Stein 2.21

$\textbf{Exercise 2.21}$ Let $f,g\in{L^1(\mathbb{R}^d)}$
(a) Prove that $f(x-y)g(y)$ is measurable in $\mathbb{R}^{2d}$

$\textit{Proof}$ : By proposition 2.3.7, with $g$ being measurable on $\mathbb{R}^d$, then the "sheet function" $\tilde{g}(x,y)=g(x)$ is measurable on $\mathbb{R}^d \times \mathbb{R}^d \cong \mathbb{R}^{2d}$ . By proposition 2.3.9, with $f$ being measurable on $\mathbb{R}^d$, then the function $\tilde{f}(x,y)=f(x-y)$ is measurable on $\mathbb{R}^d \times \mathbb{R}^d \cong \mathbb{R}^{2d}$. With $f(x-y)g(y)$ being a product of 2 measurable functions and the previous result that the product of 2 measurable functions is measurable, we get that $f(x-y)g(y)$ is measurable on $\mathbb{R}^{2d}$. $\blacksquare$
Caution: In the context of Lebesgue measure, the product of 2 measurable functions is measurable. However, it is not true in general that the product of two integrable functions is integrable. Take $f(x)=\frac{1}{\sqrt{x}}\chi_{[0,1]}$. $\int |f| = \int_0^1 \frac{dx}{\sqrt{x}}=2$ however, $\int |f \cdot f|=\int_0^1\frac{dx}{x}=+ \infty$
To get any relationship between $L^p(E)$ and $L^q(E)$ for $p,q \in{[1, \infty]}$, we need $m(E)< \infty$. This follows from Hölder's inequality and is discussed on pages 142-3 in Royden.

(b) Show that if $f,g\in{L^1(\mathbb{R}^d)}$ then $f(x-y)g(y)\in{L^1(\mathbb{R}^{2d})}$
(c) Recall the definition of the convolution of $f$ and $g$ given by $$(f \ast g)(x)=\int_{\mathbb{R}^d}f(x-y)g(y)dy $$ Show that $f \ast g$ is well-defined for a.e. $x$ (that is, $f(x-y)g(y)\in{L^1(\mathbb{R}^{d})}$ for a.e. $x$)
(d) Show that $f \ast g$ is integrable whenever $f$ and $g$ are integrable and that $$||f \ast g||_{L^1} \leq ||f||_{L^1} \cdot ||g||_{L^1}$$ with equality when $f,g \geq 0$

$\textit{Proof}$ We combine the work of items (b)-(d) together.
$(f \ast g)(x)=\int_{\mathbb{R}^d}f(x-y)g(y)dy$. We want to show that $\int_{\mathbb{R}^d}|(f \ast g )(x)|dx < \infty $ i.e. $||f \ast g||_{L^1} < \infty $.
By the triangle inequality for Lebesgue integrals, $$\int_{\mathbb{R}^d}|(f \ast g )(x)|dx=\int_{\mathbb{R}^d} \left| \int_{\mathbb{R}^d} f(x-y)g(y)dy \right| dx \leq \int_{\mathbb{R}^d} \left( \int_{\mathbb{R}^d} |f(x-y)g(y)|dy \right) dx \quad (\star)$$ By the symmetry of $x$ and $y$ in Tonelli's theorem, $$=\int_{\mathbb{R}^d} \left( \int_{\mathbb{R}^d} |f(x-y)| |g(y)|dy \right) dx \overset{T}{=} \int_{\mathbb{R}^d} \left( \int_{\mathbb{R}^d} |f(x-y)| |g(y)|dx \right) dy$$ By linearity, $$=\int_{\mathbb{R}^d} |g(y)| \left( \int_{\mathbb{R}^d} |f(x-y)| dx \right) dy= \left( \int_{\mathbb{R}^d} |g(y)| dy \right) \left( \int_{\mathbb{R}^d} |f(x-y)| dx \right) $$ Since the $L^1$ norm is translation invariant, $$=||g||_{L^1} \cdot ||f(x-y)||_{L^1}=||g||_{L^1} \cdot ||f||_{L^1}$$ We would get equality in $(\star)$ above if $f,g \geq 0$. $\blacksquare$

