Task:
I was studying these examples
- here, but it assumes here directly the linear boundedness
- here pp. 243-244, but there is something strange starting from Since each $T \in S$ ... in my proposal. The link from showing $\| Au \| < \infty$ cannot be direct to show that $\|A_{k} - A\| \to 0$. There must be some stages missing.
- one proof of the theorem here but not about the boundedness of the linear operator so not sufficient
My proposal
Let $X$ be a Banach space, $Y$ a normed space, $S \subset \mathscr{B}(X,Y)$, $D := X$ and let operators $A_{k} \in \mathscr{B}(X,Y)$ satisfy for all $u \in X$: \begin{equation*} \| A_{k} u \| < \infty, \, \text{ when } \, k \to \infty. \end{equation*} where $\| A_{k} u \|$ denotes the norm of the linear transformation $A_{k}$ from $X$ to $Y$. The sup of the $\|A_{k} u\|$ is not necessarily bounded.
Since each $T \in S$ is continuous, the inverse image of $T$ of a closed set is closed, and so \begin{equation*} E := \cap_{T \in S} T^{-1} ( \frac 12 \overline{B_{Y}}) \end{equation*} is closed in $X$. Let $x \in D$. Then there is some integer $n$ such that $Tx \in n B_{Y} \subset 2n( \frac 12 \overline{B_{Y}})$ for all $T \in S$. Because $E$ is fat and closed, $E$ must have an interior point in $X$. Let $x_{0}$ be any interior point of $E$. Then $-x_{0} + E$ is a neighborhood of $0$ in $X$. Hence, there is some $\delta > 0$ such that $\delta B_{X} \subset -x_{0} + E$. Let $T \in S$. Then $T(E) \subset \overline{B_{Y}}/2$, and since $x_{0} \in E$, we have $Tx_{0} \in \overline{B_{Y}}/2$, so that $T(\delta B_{X}) \subset -Tx_{0} + T(E) \subset \overline{B_{Y}}/2 - \overline{B_{Y}}/2 \subset \overline{B_{Y}}$. Hence $T(B_{X}) \subset \overline{B_{Y}}/\delta,$ so \begin{equation*} T(\overline{ B_{X} }) \subset T(2 B_{X}) \subset \frac 2\delta \overline{B_{Y}}. \end{equation*} Since $T \in S$ was arbitary, $\sup_{T \in S} \|T\| \leq 2/\delta$, proving \begin{equation*} \| Au \| < \infty, \end{equation*} where $A_{k} \to A$ converges strongly. Then \begin{equation*} \lim\limits_{k \to \infty} \| A_{k} u - Au \|_{Y} = 0. \, \square \end{equation*}
Addition to Svetoslav's answer about linearity
The linearity of $A$. Let $X$ be a normed space and $Y$ a Banach space. Let $M \subset X$ is a dense subspace of $X$. If $A : M \subset X \to Y$ is a bounded linear map, then there is a unique bounded linear map $\bar{A} : X \to Y$ (extension by continuity) such that $\bar{A}x = Ax$ for all $x \in M$. Moreover, $\|\bar{A}\| = \|A\|$.
For every $x \in X$, there is a sequence $(x_{n})$ in $M$ that converges to $x$. We define \begin{equation*} \bar{A}x = \lim\limits_{k \to \infty} Ax_{k}. \end{equation*} This limit exists becuase $(Ax_{k})$ is Cauchy, since $A$ is bounded and $(x_{k})$ Cauchy, and $Y$ is complete. We claim that the value of the limit does not depend on the sequence in $M$ that is used to approximate $x$. Suppose $(x_{k})$ and $(x_{k}')$ are any two sequences in $M$ that converge to $x$ and let $Ax_k'\to y'\in Y$, $Ax_k\to y\in Y$. Then \begin{equation*} \| x_{k} - x_{k}' \| \leq \| x_{k} - x \| + \| x - x_{k}' \|, \end{equation*} and, taking the limit of this expression as $k \to \infty$, we see that \begin{equation*} \lim\limits_{k \to \infty} \| x_{k} - x_{k}' \| = 0. \end{equation*} It follows that \begin{equation*} \| Ax_{k} - Ax_{k}' \| \leq \|A\| \| x_{k} - x_{k}' \| \to 0, \, \text{ as } \, k \to \infty. \end{equation*} Hence, $(Ax_{k})$ and $(Ax_{k}')$ converge to the same limit $y\equiv y'$, because $$0\leq \|y-y'\|\leq \|y-Ax_k\|+\|Ax_k-Ax_k'\|+\|Ax_k'-y'\|\to 0$$ when $k\to\infty$
The map $\bar{A}$ is an extension of $A$, meaning that $\bar{A}x = Ax$, for all $x \in M$, because if $x \in M$, we can use the constant sequence with $x_{k} = x$ for all $k$ to define $\bar{A}x$. The linearity of $\bar{A}$ follows from the linearity of $A$.
