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For $x \in (0, \infty)$, let: \begin{align*} g(x) &= \int_0^\infty \frac{1}{x+y} f(y) \, dy \\ \end{align*} Show that $g(x) \mathop{\longrightarrow}\limits_{x \to \infty} 0$ for $f \in L^p(0,\infty)$, $1 \le p \le \infty$

I'm struggling with the $p = \infty$ case, but I believe I have a solution for $1 \le p < \infty$:

Let $h(x,y) = \frac{1}{x+y}$ such that:

\begin{align*} g(x) &= \int_0^\infty \frac{1}{x+y} f(y) \, dy \\ g(x) &= \int_0^\infty f(y) h(x,y) \, dy \\ g(x) &= \lVert f \cdot h(x) \rVert_1 \\ \end{align*}

From there we can use Holder's Inequality with $q = p/(p-1)$:

\begin{align*} g(x) = \lVert f \cdot h(x) \rVert_1 &\le \lVert f \rVert_p \lVert h(x) \rVert_q \\ &= \lVert f \rVert_p \left( \int_0^\infty h(x)^q \, dy \right)^{1/q} \\ &= \lVert f \rVert_p \left( \int_0^\infty (x+y)^{-q} \, dy \right)^{1/q} \\ \end{align*}

Substitute $z=x+y$:

\begin{align*} g(x) &= \lVert f \rVert_p \left( \int_x^\infty z^{-q} \, dz \right)^{1/q} \\ &= \lVert f \rVert_p \left( \frac{1}{q-1} \cdot x^{1-q} \right)^{1/q} \\ &= \lVert f \rVert_p \left( \frac{1}{q-1} \right)^{1/q} \cdot x^{1/q-1} \\ &= \lVert f \rVert_p \left( p-1 \right)^{1-1/p} \cdot x^{-1/p} \\ \end{align*}

Take the limit of both sides:

\begin{align*} \lim\limits_{x \to \infty} g(x) &= \lim\limits_{x \to \infty} \lVert f \rVert_p \left( p-1 \right)^{1-1/p} \cdot x^{-1/p} \\ &= \lVert f \rVert_p \left( p-1 \right)^{1-1/p} \cdot \lim\limits_{x \to \infty} x^{-1/p} \\ &= 0 \\ \end{align*}

If $p=\infty$, then $q=1$, and $\lVert h(x) \rVert_1$ does not converge, and this proof does not work. Is there a different proof that works for that case?

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    $\begingroup$ If $p=\infty$, then $f$ is bounded. However in this case the integral defining $g$ may not be convergent. So it is not right to ask for $p=\infty$. $\endgroup$ – Yu Ding Apr 23 at 7:20
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The result is not true for $p =\infty$. In fact if $f \equiv 1$ then $g(x)=\infty$ for all $x$ even though $f \in L^{\infty}$.

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