(e) The Fourier transform of an integrable function $f$ is defined by $$\hat{f}(\xi)=\int_{\mathbb{R}^d} f(x)e^{-2\pi ix \xi}dx$$ Check that $\hat{f}$ is bounded and is a continuous function of $\xi$. Then, prove that $\forall \xi \in{\mathbb{C}}$ we get the identity: $$\widehat{(f \ast g)}(\xi)=\hat{f}(\xi)\hat{g}(\xi)$$ $\textit{Proof:}$ To check that $\hat{f}$ is bounded, we first observe that since $e^{-2\pi i x \xi}$ takes on points on the complex unit circle, so $|e^{-2\pi i x \xi}| \leq 1$ where $| \cdot |$ is the modulus (length in complex plane). Then, by the above observation and the triangle inequality, $$|\hat{f}(\xi)|=\left|\int_{\mathbb{R}^d} f(x)e^{-2\pi ix \xi}dx\right| \leq \int_{\mathbb{R}^d} |f(x)| \cdot |e^{-2\pi ix \xi}|dx \leq \int_{\mathbb{R}^d} |f(x)|dx = ||f||_{L^1} < \infty$$ So, $\hat{f}$ is bounded. To verify that $\hat{f}$ is a continuous function of $\xi$, we want to show that if $\xi_n \longrightarrow \xi$ then $\hat{f}(\xi_n) \longrightarrow \hat{f}(\xi) $ i.e. $\int_{\mathbb{R}^d} f(x)e^{-2\pi ix \xi_n}dx \overset{\xi_n \rightarrow \xi}{\longrightarrow} \int_{\mathbb{R}^d} f(x)e^{-2\pi ix \xi}dx$ Label the integrands $\tilde{f}(x)=f(x)e^{-2 \pi ix \xi}$ and $\tilde{f}_n(x)=f(x)e^{-2 \pi ix \xi_n}$. Then, since $f\in{L^1(\mathbb{R}^d}) \Longrightarrow f\in{\mathcal{M}(\mathbb{R}^d})$ and so $\{\tilde{f}_n\}_{n=1}^\infty$ is a sequence of measurable functions. Also, $|\tilde{f}_n| \leq |f| \in{L^1(\mathbb{R}^d)}$. And, $\tilde{f}_n \longrightarrow \tilde{f}$ a.e. since $\lim_{n \rightarrow \infty } f(x)e^{-2 \pi ix \xi_n}=f(x) \lim_{n \rightarrow \infty} e^{-2 \pi i x \xi_n} = f(x)e^{-2 \pi i x \xi}$ by the continuity of the exponential function. By the Lebesgue dominated convergence theorem, $$\int_{\mathbb{R}^d} \tilde{f}_n(x)dx \longrightarrow \int_{\mathbb{R}^d} \tilde{f}_n(x)dx \quad i.e. \quad \hat{f}(\xi_n) \longrightarrow \hat{f}(\xi)$$

By definition, the Fourier transform of the convolution of $f$ and $g$ is $$(\widehat{f \ast g})(\xi)=\int_{\mathbb{R}^d}(f \ast g)(x)e^{-2\pi i x \xi} dx=\int_{\mathbb{R}^d}\left(\int_{\mathbb{R}^d}f(x-y)g(y)dy \right)e^{-2\pi i x \xi} dx$$ Using the symmetry of $x$ and $y$ in Fubini's theorem and the linearity of the integral, $$=\int_{\mathbb{R}^d}\left(\int_{\mathbb{R}^d}f(x-y)g(y)dx \right)e^{-2\pi i x \xi} dy=\int_{\mathbb{R}^d}\left(\int_{\mathbb{R}^d}f(x-y)e^{-2\pi i x \xi}dx \right)g(y) dy$$ If we write $e^{-2 \pi i x \xi} = e^{-2 \pi i (x-y+y) \xi}=e^{-2 \pi i (x-y) \xi} e^{-2 \pi i y \xi}$ then continuing our equality chain, $$=\int_{\mathbb{R}^d}\left(\int_{\mathbb{R}^d}f(x-y)e^{-2 \pi i (x-y) \xi} e^{-2 \pi i y \xi}dx \right)g(y) dy=\left(\int_{\mathbb{R}^d}f(x-y)e^{-2 \pi i (x-y) \xi} dx \right) \left( \int_{\mathbb{R}^d} g(y)e^{-2 \pi i y \xi} dy \right)$$ By the translation invariance of the Lebesgue measure, $$=\left(\int_{\mathbb{R}^d}f(x)e^{-2 \pi i x \xi} dx \right) \left( \int_{\mathbb{R}^d} g(y)e^{-2 \pi i y \xi} dy \right)=\hat{f}(\xi)\hat{g}(\xi)$$