Suppose that \(f,g\in L^1(\mathbb R)\) have compact support, and define \[ h(x)=(f*g)(x)=\int_{\mathbb R} f(x-y)g(y)\,dy. \] Is \(h\) necessarily uniformly continuous? Prove it or give a counterexample.
Erratum: the source exam asks for a proof that \(h\) is uniformly continuous. The solution’s example shows that this fails for \(f,g\in L^1(\mathbb R)\) with compact support; it holds when one of \(f,g\) is in \(L^\infty\), or when \(f,g\in L^2\).
Solution
1. Choose compactly supported \(L^1\) functions with an integrable singularity.
Let \[ f(x)=g(x)=x^{-2/3}\mathbf 1_{(0,1)}(x). \] Then \(f,g\in L^1(\mathbb R)\) because \[ \int_0^1 x^{-2/3}\,dx=3<\infty, \] and both supports are contained in the compact interval \([0,1]\).
2. Compute the convolution near the origin.
For \(0<x<1\), the integrand is nonzero exactly when \(0<y<x\), so \[ (f*g)(x) =\int_0^x (x-y)^{-2/3}y^{-2/3}\,dy. \] With the substitution \(y=xu\), \[ (f*g)(x) =x^{-1/3}\int_0^1 (1-u)^{-2/3}u^{-2/3}\,du. \] The integral is the finite positive beta value \(B(1/3,1/3)\). Hence \[ (f*g)(x)=B(1/3,1/3)x^{-1/3} \qquad(0<x<1). \] Thus \((f*g)(x)\to\infty\) as \(x\downarrow0\). In particular \(f*g\) is not continuous, and therefore cannot be uniformly continuous.