- Let
\[ f \ast g ( x ) : = \int _ { - \infty } ^ { + \infty } f ( y ) g ( x - y ) d y \]
denote the convolution of f and \(g .\)
Let \(f , g \in L ^ { 2 } ( \mathbb { R } )\) be two square-integrable functions on R (with the usual Lebesgue measure). Show that the convolution \(f * g\) bounded continuous function on R.
Instead let \(h \in L ^ { 1 } ( \mathbb { R } )\) be fixed. Show that \(A ( f ) = f * h\) is a bounded operator \(L ^ { 1 } ( \mathbb { R } ) \to L ^ { 1 } ( \mathbb { R } )\)
Solution
Write \(\tau_hg(y)\da g(y-h)\) and \(\norm\cdot_p\) for the \(L^p(\RR)\) norm.
1. (a) For \(f,g\in L^2\), \(\abs{(f\ast g)(x)}\le\norm f_2\norm g_2\) for every \(x\).
By the Cauchy–Schwarz inequality, \(\abs{\int f(y)g(x-y)\,dy}\le\norm f_2\norm{g(x-\cdot)}_2=\norm f_2\norm g_2\).
2. (a) \(f\ast g\) is continuous.
By the Cauchy–Schwarz inequality, \(\abs{(f\ast g)(x+h)-(f\ast g)(x)}\le\norm f_2\norm{\tau_{-h}g-g}_2\), which tends to \(0\) as \(h\to0\) by continuity of translation in \(L^2\) [@Fol13].
3. (b) For \(f,h\in L^1\), \(\norm{f\ast h}_1\le\norm f_1\norm h_1\), so \(A\) is a bounded operator on \(L^1\) with \(\norm A\le\norm h_1\).
By Tonelli’s theorem, \(\int\abs{(f\ast h)(x)}\,dx\le\iint\abs{f(y)}\abs{h(x-y)}\,dy\,dx=\norm f_1\norm h_1\). The map \(A\) is linear, so this bound makes it bounded.
4. Q.E.D.
Steps 1 and 2 prove part (a), and step 3 proves part (b).