# Strong and weak limits in a dense set for a sequence of bounded operators

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.
• 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! – André Muchon Feb 24 '19 at 22:27