Homeomorphic closed copy of the unit ball of the space of bounded linear operators.

Question

Here is an example.

Let $X,Y$ be separable Banach spaces and denote by $L(X,Y)$ the Banach space of bounded linear operator $T\colon X\to Y$ with norm $$\| T\|=\sup \{\| Tx\|\colon x\in X, \| x\|\le 1\}.$$ Denote by $$L_1(X,Y)=\{T\in L(X,Y)\mid \| T\|\le 1\}$$ the unit ball. The strong topology on $L(X,Y)$ is the topology generated by the family of functions $f_x(T)=Tx,f_x\colon L(X,Y)\to Y$, for $x\in X$. It has as a basis the sets of the form $$V_{x_1,\dots,x_n;\epsilon;T}=\{S\in L(X,Y)\colon \| Sx_1-Tx_1\| < \epsilon,\dots,\| Sx_n-Tx_n\| < \epsilon\}$$ for $x_1,\dots,x_n\in X,\epsilon>0,T\in L(X,Y)$.

The unit ball $L_1(X,Y)$ with the (relative) strong topology is Polish. To see this, consider, for notational simplicity, the case of real Banach spaces, and let $D\subseteq X$ be countable dense in $X$ and closed under rational linear combinations. Consider $Y^D$ with the product topology, which is Polish, since $D$ is countable. The map $T\mapsto T|_D$ from $L_1(X,Y)$ into $Y^D$ is injective and its range is the following closed subset of $Y^D$: $$F=\{f\in Y^D\colon \forall x,y\in D,\forall p,q\in \Bbb{Q}[f(px+qy)=pf(x)+qf(y)]\}\cap\{f\in Y^D\colon \forall x\in D(\| f(x)\|\le\| x\|)\}.$$ It is easy to verify that this map is a homeomorphism of $L_1(X,Y)$ and $F$, thus $L_1(X,Y)$ with the strong topology is Polish.

This is what I found in a book. Now, I want to work out the details of the second part. As I understand it, the author can consider the case of real Banach spaces because every field of characteristic $0$ contains a copy of $\Bbb{Q}$, the prime field. I guess similarly I have to consider the finite field $\Bbb{F}_{p^n}$ when the characteristic is prime. Now, I want to show that $F$ is closed in $Y^D$. If $\operatorname{Hom}_{\Bbb{Q}}(X,Y)$ denotes the space of $\Bbb{Q}$-vector space homomorphisms from $X$ to $Y$, then I have $$F=\operatorname{Hom}_{\Bbb{Q}}(X,Y)\cap\{f\in Y^D\colon \forall x\in D(\| f(x)\|\le \| x\|)\}.$$ The last set is the preimage of the closed unit interval $[0,1]$ under the norm, which is a continuous function, thus it is closed.

However, I can't answer the following simple questions: $(1)$ how do I show $\operatorname{Hom}_{\Bbb{Q}}(X,Y)$ is closed in $Y^D$?

$(2)$ how to show the map $T\mapsto T|_D$ is a homeomorphism?

Thank you in advance for your help

Answer 1

Let's show that $T \mapsto T|_D$ is a homeomorphism.

Assume $T_n \to T$ in $L_1(X,Y)$ strongly. Then in particular for every $x \in D$ we have $T_nx \to Tx$ in $Y$ so $T_n \to T$ w.r.t the product topology on $Y^D$. Therefore, $T \mapsto T|_D$ is continuous.

To construct the inverse of $T \mapsto T|_D$, consider $f \in Y^D$. We claim that $f$ uniquely extends to an element of $L_1(X,Y)$.

For $x \in X$ take a sequence $(d_n)_n$ in $D$ such that $d_n\to x$. The sequence $(f(d_n))_n$ is Cauchy in $Y$ since $$\|f(d_m) - f(d_n)\| = \|f(d_m - d_n)\| \le \|d_m-d_n\|$$ Define $Tx := \lim_{n\to\infty} f(d_n)$. Then $T$ is a well-defined map $X\to Y$. Indeed, if $(d_n')_n$ is in $D$ such that $d_n' \to x$ as well, then the sequence $d_1, d_1', d_2, d_2', \ldots $ in $D$ also converges to $x$ so its limit is equal to $\lim_{n\to\infty} f(d_n)$ and $\lim_{n\to\infty} f(d_n')$. In particular $\lim_{n\to\infty} f(d_n) = \lim_{n\to\infty} f(d_n')$.

