# Pointwise Convergence in Banach Space Implies Convergence in Operator Norm

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?

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 $$n$$.