Prove that if \(f, g: \RR^n\to \CC\) is both measurable then \(F(x, y) \definedas f(x)\) and \(h(x, y)\definedas f(x-y) g(y)\) is measurable on \(\RR^n\cross \RR^n\).
Show that if \(f\in L^1(\RR^n) \intersect L^\infty(\RR^n)\) and \(g\in L^1(\RR^n)\), then \(f\ast g \in L^1(\RR^n) \intersect L^\infty(\RR^n)\) is well defined, and carefully show that it satisfies the following properties: \[ \norm{f\ast g}_\infty &\leq \norm{g}_1 \norm{f}_\infty \norm{f\ast g}_1 &\leq \norm{g}_1 \norm{f}_1 \norm{f\ast g}_2 &\leq \norm{g}_1 \norm{f}_2 .\]
Hint: first show \(\abs{f\ast g}^2 \leq \norm{g}_1 \qty{ \abs{f}^2 \ast \abs{g}}\).
Solution
Measurability means Lebesgue measurability; \(m\) denotes Lebesgue measure.
1. \(F(x,y) = f(x)\) and \(G(x,y) = g(y)\) are measurable on \(\RR^n \times \RR^n\).
For open \(W \subseteq \CC\), \(F^{-1}(W) = f^{-1}(W) \times \RR^n\). Write \(f^{-1}(W) = V \setminus Z\) with \(V\) a \(G_\delta\) set and \(Z\) null. Then \(V \times \RR^n\) is a \(G_\delta\) set and \(Z \times \RR^n\) is null, so \(F^{-1}(W)\) is measurable. The same argument applies to \(G\). See E-JJ746.
2. \(h(x,y) = f(x - y)g(y)\) is measurable.
Let \(T(x,y) = (x - y, y)\), an invertible linear map of \(\RR^{2n}\) with determinant \(1\). Then \(h = (F \circ T)\cdot G\). For open \(W\), \((F\circ T)^{-1}(W) = T^{-1}(F^{-1}(W))\). \(T^{-1}\) is a homeomorphism, so it maps \(G_\delta\) sets to \(G_\delta\) sets, and \(m(T^{-1}(A)) = m(A)\) for measurable \(A\), so it maps null sets to null sets. Writing \(F^{-1}(W)\) as a \(G_\delta\) set minus a null set, as in step 1, shows \((F\circ T)^{-1}(W)\) is measurable. A product of measurable functions is measurable.
3. For \(f \in L^1 \cap L^\infty\) and \(g \in L^1\), \(f\ast g(x) = \int f(x-y)g(y)\,dy\) is defined for every \(x\), and \(\|f \ast g\|_\infty \le \|g\|_1\|f\|_\infty\).
\(\int|f(x-y)||g(y)|\,dy \le \|f\|_\infty\|g\|_1 < \infty\) for every \(x\).
4. \(\|f \ast g\|_1 \le \|g\|_1\|f\|_1\).
By step 2 and Tonelli’s theorem, \(\int|f\ast g(x)|\,dx \le \iint |f(x-y)||g(y)|\,dy\,dx = \int|g(y)|\int|f(x-y)|\,dx\,dy = \|f\|_1\|g\|_1\), using translation invariance of the inner integral.
5. \(|f \ast g(x)|^2 \le \|g\|_1\,(|f|^2 \ast |g|)(x)\).
Write \(|f(x-y)g(y)| = |f(x-y)|\,|g(y)|^{1/2}\cdot|g(y)|^{1/2}\). The Cauchy–Schwarz inequality in \(y\) gives \(|f\ast g(x)|^2 \le \left(\int|f(x-y)|^2|g(y)|\,dy\right)\left(\int|g(y)|\,dy\right)\).
6. \(\|f \ast g\|_2 \le \|g\|_1\|f\|_2\).
\(|f|^2 \le \|f\|_\infty|f|\), so \(|f|^2 \in L^1\) with \(\||f|^2\|_1 = \|f\|_2^2\). Integrating step 5 and applying step 4 to \(|f|^2\) and \(|g|\) gives \(\|f \ast g\|_2^2 \le \|g\|_1\,\||f|^2 \ast |g|\|_1 \le \|g\|_1^2\|f\|_2^2\).
7. Q.E.D.
Steps 1 and 2 are part (a). Steps 3, 4 and 6 give \(f\ast g \in L^1 \cap L^\infty\) and the three inequalities of part (b).