<s>[INST] If $f \in L^p(0,\infty)$ and $g(x) = \int_0^\infty \frac{1}{x+y} f(y) \, dy$, show $\lim\limits_{x \to \infty} g(x) = 0$ 
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?
 [/INST] The result is not true for $p =\infty$. In fact if $f \equiv 1$ then $g(x)=\infty$ for all $x$ even though $f \in L^{\infty}$. 
</s>