# Unique continuous extension of $\tilde{L}: W \to Z$

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 $\|L\mid_W\| = \|\tilde{L}\|$.

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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
@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
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
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. Learning maths with you is so much fun. – Rudy the Reindeer Jul 15 '12 at 19:59