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Assume $1<p<\infty$, 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 $C_{0}((0,\infty))$.

user24367
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    yes, but for Hardy inequalities you need the function to be homogenous of degree -1. I was trying to use Holder's inequality, but $x^{-1/p}$ is not in $L_{p}$. – user24367 Mar 27 '12 at 19:52
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    Hölder sound like a good idea: $\int_0^x |f(t)|,dt \le x^{1/p} |f|{L^q([0,x])}$. So $|Tf|\infty \le |f|_q$. – martini Mar 27 '12 at 19:57
  • how do you infer the first inequality using Holder's? – user24367 Mar 27 '12 at 20:04
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    $\int_0^x |f(t)| , dt = \int_0^x |f(t)|\cdot 1, dt \le |1|{L^p([0,x])} \cdot |f|{L^q([0,x])} = (\int_0^x 1^p,dt)^{1/p} \cdot |f|{L^q([0,x])} = x^{1/p}\cdot |f|{L^q([0,x])}.$ – martini Mar 27 '12 at 20:07
  • so this proves that $Tf$ is bounded. How to infer that it goes to $0$ as $x\to \infty$. – user24367 Mar 27 '12 at 20:15
  • @user17523 Yes you are right! I did not think it over properly. – AD - Stop Putin - Mar 27 '12 at 20:42

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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.

Davide Giraudo
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