_{\infty }=1$. We have therefore determined all possible values:In particular, taking $a=0$ and $a=1$ proves nonuniqueness of norm-preserving point evaluation extensions.

For a continuous function $g$ on $[0,1]$, the L-infinity norm of its restriction to $(0,1)$ equals its supremum norm on $[0,1]$. Indeed, values arbitrarily close to a maximum occur on an interval of positive Lebesgue measure, even if the maximum lies at an endpoint. Also, continuous functions equal almost everywhere on $(0,1)$ are equal everywhere on $[0,1]$. Thus restriction embeds $C([0,1])$ isometrically as a linear subspace of $X$.

The point evaluation functional $E_{x_0}(g)=g(x_0)$ is complex-linear and satisfies $|E_{x_0}(g)|\le \Vert g{\Vert }_{\infty }$. The constant function one attains equality, so $\Vert E_{x_0}\Vert =1$. The complex Hahn-Banach theorem extends this bounded linear functional from that linear subspace to $X$, preserving its operator norm. Therefore

The Neumann series criterion states that $\Vert A\Vert <1$ for a bounded linear operator on a Banach space implieswith convergence in the operator norm. In particular, $\Vert T\Vert <1$ suffices for invertibility of $I-T$.

If $|\lambda |>\Vert T\Vert$, apply this criterion to $A=T/\lambda$. It givesThus no such $\lambda$ belongs to the spectrum, and

For each integer $j\ge 1$, if $|\lambda {|}^j>\Vert T^j\Vert$, then ${\lambda }^{j}I-T^j$ is invertible by the same argument. The commuting factorizationthen gives a bounded two-sided inverse for $\lambda I-T$, namely the displayed sum multiplied by $({\lambda }^{j}I-T^j{)}^{-1}$. Hence every $\lambda \in \sigma (T)$ satisfies $|\lambda {|}^j\le \Vert T^j\Vert$. Taking the supremum over the spectrum and then the infimum over $j$ proves

Writing $\mathcal{B}(X)$ for the Banach space of bounded linear operators on $X$, the spectrum isHere $I$ is the identity operator. Since $X$ is a Banach space, the bounded inverse theorem makes this equivalent to failure of bijectivity of $\lambda I-T$.

The spectral radius isFor a nonzero complex Banach space, the spectrum of a bounded linear operator is nonempty and compact, so the supremum is a maximum. If the zero space is allowed, its empty spectrum is assigned spectral radius zero.

When $p=\infty$, $p^{\prime }=1$ and $T_{h}f=h^{-1}{\int }_{-h}^{h}f(x)dx$. Take the bounded measurable functionIt represents an element of the Lp space $L^{\infty }((-1,1))$, with L-infinity norm at most one. Sincewhere the lower endpoint contributes zero, its symmetric Lebesgue integral givesFor $h_n=e^{-2\pi n}$ and ${\tilde{h}}_n=e^{-(2\pi n+\pi )}$, both sequences tend to zero through $(0,1)$, but $T_{h_n}f=-1$ and $T_{{\tilde{h}}_n}f=1$. Therefore

Use $D=C([-1,1])$, restricted to $(-1,1)$ and regarded as a linear subspace of $L^p((-1,1))$. The density of continuous functions in Lp on an interval says that $D$ is a dense subspace whenever $1\le p<\infty$.

For $g\in D$, the supremum norm givesIn particular $D\subseteq Y$, with zero limit. Part (c), together with $D$ being a dense subspace, implies $Y=X$. To identify the limit throughout $X$, for $f\in X$ and $g\in D$ useFirst choose $g$ in the dense subspace to make the first term arbitrarily small, then let $h\to 0^{+}$. This is convergence on a dense subspace under a uniform operator bound, and provesThe zero linear functional belongs to the continuous dual space $X^{\ast }$.

The answer isIndeed, Lebesgue measure on $E\subset {\mathbb{R}}^n$ is a sigma-finite measure, and duality of Lp spaces gives $(L^1(E){)}^{\ast }=L^{\infty }(E)$. The restricted Lebesgue measure is a non-atomic measure and $E$ has positive measure, so the Dual of L infinity is larger than L1: the canonical integral representation does not exhaust $(L^{\infty }(E){)}^{\ast }$.

For $1<p<\infty$, the Holder conjugate exponent also lies strictly between one and infinity. Duality of Lp spaces then holds in both directions, in agreement with Reflexivity of Lp spaces. For $p=\infty$, the first required identification already fails.