The fact that $\bar{A}$ is bounded follows from the inequality \begin{equation*} \| \bar{A} x \| = \lim\limits_{k \to \infty} \| A x_{k} \| \leq \lim\limits_{k \to \infty} \|A\| \|x_{k}\| = \|A\| \|x\|. \end{equation*}
It also follows that $\|\bar{A}\| \leq \|A\|$. Since $\bar{A}x = Ax$ for all $x \in M$, we have $\|\bar{A}\| = \|A\|$.
Finally, we show that $\bar{A}$ is the unique bounded linear map from $X$ to $Y$ that coincides with $A$ on $M$. Suppose that $\tilde{A}$ is another such map, and let $x$ be any point in $X$. We choose a sequence $(x_{k})$ in $M$ that converges to $x$. Then, using the continuity of $\tilde{A}$, the fact that $\tilde{A}$ is an extension of $A$, and the definition of $\bar{A}$, we see that \begin{equation*} \tilde{A}x = \lim\limits_{k\to\infty} \tilde{A}x_{k} = \lim\limits_{k\to\infty} Ax_{k} = \bar{A}x. \end{equation*}
Sources
- chapter 5, Banach spaces, here: Bounded Linear Transformation
I am not sure about how they define the strong convergence here: in Hilbert space? Which are necessities for it such that the bounded linear operator also hold. How can you prove more rigorously the boundedness of the linear operator?
The linearity of $A$ is straightforward: Let $\lambda\in\mathbb R$ and $x,y\in X$. Then we have $$A(\lambda x):=\lim\limits_{k\to\infty}{A_k(\lambda x)}=\lim\limits_{k\to\infty}{\lambda A_kx}=\lambda \lim\limits_{k\to\infty}{A_kx}=:\lambda Ax$$ $$A(x+y):=\lim\limits_{k\to\infty}{A_k(x+y)}=\lim\limits_{k\to\infty}{A_kx}+\lim\limits_{k\to\infty}{A_ky}=:Ax+Ay$$
Boundedness: You have that for each fixed $u\in X:$ $\|A_ku-Au\|_Y\to 0$. This means that $\|A_ku\|_Y\to \|Au\|_Y$ (here $\|Au\|_Y$ is some finite number ). Therefore $\exists M_u :\|A_ku\|_Y\leq M_u,\,\,\forall k\in\mathbb N$, i.e the quantity $\|A_ku\|_Y$ is uniformly bounded in $k$ for each fixed $u\in X$. Now, from the uniform boundedness principle (Banach–Steinhaus theorem) we get that $\|A_k\|$ (operator norm) is uniformly bounded in $k$, i.e $\exists M=const:\|A_k\|\leq M,\,\,\forall k\in\mathbb N$.
Using this and the triangle inequality, we get: $$\|Au\|_Y-\|A_ku\|_Y\leq \|A_ku-Au\|_Y$$ Let $\epsilon>0$ is arbitrary. Then $$\|Au\|_Y\leq \|A_ku\|_Y + \|A_ku-Au\|_Y\leq \|A_k\|\|u\|_X+\|A_ku-Au\|_Y\leq M\|u\|_X+\epsilon$$ for big enough $k$. This means that $\|Au\|_Y\leq M\|u\|_X+\epsilon$ for each $\epsilon>0$, or equivalently $\|Au\|_Y\leq M\|u\|_X$ which shows the boundedness of $A$.