$B(V,W)$ is complete if $W$ is

Let $B(V,W)$ be the space of bounded linear maps from $V$ to $W$. Then it is complete with respect to the operator norm. Can you tell me if my proof is correct? Thanks.

It's easy to verify that the operator defines a norm. Let $T_n$ be Cauchy in $B(V,W)$ with respect to $\|\cdot\|$. Let $\varepsilon > 0$. We want to show that $T_n \to T$ for some $T \in B(V,W)$.

We have that $T_n v$ is a Cauchy sequence in $W$ since $\|T_nv - T_mv\|_W \leq \|T_n-T_m\| \|v\|_V < \varepsilon$ for $n,m$ large enough since $T_n$ is Cauchy with respect to $\|\cdot\|$ by assumption. Since $W$ is complete, the pointwise limit $Tv$ of $T_nv$ is in $W$ for all $v$. Now we need to show that $T$ is linear, bounded and $T_n \to T$ in the operator norm.

(i) $T(\alpha v + \beta w) = \lim_{n \to \infty} T_n(\alpha v + \beta w) = \alpha \lim_{n \to \infty} T_nv + \beta \lim_{n \to \infty} T_n w = \alpha T v + \beta T w$

(ii) $\|Tv\| = \sup_{\|v\| \leq 1} \|Tv + T_n v - T_n v\| \leq \sup_{\|v\| \leq 1} \|Tv - T_n v\| + \| T_n v\|$

(iii) $\|(T-T_n)v \|_W < \varepsilon$ for all $v$ if $n$ large enough, hence for $n$ large enough, $\|T-T_n\| < \varepsilon$

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I'm quite sure it's correct but I need to have you check it anyway since I might be missing something. – Rudy the Reindeer Jul 14 '12 at 17:23
What do you mean w.r.t? – user29999 Jul 14 '12 at 18:04
A converse: math.stackexchange.com/q/1674 – Jonas Meyer Jul 14 '12 at 19:21
@user29999 w.r.t. = with respect to – Rudy the Reindeer Jul 14 '12 at 20:21
@JonasMeyer That's great, thank you very much for the link! – Rudy the Reindeer Jul 14 '12 at 20:40

3 Answers

In your $(ii)$ the first term should be $$||T||:=\sup_{||v||\leq 1}||Tv||$$ instead of $||Tv||$.
And your $(iii)$ should be $||(T_m-T_n)v||\leq \varepsilon ||v||$ for all $v\in V$ and for $m,n\geq \nu_{\varepsilon},$ therefore $$||(T-T_n)v||\leq \varepsilon||v||, \forall v\in V,\forall n\geq\nu_{\varepsilon}$$

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Right! (ii) was actually just a typo : ) (not a proper mistake). Thank you! – Rudy the Reindeer Jul 15 '12 at 9:00
And from $$||(T-T_n)v||\leq \varepsilon||v||, \forall v\in V,\forall n\geq\nu_{\varepsilon}$$ we get $$||(T-T_n)|| = \sup_{\|v\| \leq 1}||(T-T_n)v|| \leq \varepsilon$$ – Rudy the Reindeer Jul 15 '12 at 9:05
@MattN. It was my pleasure to be useful to you. Bye. – Giuseppe Tortorella Jul 15 '12 at 9:20

I'm not convinced from your last point (the third). You can prove the convergence simply basing on the fact that $T_n$ is a Cauchy sequence in the operator norm, helping you with the pointwise convergence result that you just obtained. You can in fact fix $m$ (in the Cauchy sequence definition) and let $n$ tend to infinity, using the continuity of the norm and the linearity of the limit. The rest is ok, but you can note that a nice proof of the boundedness of the limit can be obtained as a Banach-Steinhaus theorem consequence. But I just noted that you supposed only the completeness of the codomain, so Banach-Steinhaus is not applicable.

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Correct in (iii)

(iii) $\|(T -T_n)v\|_{W} < \varepsilon \forall v; \|v\|_V \le 1$ for $n>>1$.

in (ii) Giuseppe said above.

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