Proving Ito Isometry using Functional Analysis I would like to know whether it is possible to give a proof of Ito Isometry using a tool which I like to call "the functional analysis"-way. 
Let me explain the settings first. What we did is the following. Let $(\Omega,\mathcal F,(\mathcal F_t)_{t\geq 0},\mathbb P)$ be a filtered probability space. We defined a space 
\begin{align}
\mathcal M:=\{ M=(M)_{t\geq 0}\ | \ M \text{ is } \mathcal F_t\text{-martingale with continuous paths},\ M_0=0, M \text{ is bounded in } L^2(\mathbb P)\} 
\end{align}
We said for all $M\in\mathcal M$ we have the existence of $M_\infty^2$ and $\langle M\rangle_\infty$, where $\langle \cdot\rangle $ is the quadratic variation process. Now we may define an innerproduct on $\mathcal M$ as follows:
\begin{align}
(M,N):=\mathbb E[\langle M,N\rangle_\infty]
\end{align}
The space $\mathcal M$ with the given innnerproduct is a Hilbert space. 
We say $H\in L^2(M)$ iff $H$ is progressively measurable and
\begin{align}
\|H\|_{L^2(M)}^2:=\mathbb E\left[ \int^\infty_0 H_s^2\,d\langle M\rangle_s\right]<\infty
\end{align}
Let us fix $H\in L^2(M)$ and define a functional $f_H$ on $\mathcal M$ as follows
\begin{align}
f_H(N)=\mathbb E\left[\int^\infty_0 H_s\,d\langle M,N\rangle_s \right]
\end{align}
Using Kunita Watanabe we can get
\begin{align}
|f_H(N)|\leq \|H\|_{L^2(M)}\|N\|_{\mathcal M}
\end{align}
We can use Riesz to get an element which we will call $H\bullet M\in\mathcal M$ the integral of $H$ with respect to  $M$, and
\begin{align}
f_H(N)=(H\bullet M, N)
\end{align}
but more importantly is that we also get for free that the operator norm of $f_H$ satisfies $\|f_H\|=\|H\bullet M\|_{\mathcal M}$. 

Problem. Prove that $\|H\|_{L^2(M)}=\|H\bullet M\|_{\mathcal M}$.

I can prove this using properties of $H\bullet M$ and Lebesgue-Stieltjes integral. But I want to prove it using $\|f_H\|=\|H\|_{L^2(M)}$.

Question. How to prove that $\|f_H\|=\|H\|_{L^2(M)}$ not knowing about the integral $H\bullet M$ and everything related to that? I mean only by looking at the functional $f$ itself.

I already have the inequality $\|f_H\|\leq \|H\|_{L^2(M)}$. So we either need to find $N$ for which $|f_H(N)|=\|H\|_{L^2(M)}\|N\|_{\mathcal M}$ or a sequence of $X^{(n)}$ for which $\|X^{(n)}\|_{\mathcal M}=1$ and 
\begin{align}
|f_H(X^{(n)})|\to \|H\|_{L^2(M)}
\end{align}
I want to do this because I believe that it may increase my knowledge about elements of $\mathcal M$ and dealing with such operators on that space. 
The answer that I'm seeking is from the following perspective. Consider yourself as someone who never saw ito integral and just knows the space 
$\mathcal M$ and $L^2(M)$ and the functional $f_H$. How would you then find the norm of $f_H$? 
Edit. I would be also satisfied if someone provides a solution for a possible simple case where $M$ is a (nicely, but not too trivial) stopped Brownian motion, for example $B_{t\wedge 1}$ or so. 
 A: This answer may or may not fit with what you are looking for. In the spirit of not knowing of the Ito integral, I use only the definition of $H \bullet M$ via $f_H$ given in the question and some martingale techniques (which are certainly more elementary than Kunita-Watanabe). However, my approach will be to prove the property of $H \bullet M$ you would want to use to show the Ito isometry if you were allowed to construct the Ito integral via simple functions and so obviously benefits from the hindsight of knowing that construction. 
To make it clear when I am using simple functions, I will write $H \bullet M$ for the element given by the Riesz Representation Theorem applied to $f_H$ and $\int H dM$ for the Ito integral as constructed by approximation by simple functions. I will sometimes write $H \cdot \langle M,N \rangle$ for Stieltjes integration.
The key to proving the Ito isometry for $\int H dM$ is the realisation that (first for simple integrands and then by a continuity argument, for all integrands) $$\bigg \langle \int H dM, N \bigg \rangle = H \cdot \langle M,N \rangle.$$
It is possible to prove that $H \bullet M$ has the same property without ever knowing this other construction of the Ito integral. 
First, notice that $$\mathbb{E}[(H \bullet M)_\infty N_\infty] = (H \bullet M, N) = \mathbb{E}[(H \cdot \langle M,N \rangle)_\infty].$$ We will extend this relation to hold at stopping times. Fix a stopping time $\tau$. We have
\begin{align}
\mathbb{E}[(H \bullet M)_\tau N_\tau] &= \mathbb{E}[\mathbb{E}[(H \bullet M)_\infty \mid \mathcal{F}_\tau] N_\tau]
\\& = \mathbb{E}[(H \bullet M)_\infty N_\tau]
\\& = \mathbb{E}[(H \bullet M)_\infty N_\infty^\tau] \text{ where } N_t^\tau = N_{t \wedge \tau}
\\& = \mathbb{E}[(H \cdot \langle M, N^\tau \rangle)_\infty]
\\& = \mathbb{E}[(H \cdot \langle M, N \rangle)_\tau]
\end{align}
where the last line is a straightforward consequence of the definition of the Stieltjes integral. This already implies that $(H \bullet M)N - H \cdot \langle M,N\rangle$ is a martingale and so by uniqueness of the quadratic variation, $H \cdot \langle M,N\rangle = \langle H \bullet M, N \rangle$. Now we are done. Indeed, we have
$$|f_H(H \bullet M)| = \mathbb{E}[\langle H \bullet M, H \bullet M \rangle_\infty] = \mathbb{E}[(H^2 \cdot \langle M \rangle)_\infty = \|H\|_{L^2(M)}^2$$
and so $\|f_H\| \geq \|H\|_{L^2(M)}$.

Remark: The use of hindsight here isn't even particularly overwhelming. I write $I(H) = M \bullet H$ so we don't get led on by notation. It is natural to try plugging $I(H)$ back into $f_H$. From the definition of $f_H$, we know that 
$$\mathbb{E}\bigg[\int_0^\infty d \langle I(H),I(H) \rangle \bigg] = \mathbb{E} \bigg[ \int_0^\infty H_s d \langle I(H),M \rangle_s \bigg]$$
i.e. we can take one $H$ out of the quadratic variation and into the integral by replacing $I(H)$ with $M$. We'd like to prove that 
$$\mathbb{E} \bigg[ \int_0^\infty H_s^2 d \langle M \rangle_s \bigg] = \mathbb{E}\bigg[\int_0^\infty d \langle I(H),I(H) \rangle \bigg] $$
i.e. we'd like to pull the other $H$ out of the stochastic integral and replace it with $M$. Knowing about stieltjes integration, this means we even might hope that $\langle I(H), M \rangle = H \cdot \langle M \rangle$. The above is a proof of this last fact about $I(H)$. The key point is that at no point did we need to know what the Ito integral was and we especially didn't need to argue via approximation by simple integrands.
