Extending an equality from a dense subspace - possible mistake in a proof. I am reading K.Gröchenig's Introduction to Time-Frequency Analysis , Theorem 3.2.1. and the proof seems to be incorrect as it is presented.
The whole context is actually not important, but I will report it anyway. The map
$$V_gf(x,\xi):=\int_{\mathbb{R}^d}f(t)\overline{g(t-x)}e^{-2\pi it\cdot \xi}dt,\qquad f,g\in L^2(\mathbb{R}^d) $$
is called the Short Time Fourier Transform (STFT) of window $g$. The theorem considered reads

Let $f_1,f_2,g_1,g_2\in L^2(\mathbb{R}^d)$. Then $V_{g_j}f_j\in L^2(\mathbb{R}^{2d})$ for $j=1,2$ and
  $$\left\langle V_{g_1}f_1,V_{g_2}f_2\right\rangle=\left\langle
 f_1,f_2\right\rangle \overline{\left\langle
 g_1,g_2\right\rangle}$$

The proof Gröchenig proposes is as follows. The equality above is first proved for $g_1,g_2\in (L^1\cap L^{\infty})(\mathbb{R}^d)$, which is dense in $L^2(\mathbb{R}^d)$ (this allows one to use Parseval's equality and Fubini's theorem). However, to extend the equality to $g_1,g_2\in L^2(\mathbb{R}^d)$, he proceeds as follows

With $g_1\in L^1\cap L^{\infty}$ fixed, the mapping $g_2\mapsto \left\langle V_{g_1}f_1,V_{g_2}f_2\right\rangle$ is a linear functional that coincides with $\left\langle f_1,f_2\right\rangle \overline{\left\langle g_1,g_2\right\rangle}$ on the dense subspace $L^1\cap L^{\infty}$. It is therefore bounded and extends to all $g_2\in L^2(\mathbb{R}^d)$.

Up to this point everything seems fine. But now:

In the same way, for arbitrary $f_1,f_2$ and $g_2\in L^2(\mathbb{R}^d)$, the conjugate linear functional $g_1\mapsto \left\langle V_{g_1}f_1,V_{g_2}f_2\right\rangle $ equals $\left\langle f_1,f_2\right\rangle \overline{\left\langle g_1,g_2\right\rangle}$ on $L^1\cap L^{\infty}$ and extends to all of $L^2$. The orthogonality relations are therefore established for all $f_j,g_j\in L^2(\mathbb{R}^d)$.

I believe that this proof is not correct. Reason: a priori the linear operator $g_2\mapsto \left\langle V_{g_1}f_1,V_{g_2}f_2\right\rangle$ is not bounded on $L^2(\mathbb{R}^d)$, as we do not even know if $V_{g_i}(f_i)\in L^2(\mathbb{R}^{2d})$ yet (it is part of the thesis), so the inner product might not even make sense. Therefore, even if we have extended the linear operator  $g_2\mapsto \left\langle V_{g_1}f_1,V_{g_2}f_2\right\rangle$ , which is bounded on $(L^1\cap L^{\infty})(\mathbb{R}^d)$, on the whole $L^2(\mathbb{R}^d)$, the resulting operator will not necessarily be still given by the expression $g_2\mapsto \left\langle V_{g_1}f_1,V_{g_2}f_2\right\rangle$. A priori, the operator $g_2\mapsto \left\langle V_{g_1}f_1,V_{g_2}f_2\right\rangle$  is unbounded on $L^2(\mathbb{R}^d)$ (despite being bounded on $(L^1\cap L^{\infty})(\mathbb{R}^d)$) and the extension we have just found is different - they only agree on $(L^1\cap L^{\infty})(\mathbb{R}^d)$. Hence we cannot deduce that the equality we are looking to prove is preserved by the extension.
What do you think, MSE?
 A: You are completely right that the given proof is not completely correct.
However, it can be salvaged by the following additional argument: In the following, I will use the notations $L_x f (y) = f(y-x)$ and $M_\xi f (x) = e^{2\pi i \langle x, \xi \rangle} f(x)$ for the translation and modulation of the function $f$, respectively. With this notation, you have
$$
V_g f (x,\xi) = \langle f, M_\xi L_x g \rangle \, ,
$$
which is (for fixed $x,\xi$) jointly continuous as a function of $f,g \in L^2 (\mathbb{R})$. This uses that $M_\xi : L^2 \to L^2$ and $L_x : L^2 \to L^2$ are continuous (in fact isometric) linear maps.
Thus, even though we do not know yet that $V_g f \in L^2$, and in particular not that $(f,g) \mapsto V_g f$ is continuous as a map of $L^2 \times L^2$ into $L^2$, we know that the pointwise evaluations of the short time Fourier transform are continuous as functions of $f,g \in L^2$.
This is useful as follows: On a dense subset of $L^2$, you know that $\langle V_{g_1} f_1 , V_{g_2} f_2 \rangle = \langle f_1, f_2 \rangle \langle g_2, g_1 \rangle$. But this implies in particular that $\|V_g f \|_{L^2} = \|f\|_{L^2} \|g\|_{L^2}$ for $f,g$ in this dense subspace.
Now, let $f,g \in L^2$ be arbitrary, and choose sequences $f_n, g_n$ from the dense subspace that converge to $f,g$, respectively. As we saw above, this implies $V_{g_n} f_n \to V_g f$ pointwise. Using Fatou's Lemma, this easily implies
$$
\|V_g f \|_{L^2} \leq \liminf_{n \to \infty} \|V_{g_n} f_n\|_{L^2} = \liminf_n \|f_n\|_{L^2} \|g_n\|_{L^2} = \|f\|_{L^2} \|g\|_{L^2} \, .
\tag{$\ast$}
$$
In particular, $V_g f \in L^2 (\Bbb{R}^{2n})$. Note that this holds for arbitrary $f,g \in L^2$.
Finally, it is not hard to see that the estimate $(\ast)$ implies that $L^2 \times L^2 \to L^2, (f,g) \mapsto V_g f$ is a continuous sesquilinear map. Indeed, if $f_n \to f$ and $g_n \to g$, then
\begin{align*}
\| V_{g_n} f_n - V_g f\|_{L^2}
& \leq \|V_{g_n} f_n - V_{g_n} f\|_{L^2} + \|V_{g_n} f - V_{g} f\|_{L^2} \\
& \leq \|g_n\|_{L^2} \cdot \|f_n - f\|_{L^2} + \|f\|_{L^2} \cdot \|g_n - g\|_{L^2}
\to 0 \, .
\end{align*}
Finally, since you now know that the map $(f,g) \mapsto V_g f$ is continuous as a map of $L^2 \times L^2$ into $L^2$, and since you know that the identity to be proved holds on a dense subset (where the right-hand side is also continuous), the desired identity holds for all $f_1,f_2, g_1, g_2 \in L^2$.
