3
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.