# Theorem 9.13 in Rudin's Real and Complex Analysis

9.13 Theorem One can associate to each $$f \in L^{2}$$ a function $$\hat{f} \in L^{2}$$ so that the following properties hold:

(a) If $$f \in L^{1} \cap L^{2},$$ then $$\hat{f}$$ is the previously defined Fourier transform of $$f .$$

(b) For every $$f \in L^{2},\|\hat{f}\|_{2}=\|f\|_{2}$$.

(c) The mapping $$f \rightarrow \hat{f}$$ is a Hilbert space isomorphism of $$L^{2}$$ onto $$L^{2}$$.

(d) The following symmetric relation exists between $$f$$ and $$\hat{f}:$$ If $$\varphi_{A}(t)=\int_{-A}^{A} f(x) e^{-i x t} d m(x) \quad \text { and } \quad \psi_{A}(x)=\int_{-A}^{A} \hat{f}(t) e^{i x t} d m(t),$$ then $$\left\|\varphi_{A}-\hat{f}\right\|_{2} \rightarrow 0 \text { and }\left\|\psi_{A}-f\right\|_{2} \rightarrow 0 \text { as } A \rightarrow \infty$$.

To prove $$(d),$$ let $$k_{A}$$ be the characteristic function of $$[-A, A] .$$ Then $$k_{A} f \in L^{1} \cap L^{2}$$ if $$f \in L^{2},$$ and $$\varphi_{A}=\left(k_{A} f\right)^{\wedge}$$ since $$\left\|f-k_{A} f\right\|_{2} \rightarrow 0$$ as $$A \rightarrow \infty,$$ it follows from $$(b)$$ that $$\left\|\hat{f}-\varphi_{A}\right\|_{2}=\|\left(f-k_{A} f\right) \hat{\|}_{2} \rightarrow 0$$ as $$A \rightarrow \infty$$.

The other half of $$(d)$$ is proved the same way.$$\qquad\qquad\qquad\qquad ////$$

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This is Theorem 9.13 in Rudin's Real and Complex Analysis, and a part of its proof. I have understood the first half of the proof of (d), but I can't see how the other half of (d) follows. Am I missing something? Thanks in advance.

Let $$\check{f}:=\mathscr{F}^{-1}(f)$$ denote the inverse Fourier transform. By (c) it follows that $$||f||_2=||\check{f}||_2$$ for all $$f\in L^2$$. Observe that $$k_A \hat{f}\in L^1\cap L^2$$ and $$\psi_A=(k_A \hat{f})^\vee$$ for all $$f\in L^2$$. It follows that $$||f-\psi_A||_2=||\hat{f}^\vee-(k_A \hat{f})^\vee||_2=||\hat{f}-(k_A \hat{f})||_2\rightarrow 0$$ as $$A\rightarrow \infty$$.
Before someone jumps at me for forgetting the nuisance factor $$2\pi$$, please observe that the measure $$dm$$ in Rudin's book is a scaling of the Lebesgue measure so that the factor is not needed in the argument above.
• How do you know $\psi_A = (k_A \hat{f})^\vee$? In other words, it is equivalent to show $\hat{\psi}_A = k_A \hat{f}$. Since we don't know $\psi_A \in L^1$, we might not even be able to use $\hat{\psi}_A = \int_{-\infty}^{\infty} \psi_A(x)e^{-ixt}\,dm(x)$. Jun 23, 2020 at 12:28
• @withgrace1040: I use that $k_A \hat{f}\in L^1$. This implies that the inverse Fourier transform hereof, that is $x\mapsto\big(k_A \hat{f}\big)^\vee(x)$, is given by the integral $\int_{-\infty}^\infty k_A(t) \hat{f}(t) e^{itx} dt$. Jun 23, 2020 at 14:16
• I think if you use the inverse Fourier transform, then you cannot use $f = \hat{f}^\vee$ in the first equality, which is the one this theorem aims to show eventually. Jun 23, 2020 at 14:50
• @withgrace1040: Observe that I am not actually using the identity $f(x)=\int_{-\infty}^\infty \hat{f}(t) e^{itx} dm(t)$, which I agree with you is not correct (at least not in general for $f\in L^2$). Instead, I am using (c) from Rudin's Theorem to justify $f=\hat{f}^\vee$ Jun 23, 2020 at 22:07
• @withgrace1040: At the outset, the inverse Fourier transform is defined as the operator $\mathcal{F}^{-1}: L^1\rightarrow L^\infty$ with $\mathcal{F}^{-1}(f)(t):=f^\vee(t):=\int f(x) e^{itx}dx$. Now (c) tells me that this operator extends to an isomorphism $\mathcal{F}^{-1}:L^2\rightarrow L^2$. Since I only know $f\in L^2$ (and not necessarily in $L^1$), I can only apply the extension of $\mathcal{F}^{-1}$ to $f$. In contrast, I know that $k_A\hat{f}\in L^1$, which means I can use the original integral representation of $\mathcal{F}^{-1}$ to $k_A\hat{f}$. Jun 24, 2020 at 12:01