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    The convolution of two $L^2(\mathbb R)$ functions is continuous

    P-8XT20

    Area
    Real Analysis
    Topics
    Convolution
    Status
    draft
    1. Define

    \[ f \ast g ( x ) : = \int _ { - \infty } ^ { \infty } f ( y ) g ( x - y ) \ : d y . \]

    Prove that if \(f , g \in L ^ { 2 } ( \mathbb { R } )\) , then \(f * g\) is a continuous function on \(\mathbb { R }\)

    Solution

    1. \(f\ast g\) is defined everywhere and bounded by \(\norm f_2\norm g_2\).

    By the Cauchy–Schwarz inequality, \(\abs{\int f(y)g(x-y)\,dy}\le\norm f_2\norm g_2\).

    2. Q.E.D.

    By the Cauchy–Schwarz inequality, \(\abs{(f\ast g)(x+h)-(f\ast g)(x)}\le\norm f_2\norm{g(\cdot+h)-g}_2\), which tends to \(0\) as \(h\to0\) by continuity of translation in \(L^2\) [@Fol13].