2
$\begingroup$

$(X, \|•\|) $and $(Y, \|•\|') $ be two normed space and $\begin{align} {\scr{B}}{(X, Y) }&=\{T\in {\scr{L}}{(X,Y)} : T \text { is bounded } \}\end{align}$

$\|T\|_{op}=\sup_{\|x\|\le1}\|Tx\|' $

I want to understand the convergence of sequences in the space ${(\scr{B}}{(X, Y) }, \|•\|_{op}) $

$T_n\to T$ pointwise on $X$ i.e $\forall x\in X, T_n x \to Tx$$ \text { in the space }\space \space$$ (Y, \|•\|') $.

Question : Does this implies $T_n\to T $ in the space ${(\scr{B}}{(X, Y) }, \|•\|_{op}) $?

  1. $T_n\to T $ pointwise implies $T\in {\scr{L}}{(X, Y) }$

Proof: $\forall$ $x, y\in X$ and $\lambda \in \mathbb{K}$

$\begin{align}T(x+\lambda y) &=\lim_{n\to\infty} T_n(x+\lambda y)\\&=\lim_{n\to\infty} T_nx +\lambda T_ny\\&=\lim_{n\to \infty} T_nx+\lambda \lim_{n\to\infty}T_ny\\&=Tx+\lambda Ty \end{align}$

  1. $T$ may not be continuous.

For an example, choose $ (\ell_1, \|•\|_{\infty}) $ and

$(f_n) \in (\ell_1,\|•\|_{\infty})^{*}$ defined by $$f_n(x) =\sum_{k=1}^{n} x_k$$

Where $x=(x_1, x_2,...) \in\ell_1$

Then, $f_n\to f$ pointwise where$$f(x) =\sum_{k\in \Bbb{N}} x_k$$

But $f\notin (\ell_1,\|•\|_{\infty})^{*}$

Here we need $(X, \|•\|) $ to be a Banach space to ensure that $T\in {\scr{B}}{(X, Y) }$ [Banach-Steinhaus Theorem]

  1. Suppose in addition, $(X, \|•\|) $ is a Banach space and then $T\in {\scr{B}}{(X, Y) }$ but still $(T_n)$ may not converge to $T$ in the space ${(\scr{B}}{(X, Y) }, \|•\|_{op}) $

For an example, choose $(c_0, \|•\|_{\infty})$ and $(f_n) $ defined by $$f_n(x) =x_n$$ where $x=(x_1, x_2,...) \in c_0$ i.e converges to $0$.

Then, $f_n\to 0 $ pointwise but $f_n \nrightarrow 0$ in the Dual space.

My question : What will be the next step to ensure that the pointwise convergence implies convergence in the operator norm?

$\endgroup$
1
  • 2
    $\begingroup$ There is no useful sufficient condition for that. $\endgroup$ Commented Apr 27, 2022 at 8:24

1 Answer 1

4
$\begingroup$

As Kavi mentioned, there are no good answers to your question. Pointwise convergence is strictly weaker than convergence in $\|\cdot\|_{op}$, so you will need to add fairly strong hypotheses to the former in order to get the latter.

The following are examples of some of these additional hypotheses:

  1. The domain $X$ is finite-dimensional.

  2. All of the $T_n$ lie in some fixed finite-dimensional subspace of ${\scr B}(X, Y)$.

  3. $\{T_n\}_n$ is a Cauchy sequence for $\|\cdot\|_{op}$.

  4. When verifying that $T_n(x)→T(x)$, that is, that for every $ε>0$ there is some $n_0$ such that $n≥n_0 \Rightarrow \|T_n(x)-T(x)\|<\epsilon$, you are able to exhibit $n_0$ depending only on $\epsilon$ and $\|x\|$ (rather than $\epsilon$ and $x$).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .