A question on the extension of of integrants from simple processes t0 $L^2$?

Question

I have a question. While defining the Stochastic integral w.r.t to the Brownian Motion we begin with simple processes which are adapted and left continuous and then extend it to the square integrable processes $L^2(R_+ \times \Omega,\mathcal{P_r},ds \times\mathbb{P})$ which are progressive . Let $\mathcal{E}$ refer to the set of all simple processes

Now clearly $\mathcal{E}\subset L^2(R_+ \times \Omega,\mathcal{P_r},ds \times\mathbb{P}) $ and the map $$I:\phi \mapsto \int_ 0^{\infty} \phi_s dB_s $$ is a linear isometry from E (endowed with the norm $\|.\|_{L^2(R_+ \times \Omega,\mathcal{P_r},ds \times\mathbb{P})})$ into $L^2(\Omega,\mathcal{F},\mathbb{P})$

Now the claim is that I can be uniquely extended to a linear isometry from the closure of $\mathcal{E}$ into $L^2(\Omega,\mathcal{F},\mathbb{P})$.

Is it because as $\bar{\mathcal{E}}$ is dense in $L^2(R_+ \times \Omega,\mathcal{P_r},ds \times\mathbb{P})$ and since $\phi$ is a bounded linear operator(I am not so sure about it as i have not studied any functional analysis) we can apply the BLt theorem and say that the operator is closable and uniquely extended to the new class of integrands?

Thank you

Answer 1

Lemma: Let $(X,\|\cdot\|_X)$ and $(Y,\|\cdot\|_Y)$ be Banach spaces and $D \subseteq X$ dense. If $f:D \to Y$ is an isometry, i.e. $$\|f(x)-f(y)\|_Y = \|x-y\|_X$$ for all $x,y \in D$, then $f$ admits a unique continuous extension which is again an isometry.

Proof: For any $x \in X$, we can choose $(x_n)_n \subseteq D$ such that $x_n \to x$. Then it follows from the fact that $f$ is an isometry on $D$ that $(f(x_n))_{n \in \mathbb{N}}$ is a Cauchy sequence. Since $Y$ is complete, there exists $$\tilde{f}(x) := \lim_{n \to \infty} f(x_n).$$

1. $\tilde{f}$ is well-defined (i.e. does not depend on the approximation sequence): Let $(x_n)_n \subseteq D$ and $(\tilde{x}_n)_n \subseteq D$ such that $x_n \to x$ and $\tilde{x}_n \to x$. Using again that $f$ is an isometry on $D$, we get $$\|f(x_n)-f(\tilde{x}_n)\|_Y = \|x_n-\tilde{x}_n\|_X \to 0,$$ hence, $$\lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} \tilde{f}(x_n).$$ In particular, $f(x) = \tilde{f}(x)$ for all $x \in D$.
2. $\tilde{f}$ is an isometry: Let $x,y \in X$ and $(x_n)_n \subseteq D$, $(y_n)_n \subseteq D$ such that $x_n \to x$, $y_n \to y$. Then $$\|\tilde{f}(x)-\tilde{f}(y)\|_Y = \lim_{n \to \infty} \|f(x_n)-f(y_n)\|_Y = \lim_{n \to \infty} \|x_n-y_n\|_X = \|x-y\|_X.$$

Now let $g$ be an arbitrary continuous extension of $f$. Then, as $f(x_n)=g(x_n)$ for any $x_n \in D$, it follows easily from the continuity of $g$ and $\tilde{f}$ that $\tilde{f}(x)=g(x)$ for any $x \in X$. This finishes the proof.

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Apply the above Lemma for $$\begin{align*} X &:= L^2([0,\infty) \times \Omega, \mathcal{P}, ds \otimes \mathbb{P}) \\ D &:= \mathcal{E} \\ Y &:= L^2(\Omega,\mathbb{P}), \\ f(\phi) &:= \int_0^{\infty} \phi_s \, dB_s. \end{align*}$$