I am working through Lax's "Functional Analysis", and I'm trying to prove Theorem 6 of section 15.2, dealing with weak and strong convergence in the operator space. We denote by $s-\lim$ the fact that a sequence converges strongly (in norm), and by $w-\lim$ the weak convergence. Specifically, the theorem states:

Let X, U be Banach spaces, $M_n$ a sequence of linear maps in $\textit{L}(X,U)$ uniformly bounded in norm: $\mid M_n\mid\leq c$ for all $n$. Suppose further that $s-\lim M_nx$ exists for a dense set of $x$ in $X$. Then, $M_n$ converges strongly, i.e., the $s-\lim$ exists for all $x \in X$.

I need to prove this result, and also an analogous one for weak convergence.

My attempt at the strong convergence goes as follows: Let $D\subset X$ be the dense set for which $s-\lim M_nx$ exists. If $x \in X-D$, since D is dense, there exists a sequence $(x_j)_{j\in \mathbb{N}} \subset D$ such that $x_j \to x$ as $j\to\infty$. For each $j\in\mathbb{N}$, let $y_j = s-\lim_{n\to\infty} M_n(x_j)$. Then, we claim that the sequence $(y_j)_{j\in\mathbb{N}}$ is Cauchy in U. This follows from the fact that $(x_j)_{j\in\mathbb{N}}$ is Cauchy and $(M_n)$ is uniformly bounded in norm:

$\mid y_j - y_k \mid = \mid \lim M_n(x_j) - \lim M_n(x_k) \mid = \mid \lim M_n(x_j-x_k) \mid = \lim \mid M_n(x_j-x_k)\mid \leq \liminf \mid M_n \mid \mid x_j - x_k\mid \leq c\mid x_j-x_k\mid$

Hence, since U is Banach, $y_j \to y$ as $j\to\infty$ for some $y\in U$. Now, we claim that $M_nx \to y$ as $n\to\infty$. This follows since:

$\mid M_nx-y \mid \leq \mid M_nx-M_nx_j\mid + \mid M_nx_j-y_j\mid + \mid y_j-y \mid$.

Since the above is true $\forall j \in \mathbb{N}$ and we have that $x_j \to x$ and $y_j \to y$, given $\varepsilon>0$ we find an $j_0$ such that $\mid x_j - x \mid < \frac{\varepsilon}{3c}$ and $\mid y_j -y \mid < \frac{\varepsilon}{3}$ for all $j\geq j_0$. Then, we choose $n$ large enough such that $\mid M_nx_{j_0} - y_{j_0} \mid < \frac{\varepsilon}{3}$, and the result follows.

Firstly, I don't see where we use the completeness of X. Futhermore, I'm not sure this proof generalizes for weak convergence.

For weak convergence, the weak limit would exist for a dense set $D\subset X$. Then, we proceed in the same manner: if $x\in X-D$, there exists a sequence such that $x_j\to x$ as $j\to\infty$. Let $l \in U'$ be non-zero. Since the weak limit exists in D, we have that $M_nx_j$ converges weakly to $y_j \in U$. Hence, $l(M_nx_j)\to l(y_j)$ for some $y_j \in U$. We claim that $(l(y_j))_j$ is Cauchy in $\mathbb{K}$. Again, this follows since:

$\mid l(y_j) - l(y_k) \mid = \mid \lim_n l(M_nx_j) - \lim_n l(M_nx_k) \mid \leq \liminf \mid l \mid \mid M_n \mid \mid x_j - x_k \mid $

Hence, $l(y_j)\to l(y)$ for a certain $y\in U$, again, this is because $l$ is surjective. Finally, we claim that $l(M_nx) \to l(y)$, and the proof is analogous to the case above. However, I don't think that this is enough to conclude that the weak limit $w-\lim M_nx$ exists. This is because our choice of $y$ depends very much on the functional $l$ we chose to start with. Assuming that my proof on the strong limit is right, is there a better way to generalize it to prove the statement about weak limits? If not, how can I approach this problem?


It seems that the assumption that $X$ is a Banach space is not needed in the first part.

For the second one, we assume that for all $x\in D$, there exists a $y\in U$ such that $M_nx\to y$ weakly.

Let $x\in X$. Let $x_i\in D$ such that $\left\lVert x-x_i\right\rVert\leqslant i^{-1}$. We know that there exists $y_i\in U$ such that $M_nx_i\to y_i$ weakly. We can prove that the sequence $\left(y_i\right)_{i\geqslant 1}$ is Cauchy. Indeed, if $i,j$ are integers, then $\left(M_n\left(x_i-x_j\right)\right)_{n\geqslant 1}$ converges weakly to $y_i-y_j$ hence $$ \left\lVert y_i-y_j\right\rVert_U\leqslant \liminf_{n\to+\infty}\left\lVert M_n\left(x_i-x_j\right)\right\rVert_U\leqslant c \left\lVert x_i-x_j \right\rVert_X\leqslant ci^{-1}+cj^{-1}. $$ Now we have to prove that $\left(M_n\left(x\right)\right)_{n\geqslant 1}$ converges weakly to $y$. Pick $\ell\in U'$. Then $$ \left\lvert \ell\left(M_n\left(x\right)\right)- \ell\left(y\right) \right\rvert \leqslant \left\lvert \ell\left(M_n\left(x_i\right)\right)- \ell\left(y_i\right) \right\rvert+\left\lvert \ell\left(M_n\left(x\right)\right)- \ell\left(M_n\left(x_i\right)\right)\right\rvert+\left\lvert \ell\left(y\right)- \ell\left(y_i\right) \right\rvert.$$ The second term of the right hand side does not exceed $\left\lVert \ell\right\rVert_{U'}ci^{-1}$ hence we deduce that for all fixed $i$, $$ \limsup_{n\to +\infty}\left\lvert \ell\left(M_n\left(x\right)\right)- \ell\left(y\right) \right\rvert \leqslant \left\lVert \ell\right\rVert_{U'}ci^{-1}+\left\lvert \ell\left(y\right)- \ell\left(y_i\right) \right\rvert. $$ Since $i$ is arbitrary and the right hand side goes to zero, we get the wanted result.

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  • $\begingroup$ In the end I also came up with a similar argument. I was able to prove that the operator $M$ on $D$ defined by $Mx = w-\lim M_nx$ is continuous. After that, I could prove that the extension of $M$ to the whole space $X$ agrees with the weak limit even if $x \in X-D$. But, after all, continuity of $M$ is amounts to the same as the fact that $(y_j)$ is Cauchy. Anyway, your solution is much cleaner, thank you! $\endgroup$ – André Muchon Feb 24 '19 at 22:27

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