Past exam of the functional analysis II course of the University of Oxford / 2023 / 2

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.

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 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.

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.

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 $Y$.

Let $J:C([0,1])\to L^1(0,1)$ send a continuous function to its almost everywhere equivalence class. The domain with the supremum norm and the target with the L1 norm are Banach spaces, andThus $J$ is a bounded linear operator, and the requested vector subspace is $J(C([0,1]))$.

To show this image of a linear map is not closed, consider the continuous functionsand $g=1_{(1/2,1)}$. Their L1 norm difference is the area of one triangle:But $g$ has no continuous function representative. Indeed, a continuous function equal to zero almost everywhere on $(0,1/2)$ must be identically zero there: any nonzero value would persist on an open interval of positive Lebesgue measure. Likewise any continuous representative would be identically one on $(1/2,1)$. Continuity at $1/2$ makes these requirements incompatible. Therefore $g\notin J(C([0,1]))$, although it lies in its closure.

The preceding operator range category dichotomy proves continuous functions are meagre in L1: