How do you extend an isometry from a dense set?

Question

Question

Assume you have a $W \subset V$ such that $W$ is dense in $V$, where $W,V$ are metric spaces. You have a function $f: W \mapsto H$, where $H$ is another metric space. Also, $\lVert w\rVert_W = \lVert f(w)\rVert_H$ (ie $f$ is an isometry). My questions are:

• How do you "naturally extend" $f$ from $W$ to $V$ in an unique way, such that $f$ is still an isometry on $V$?
• What happens if $f$ is just a function, not necessarily an isometry?
• Do you need $W,V$ to be complete for this to work?

If in this general setting this is not possible, please refer below to the actual problem I have.

Thoughts

The obvious thing would be to find a sequence $(w_i)_{i\ge 0} \subset W$ that tends in the $V$-norm to $v \in V$ and define $$f(v) = \lim_{n \to \infty} f(w_n)$$

But I have already trouble proving that it's an isometry; in fact doing

$$\lVert f(v) \rVert _V =\lVert \lim_{n \to \infty}f(w_n) \rVert _V \mathop{=}^? \lim_{n \to \infty} \lVert f(w_n) \rVert _V = \lim_{n \to \infty} \lVert w_n \rVert _V = \lVert v \rVert _V$$

But how to justify bringing the limit outside?

To check that is well defined one would also need to check that if $w_n, v_n $ both tend in the $V$-norm to $v$, then

$$f(v) = \lim_{n \to \infty} f(w_n) = \lim_{n \to \infty} f(v_n)$$

It's easy to show that $\lVert v_n - w_n \rVert_V \to 0$, which implies $$\lVert f(w_n) - f(v_n) \rVert_V \to 0$$, but that does not imply pointwise convergence (as in the limit above). Can you provide some help?

Actual background

The actual problem is: I have defined the stochastic integral on the space $\Lambda_1$ of simple processes. I view $\Lambda_1$ as a subspace of $L^2(\Omega \times \mathbb R_+, \mathcal G \otimes \mathcal B(\mathbb R_+), dP \times ds)$.

Then I show that $\Lambda_1$ is dense in $\Lambda_2 = L^2(\Omega \times \mathbb R_+, \mathcal P, dP \times ds)$ where $\mathcal P$ is the sigma algebra generated by progressively measurable sets. The function $f$ in this context would be the function $$K \in \Lambda_1 \mapsto \int_0^\infty K_s dB_s \in L^2(\Omega, \mathcal G, dP)$$

which is an isometry, and I want to extend it as such from $\Lambda_1$ to $\Lambda_2$ ($B_s$ is a brownian motion)

Answer 1

1. If $\displaystyle \ell:= \lim_{n\to \infty}f(v_n)$, then reverse triangle inequality gives: $$|\Vert f(v_n)\Vert-\Vert \ell\Vert|\leq \Vert f(v_n)-\ell \Vert$$ so $\displaystyle \lim_{n\to \infty} \Vert f(v_n)\Vert = \Vert \ell\Vert =\Vert \lim_{n\to\infty} f(v_n)\Vert$.

2. How to prove that $(f(v_n))$ converges? If $H$ is complete, then note that $(v_n)$ converges thus it is a Cauchy sequence. Hence $(f(v_n))$ is a Cauchy sequence because $f$ is an isometry and it converges since $H$ is complete. I don't know what to do if $H$ is not complete.

3. Now, if $(v_n)$ and $(w_n)$ converge to a same $v$, then as you noted, $\Vert f(v_n)-f(w_n)\Vert$ tends to 0. Since $(f(v_n))$ converges to some $\ell$ and $(f(w_n))$ converges to some $\ell'$, continuity of the norm (see 1. above) shows that $$0= \lim_{n\to\infty} \Vert f(v_n)-f(w_n)\Vert =\Vert \ell-\ell'\Vert$$ thus $\ell=\ell'$.