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Assume $1, let $Tf(x) = x^{-1/p}\int_{0}^{x}f(t)dt$. If $p^{-1}+q^{-1} = 1$, then $T$ is a bounded linear map from $L^{q}((0,\infty))$ to $C_{0}((0,\infty))$.

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    @user17523 Yes you are right! I did not think it over properly.2012-03-27

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For the continuity, as @martini showed it's a consequence of Hölder's inequality: $Tf(x)=x^{-1/p}\int_{[0,x]}|f(t)|dt\leq x^{-1/p}x^{1/p}\lVert f\rVert_{L^q}=\lVert f\rVert_{L^q}$ so $T$ is bounded. To show that the functions in the range of $T$ converge to $0$ at $+\infty$, let $f\in L^q$. Let $f_n:=f\cdot \chi_{(0,n)}$. By the monotone converge theorem, $f_n$ converges to $f$ in $L^q$. So for a fixed $x$, using the last inequality, $|Tf(x)|=|T(f-f_n)(x)+T(f_n)(x)|\leq \lVert f-f_n\rVert_{L^q}+|T(f_n)(x)|.$ Let $\varepsilon>0$. We fix $n_0$ such that $\lVert f-f_{n_0}\rVert_{L^q}\leq \varepsilon$. We have for $x\geq n_0$ that $T(f_n)(x)=x^{-1/p}\int_0^{n_0}f(x)dx$ so which gives the wanted result.