By considering a constant sequence it follows that $T$ extends $f$.

For $x \in X$ take a sequence $(d_n)_n$ in $D$ such that $d_n \to x$. We have

$$\|Tx\| = \left\|\lim_{n\to\infty}f(d_n)\right\| = \lim_{n\to\infty}\|f(d_n)\| \le \lim_{n\to\infty} \|d_n\| = \|x\|$$

We claim that $T$ is linear. For $x,y \in X$ take sequences in $D$ such that $d_n \to x$, $d_n' \to y$. Then $d_n + d_n' \to x+y$ so $$T(x+y) = \lim_{n\to\infty} f(d_n + d_n') = \lim_{n\to\infty} f(d_n) + \lim_{n\to\infty} f(d_n') = Tx+Ty$$

For $x \in X$ take a sequence $(d_n)_n$ in $D$ such that $d_n \to x$. For $q \in \mathbb{Q}$ we have that $qd_n \to qx$ so $$T(qx) = \lim_{n\to\infty} f(qd_n) = q\lim_{n\to\infty} f(d_n) = qTx$$ so $T$ is $\mathbb{Q}$-linear on $X$.

For $x \in X$ we have that the maps $\mathbb{R} \to \mathbb{R}$ given by $\lambda \to T(\lambda x)$ and $\lambda \to \lambda Tx$ are continuous:

$$\|T(\lambda x) - T(\mu x)\| = \left\|T\left((\lambda - \mu)x\right)\right\| \le \|\lambda - \mu\|\|x\|$$ $$\|\lambda Tx - \mu Tx\| = \|\lambda - \mu\| \|Tx\|$$

They also conincide on $\mathbb{Q}$ so they are equal. We conclude that $T$ is linear, and of norm $\le 1$ so $T \in L_1(X,Y)$. Clearly $T$ is a unique extension of $f$ by density of $D$ in $X$.

Now assume that $f_n \to f $ in $Y^D$. Then the induced maps $T_n, T$ satisfy $T_nx \to Tx$ for all $x \in D$. For $x \in X$ and $\varepsilon > 0$ take $d \in D$ such that $\|x-d\| \le \frac\varepsilon2$. Since $T_n \in L_1(X,Y)$ we have \begin{align} \|T_mx - T_nx\| &= \|T_mx - T_md\| + \|T_md - T_nd\| + \|T_nd - T_nx\|\\ &\le \|x-y\| + \|T_md - T_nd\| + \|x-y\|\\ &\le \varepsilon + \|T_md - T_nd\|\\ &\xrightarrow{m,n\to\infty} \varepsilon \end{align}

so $(T_nx)_n$ converges to an element of $Y$. If we define a linear map $T' : X\to Y$ with $T'x := \lim_{n\to\infty} T_nx$, we get that $T' \in L_1(X,Y)$ so it is continuous. Since $T$ and $T'$ coincide on a dense subset $D$, we conclude $T' = T$, so $T_n \to T' = T$ in $L_1(X,Y)$.

Therefore $T \mapsto T|_D$ is a homeomorphism.

It remains to prove that $F$ is closed in $Y^D$. Let $(f_n)_n$ be a sequence in $F$ such that $f_n \to f \in Y^D$. We claim that $f \in F$.

For $x \in D$ we have $$\|f(x)\| = \left\|\lim_{n\to\infty}f_n(x)\right\| \le \limsup_{n\to\infty}\left\|f_n(x)\right\| \le \|x\|$$

For $x,y \in D$, $p,q \in \mathbb{Q}$ we have $$f(px+qy) = \lim_{n\to\infty} f_n(px+qy) = \lim_{n\to\infty} (pf_n(x) + qf_n(y)) = p\lim_{n\to\infty} f_n(x) + q\lim_{n\to\infty} g_n(y) = pf(x) + qf(y)$$

so $f \in F$.