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Can you read my proof and tell me if it's correct? Thank you!

Let $W$ be a dense subset of a normed vector space $V$ and let $\tilde{L}: W \to Z$ be a bounded linear operator into a Banach space. Then $\tilde{L}$ uniquely and continuously extends to $L : V \to Z$ and $\|L\| = \|\tilde{L}\|$.

Proof

Define $L : V \to Z$ as $$ \begin{cases} \tilde{L}(v) & v \in W \\ \lim_{n \to \infty} \tilde{L}(v_n) & v \in V - W, \lim_{n \to \infty} v_n = v \end{cases}$$

Note: Since $W$ is dense in $V$ we can find $v_n$ in $W$ s.t. $v_n \to v$ for each $v$ in $V$. Then

(i) $L$ is well-defined: let $v_n \to v$ and $v_n^\prime \to v$ be two different sequences with the same limit. Then $\lim_{n \to \infty} \tilde{L}(v_n) = \lim_{n \to \infty} \tilde{L}(v_n^\prime)$ since $\lim_{n \to \infty} \tilde{L}(v_n) - \lim_{n \to \infty} \tilde{L}(v_n^\prime) = \lim_{n \to \infty} \tilde{L}(v_n - v_n^\prime) = \lim_{n \to \infty} \tilde{L}(\frac{v_n - v_n^\prime}{\|v_n - v_n^\prime\|}) \|v_n - v_n^\prime\| \leq \lim_{n \to \infty} \|\tilde{L} \| \|v_n - v_n^\prime\|\leq \|\tilde{L} \| \varepsilon $ for $n$ large enough since $v_n - v_n^\prime \to 0$ by assumption.

(ii) $L$ is linear: $L(av_1 + bv_2) = \lim_{n \to \infty} \tilde{L} a v_{1n} + b v_{2n} = a \lim_{n \to \infty} \tilde{L} v_{1n} + b \lim_{n \to \infty} \tilde{L} v_{2n} = a \tilde{L} v_1 + b \tilde{L} v_2$.

(iii) $L$ is bounded: $\|Lv\|_Z = \|\lim_{n\to \infty} Lv_n\|_Z \leq \|\lim_{n\to \infty} Lv_n - Lv_N\|_Z + \|Lv_N \| \leq \varepsilon + \| \tilde{L} v_n\|_Z \leq \varepsilon + \|\tilde{L}\|$. Hence $\|L\| \leq \varepsilon + \|\tilde{L}\|$. Let $\varepsilon \to 0$.

To see that $\|L\| \geq \|\tilde{L}\|$, observe that $\|L\mid_W\| = \|\tilde{L}\|$.

Rudy the Reindeer
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    Hm... 1. How exactly is $L$ defined (what assumption do you impose on the sequence $v_n$?) 2. Why do all the limits you write down exist in the first place? 3. Why don't you apply this? – t.b. Jul 15 '12 at 12:14
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    @t.b. You seem to know my post history better than I do! : / I will have to think about 2. & 3. (after finishing something else first). As for 1.: I forgot to write the assumption. Probably because I thought it was "obvious" : ) – Rudy the Reindeer Jul 15 '12 at 12:26
  • @t.b. I'm back. As for 2.: $Z$ has to be a Banach space. As for 3.: Because then you would've told me that it's the lazy version and I should do it by hand. (I tried to find the thread where you tell me instead of using the uniform limit theorem I should do it by hand but couldn't find it anymore.) On a more serious note: For one I didn't think of it. Also, I'm working through the lecture notes and I'm practicing the stuff in there. Doing the almost same proof by hand again helps me get better. – Rudy the Reindeer Jul 15 '12 at 15:06
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    Okay, Frächdachs :) I think you have this in mind. Still, I miss the important bit: since $v_n$ is a Cauchy sequence in $W$ and since $\tilde{L}$ is Lipschitz (hence uniformly continuous), we have that $\tilde{L}v_n$ is Cauchy in $Z$, hence $\tilde{L}v_n$ is convergent. – t.b. Jul 15 '12 at 15:17
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    I forgot: in the proof of (i) there are some norms missing. Also, I'd use directly that $\lVert \tilde L(v_n-v_n') \rVert \leq \lVert \tilde L \rVert \lVert v_n-v_n' \rVert$ there (without dividing out the norm, which smells like dividing by zero). – t.b. Jul 15 '12 at 15:22
  • @t.b. : ) No, that's the thread where you tell me that I should redo the argument by hand without using Arzelà-Ascoli. But there is another one where I wrote a proof that ends "apply the uniform limit theorem to get that the limit is continuous" and you tell me to do the argument in the proof of the uniform limit theorem instead of (lazily)(<- you did not say that) applying the theorem. – Rudy the Reindeer Jul 15 '12 at 15:37
  • @t.b. Thank you. So you are saying that (i) is correct but I should prove it in a slightly nicer way? – Rudy the Reindeer Jul 15 '12 at 15:38
  • The shortest thing I proved using division by $|v|$ is $|Tv|W \leq |T| |v|_V$: $$ |Tv|_W = |T \frac{v}{|v|_V}|_W |v|_V \leq \sup{|v|_V \leq 1} |Tv|_W |v|_V = |T| |v|_V$$ – Rudy the Reindeer Jul 15 '12 at 16:37
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    @t.b. Learning maths with you is so much fun. – Rudy the Reindeer Jul 15 '12 at 19:59

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