Folland - Real Analysis, Plancherel theorem proof

Question

I have a problem in understanding the proof of Plancherel Theorem in Folland - Real Analysis.

Theorem. If $f \in L^1 \cap L^2$, then $\hat{f} \in L^2$; and $\mathcal{F}|(L^1 \cap L^2)$ extend uniquely to a unitary isomorphism on $L^2$.

The proof goes as follows: First define $ X := \{f \in L^1: \hat{f} \in L^1\} $.

It is easy to show that $X \subset L^2$ and $X$ is dense in $L^2$. At this point, Folland shows that $\mathcal{F}|X$ preserves the $L^2$ inner product. Moreover, it is bijective by the Inversion Theorem ( we can use it since $f \in L^1$ and $\hat{f} \in L^1$ ) .

$\mathcal{F}|X$ extends by continuity to a unitary isomorphism on $L^2$ (that I call $\mathcal{F}_{ext}) $. It remains to show that this extension agrees with $\mathcal{F}$ on all of $L^1 \cap L^2$.

To do so, take $f\in L^1 \cap L^2$ and consider $g(x) = e^{-\pi|x|^2}$. We have $f*g_t \in L^1$ by Young inequality and $\hat{(f*g_t)} \in L^1$ since $\hat{f}$ is bounded. ( I omitted the details ) .

Therefore $f*g_t \in X$ and we know, by properties of convolutions, that $f*g_t \to f$ in both the $L^1$ and $L^2$ norms.

Therefore, $\mathcal{F}_{ext}(f) = \lim_{t \to 0} \mathcal{F}(f*t_t)$. It remains to show that: $$ \lim_{t \to 0} \mathcal{F}(f*t_t) = \hat{f} \mbox{ in } L^2 $$ Folland claims that $\hat{f*g_t} \to \hat{f}$ both uniformly and in $L^2$ norm, so we are done. However, I don't understand how to prove the convergence in $L^2$. This is what I tried: $$ \|\hat{f*g_t} - \hat{f} \|_{L^2} = \lim_{t \to 0 } \int |\hat{f}(\xi)e^{-\pi t^2 |\xi|^2} - \hat{f}(\xi)|^2 d\xi = \lim_{t \to 0 } \int | \hat{f}(\xi)|^2 |e^{-\pi t^2 |\xi|^2}-1|^2 d\xi $$ Now the integrand converges pointwise to $0$ as $t \to 0$. I thought of using the Dominated Convergence Theorem to conclude, but I would need $\hat{f} \in L^2$ to use it (right ? ). If $f$ was in $X$, we would obviously have $\hat{f} \in L^2$ since, by the above, $\mathcal{F}|X$ preserves the $L^2$ norm and $X \subset L^2$. However, here we are dealing with $f \in L^1 \cap L^2 $. How can I conclude? Thanks for your help.

Answer 1

I am writing $f_n$ for $f*g_{1/n}$. The sequence $\{\hat f_n\}$ is Cauchy in $L^2$, since $\{f_n\}$ is Cauchy in $L^2$ (it converges to $f$) and $\|\hat f_n-\hat f_m\|_2=\|f_n-f_m\|_2$. Since $L^2$ is complete, there is $g\in L^2$ so that $\hat f_n\to g$ in $L^2$. Use that $\hat f_n\to\hat f$ uniformly to conclude that $g=\hat f$ a.e.

Answer 2

Let $\widetilde{f}(x)=\overline{f(-x)}$ and $\phi=f\ast\widetilde{f}$. Then $\phi\in L^1$, $\phi$ is continuous and $\widehat{\phi}=|\widehat{f}|^2\ge0$. Since $\phi\in L^1$, $$ \int\phi(x)\,g_t(x)\,dx=\int\widehat{\phi}(\xi)\,\widehat{g_t}(\xi)\,d\xi=\int\widehat{\phi}(\xi)\,g(t\,\xi)\,d\xi. $$ Since $\phi$ is continuous, the leftmost integral converges to $\phi(0)$ as $t\to0$. Then by Fatou's lemma $$ \phi(0)=\lim_{t\to0}\int\widehat{\phi}(\xi)\,g(t\,\xi)\,d\xi\ge\int\lim_{t\to0}\widehat{\phi}(\xi)\,g(t\,\xi)\,d\xi=\int\widehat{\phi}(\xi)\,d\xi. $$