# 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?

• 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 $p=\infty$. – Yu Ding Apr 23 at 7:20

The result is not true for $$p =\infty$$. In fact if $$f \equiv 1$$ then $$g(x)=\infty$$ for all $$x$$ even though $$f \in L^{\infty}$$.