Let $\epsilon>0$, $f\in L^1(R)\cap L^2(R)$ and define
$$g_\varepsilon(x) = \frac{1}{\varepsilon\sqrt{2\pi}}e^{-x^2/(2\varepsilon^2)}.$$
Show that
$$\lim_{\varepsilon\to 0}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x)\overline{f(x+t)}g_\varepsilon(t) dt dx = \Vert f\Vert^2_{L^2}.$$
My attempt:
\begin{align*} \lim_{\varepsilon\to 0}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x)\overline{f(x+t)}g_\varepsilon(t) dt dx &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x)\overline{f(x+t)}\lim_{\varepsilon\to 0}g_\varepsilon(t) dt dx \end{align*}
and
$$\lim_{\varepsilon\to 0} g_\varepsilon(t) = \lim_{\varepsilon\to 0}\frac{1}{\varepsilon\sqrt{2\pi}}e^{-x^2/(2\varepsilon^2)} = 0,$$
which means the integral is $0$. How does one resolve this?
The problem in your attempt is that you can not swap the limit and the integration in the third line.
I will present a sketch of the proof. We have that $$ \int_\mathbb{R} f(x+t) g_\varepsilon (t) dt = \int_\mathbb{R} f(t) g_\varepsilon (t-x) dt = \int_\mathbb{R} f(t) g_\varepsilon (x-t) dt =f * g_\varepsilon (x). $$ Where $f*g_\varepsilon$ is the convolution. We also have $f*g_\varepsilon \in L^2$. Then the identity we want to prove reduces (with the scalar product $\langle \cdot, \cdot \rangle$ anti-linear in the first component) to showing $$\lim_{\varepsilon \to 0 }\langle f * g_\varepsilon, f\rangle = \| f \|^2.$$ Now since the Fourier transform $\mathcal{F}$ is unitary we have $$\lim_{\varepsilon \to 0 }\langle f * g_\varepsilon, f\rangle =\lim_{\varepsilon \to 0 }\langle \mathcal{F} (f * g_\varepsilon),\mathcal{F} f\rangle .$$ From the convolution theorem (using the fact that $f \in L^1 \cap L^2)$ we have $$ \mathcal{F} (f * g_\varepsilon) = (2\pi)^{1/2} \mathcal{F} (f) \mathcal{F}(g_\varepsilon). $$ We can compute $\mathcal{F}(g_\varepsilon)$ and show that it converges pointwise from below to $1/(2\pi)^{1/2}$ as $\varepsilon \to 0$. Applying dominated convergence with $4 |\mathcal{F}(f)|^2 $ as the dominator shows that $\mathcal{F}(f * g_\varepsilon) \to \mathcal{F}(f)$ in $L^2$. Continuity of the map $\langle \cdot,\mathcal{F} f \rangle$ concludes the proof.