Suppose that \(1<p<\infty\) and define \[ Tf(x)=x^{-1/p}\int_0^x f(t)\,dt, \qquad x>0. \] Show that \(T\) is a bounded map from \(L^q((0,\infty))\) to \(C^0((0,\infty))\), where \(q\) is conjugate to \(p\): \[ \frac1p+\frac1q=1. \]
Solution
1. \(Tf\) is continuous on \((0,\infty)\).
If \(f\in L^q((0,\infty))\), then \(f\in L^1_{\mathrm{loc}}((0,\infty))\) by Hölder’s inequality. Hence \[ F(x):=\int_0^x f(t)\,dt \] is absolutely continuous, and therefore continuous, on every compact subinterval of \((0,\infty)\). Since \(x\mapsto x^{-1/p}\) is continuous on \((0,\infty)\), the product \[ Tf(x)=x^{-1/p}F(x) \] is continuous on \((0,\infty)\).
2. \(T\) is bounded, with operator norm at most \(1\) for the supremum norm.
For every \(x>0\), Hölder’s inequality on \((0,x)\) gives \[ \left|\int_0^x f(t)\,dt\right| \le \|f\|_{L^q(0,x)}\,\|1\|_{L^p(0,x)} =x^{1/p}\|f\|_{L^q(0,x)}. \] Therefore \[ |Tf(x)| \le \|f\|_{L^q(0,x)} \le \|f\|_{L^q(0,\infty)}. \] Taking the supremum over \(x>0\) yields \[ \|Tf\|_\infty\le\|f\|_q. \] Thus \(T:L^q((0,\infty))\to C_b((0,\infty))\subset C^0((0,\infty))\) is a bounded linear map.