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Assume that $(a_n : V \rightarrow W, n \geq 0)$ is a sequence of continuous linear maps with $V$ is Banach space, $W$ a normed space such that $(a_n(v))_{n \geq 0}$ is convergent for any $v \leq V$. Prove $(a_n)_{n\geq 0}$ converges in $B(V,W)$ to a continuous linear map for the norm topology of $B(V,W)$.

$B(V,W)$ is defined to be the space of continuous, bounded linear maps between $V$ and $W$, and the operator norm is defined be: $\|a\| = \sup _{\|v\| = 1}\|a(v)\|$.

My Attempt

Since $a_n(v)$ converges pointwise for all $v\in V$, we can define a function $a: V \rightarrow W$ where $a(v) = \lim _{n\rightarrow \infty} a_n(v)$. First prove that $a$ is linear: $$\begin{align} \|a(\lambda v + w) - \lambda a(v) - a(w) \| &\leq \|a(\lambda v + w) - a_n(\lambda v + w)\| + \|a_n(\lambda v + w) - \lambda a(v) - a(w) \| \\ &\leq \|a(\lambda v + w) - a_n(\lambda v + w)\| + |\lambda| \|a(v) - a_n(v)\| + \| a(w) - a_n(w) \| \end{align}$$ Since $a_n(v)$ converges pointwise to $a(v)$ for all $v \in V$, the above converges to $0$, thus $a$ is linear.

To prove continuity, a previous theorem stated that if $A \subset B(V,W)$ and for any $v \in V$ we have $\sup _{a\in A} \|a(v)\| < \infty$, then $\sup _{a\in A} \|a\| < \infty$. Since the sequence $(a_n(v))_n$ converges pointwise, for all $v \in V$, $a_n(v)$ is bounded and hence satisfies the condition for the previous theorem. Hence we can define $M = \sup _n \|a_n\|$ and hence obtain that $\|a(v)\| \leq \|v\| M$. Therefore $a$ is bounded and hence continuous.

Now, the part I am struggling with is how to prove that $(a_n)_n$ converges to $a$ using the operator norm. I am not really sure how to approach this nor how I can use the property that $V$ is a Banach space to prove this.

Could anyone please point me in the right direction to prove this last part. Thank you.

Edit

The question asks to show $(a_n)_n$ converges to $a$ "for the norm topology of $B(V,W)$". I am not sure what that last bit means. I initially assumed it meant using operator norm, however I can now see from @Gae. S. answer, that is not true. Am I misinterpreting the question, and if so, could someone please explain what that last phrase means?

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This is false. Consider $T:\ell^2\to\ell^2$ defined by $[Tv]_j=v_{j+1}$ and consider the sequence $\{T^n\}_{n\in\Bbb N}$. $T^nv\to 0$ for all $v\in\ell^2$, yet $\lVert T^n\rVert=1$ for all $n$.

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  • $\begingroup$ Thank you for your answer, that makes sense. I have edited my question to show the original statement that I was asked to prove. Could you please see if I have misinterpreted the question (which I probably have seeing that it is not true)? Thanks. $\endgroup$ – Nanoputian May 15 at 22:37
  • $\begingroup$ @Nanoputian I don't see a difference. $\endgroup$ – Gae. S. May 16 at 14:34

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