Past exam of the functional Analysis and Banach and C*-Algebras course of the University of Oxford / 2007 / A

The triangular kernel is nonnegative, has integral one, support in $[-\delta ,\delta ]$, and supremum $1/\delta$. Thus the convolution is defined at every point and

The kernel is globally Lipschitz with constant $1/{\delta }^2$: this holds on each of its linear pieces, and summing across the finitely many intervening breakpoints proves it across pieces as well. Consequently

Since the kernel has integral one, The Tonelli theorem and the triangle inequality giveThis proves the stated convolution approximation estimate.

For the unlabelled compactness assertion let $M$ bound the norms of $S$, and fix $\eta >0$. Choose the kernel width $\delta$ smaller than the translation radius supplied for $\eta$, so that the same bound holds for all shifts of size at most this width. The preceding estimate gives $\Vert f\ast g-f{\Vert }_1<\eta$ uniformly. The smoothed family has uniform bound $M/\delta$ and uniform Lipschitz constant $M/{\delta }^2$. Choose $K$ with all original tails outside $[-K,K]$ below $\eta$, and put $L=K+\delta$. The Tonelli theorem givesOn $[-L,L]$ the smoothed family is relatively compact in the uniform norm by the Arzelà-Ascoli theorem. Take a finite uniform net with error below $\eta /(2L)$ and extend its centers by zero outside that interval. It is a finite $L^1$ net for the smoothed-and-truncated family with error $\eta$. Each original $f$ is within another $2\eta$ of its corresponding truncated smoothing. Thus $S$ is totally bounded in $L^1$. The displayed (a),(b) are the two hypotheses used together in this argument.

For the final family, Young convolution inequality gives $\Vert f_0\ast f{\Vert }_1\le \Vert f_0{\Vert }_1$. Its translations satisfywhich tends to zero uniformly over the family. Since $f$ vanishes outside $[-1,1]$,uniformly as well. Both compactness hypotheses apply, proving $S_0$ is totally bounded.

Regard the spaces as real and set $u=\mathrm{Re}\varphi$. Its real norm equals $\Vert \varphi \Vert$, since rotating any test vector makes its functional value real and nonnegative without changing its norm. By the allowed real Hahn-Banach theorem extend $u$ to a real-linear $v:X\to \mathbb{R}$ with that norm, and define $\psi (x)=v(x)-iv(ix)$. Real linearity gives $\psi (ix)=i\psi (x)$, so $\psi$ is complex linear. On $Y$ this formula recovers $\varphi$. Also $\mathrm{Re}\psi =v$; rotating a test vector again gives $|\psi (x)|\le \Vert v\Vert \Vert x\Vert$. The reverse norm inequality follows by restriction to $Y$, proving a norm-preserving complex extension.

We use two standard consequences: a point outside a closed subspace can be separated by a bounded functional annihilating that subspace, and the norm of a vector is the supremum of its pairings with dual unit-ball functionals. If $\psi \in X^{\ast }$ annihilates $Y$, the scalar holomorphic function $\psi \circ f$ vanishes at the prescribed sequence tending to zero. Its identity theorem makes it identically zero. Separation then gives $f(z)\in Y$ for every $z\in D$.

For $\varphi \in Y^{\ast }$ extend to $X^{\ast }$ with the same norm and put $F=\varphi \circ f$. Then $F$ is holomorphic with $|F|\le \Vert \varphi \Vert$. For the understood nonzero increments $h,k$, the Cauchy integral formula on $|\zeta |=r$ givesTaking absolute values, bounding the contour length by $2\pi r$, and each denominator below by $r-|z|$, $r-|z|-|h|$, $r-|z|-|k|$ proves exactly the stated estimate.

Fix $z,r$ and restrict $|h|,|k|<(r-|z|)/2$. Taking the supremum over $\Vert \varphi \Vert \le 1$ bounds the norm of the difference of the two $Y$-valued difference quotients by a fixed constant times $|h-k|$. They are therefore Cauchy as $h\to 0$. Since $Y$ is a closed subspace of a Banach space, it is complete. Consequently the derivative exists as a norm limit in $Y$. The labelled (a)-(c) are the simultaneous hypotheses, not separate claims to prove.

A linear operator has closed graph when $\{(x,Tx):x\in X\}$ is closed in $X\times Y$. The closed graph theorem states that an everywhere-defined linear operator between Banach spaces is bounded if and only if it has closed graph. Boundedness gives the closed-graph property by continuity. Conversely a closed graph is Banach in the product norm $\Vert (x,y)\Vert =\Vert x\Vert +\Vert y\Vert$. Projection of the graph onto $X$ is a bounded linear bijection, so the open mapping theorem makes its inverse bounded. Composing that inverse with the second-coordinate projection proves $T$ is bounded.

For the composition assertion, suppose $x_n\to x$ and $T{x}_n\to y$. Boundedness of $S$ gives $ST{x}_n\to Sy$, and boundedness of $ST$ gives $ST{x}_n\to STx$. Injectivity of $S$ forces $y=Tx$. Hence $T$ has closed graph and is bounded by the theorem.

For the derivative bound take $Y=C^1[0,1]$ with norm $\Vert v{\Vert }_{C^1}=\Vert v{\Vert }_{\infty }+\Vert v^{\prime }{\Vert }_{\infty }$, and $Z=C[0,1]$ with its supremum norm; both are Banach. Regard $T(x)=Ux$ as a map into $Y$, and let $S:Y\to Z$ be inclusion. This inclusion is bounded and injective, and $ST=U$ is bounded by hypothesis. The preceding result gives a constant $c$ with $\Vert Ux{\Vert }_{C^1}\le c\Vert x\Vert$. In particular $|(Ux{)}^{\prime }(t)|\le c\Vert x\Vert$ uniformly in $t$.

A bounded linear operator is compact when the closure of the image of its unit ball is compact. Equivalently every bounded sequence has a subsequence whose images converge; on a Banach target it also suffices that the image of its unit ball be totally bounded.

For a finite-rank operator that image is bounded in a finite-dimensional subspace, whose closed bounded sets are compact. Thus every finite-rank operator is compact.

Given a bounded sequence, first pass to a subsequence on which the images under $T_1$ converge. From that subsequence extract another on which the images under $T_2$ converge. Both image sequences, and hence their sum, converge on the latter subsequence. Thus $T_1+T_2$ is compact.

Extract a subsequence whose $T$ images converge, and apply continuity of the bounded operator $S$ to those images. This proves $ST$ is compact. The reverse composition $TS$ is compact as well, since $S$ sends a bounded sequence to a bounded sequence to which compactness of $T$ applies.

Suppose $T_n$ are compact and $\Vert T_n-T\Vert \to 0$. For any $\epsilon >0$ take $n$ with $\Vert T_n-T\Vert <\epsilon$. A finite $\epsilon$-net of $T_n$ of the unit ball is then a finite $2\epsilon$-net of its $T$ image. That image is totally bounded, so $T$ is compact by the stated criterion. Hence $\mathcal{K}(X)$ is norm closed in $B(X)$.

Compact. Let $P_N$ keep the first $N$ coordinates. The operator $P_{N}T$ has finite rank and $\Vert T-P_{N}T\Vert \le 1/(N+1)$, since the remaining diagonal coefficients are at most that number. Norm-closedness of compact operators proves the claim.

Not compact. For coordinate unit vectors $e_j$ with $j\ge 2$, the images are $e_{j-1}$. Distinct such images have ${\ell }^1$ distance two, so no image subsequence converges.

Compact. This operator is $BD$, where $D$ is the compact diagonal operator in (a) and $B$ is the bounded backward shift in (b). The composition property applies. Equivalently its tails have the same vanishing-coefficient finite-rank approximation.

Compact. Its column indexed by $m$ has entries $n!/m!$ for $1\le n<m$, and zero otherwise. The sum of that column is $c_m=(1!+\cdots +(m-1)!)/m!\le 2/m$ for $m\ge 2$, by the elementary induction $1!+\cdots +k!\le 2k!$. For operators from ${\ell }^1$ the supremum of the absolute column sums is the operator norm: one inequality is the triangle inequality, and coordinate unit vectors give the reverse. Keeping just the first $N$ input columns is finite rank and its remainder norm is at most ${\sup }_{m>N}{c}_m\le 2/(N+1)$. Hence it converges to $T$ in norm, proving compactness.