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I was working on an old practice qual and I came across this problem. I have a solution, but it's rather...convoluted...and I feel like there should be a simple way of using the dominated convergence theorem to solve it. Anyways, here's the problem:


Let $f: [0,\infty) \rightarrow \mathbb{R}$ be Lebesgue integrable and suppose that $\lim_{x\rightarrow 0} f(x) = 2016$. Show that $$\lim_{n\rightarrow \infty} \int_0^\infty nf(x)e^{-nx} \, dx = 2016$$


If you change variables, you get that the integral is equal to $\int_0^\infty f(\frac{x}{n})e^{-x}\, dx$, and so the claim is easy to prove if you know that $f$ is bounded. In particular, if $f$ is continuous with compact support, the problem is solved. However, despite the fact that we can prove the claim for a dense subset of $L^1$, it doesn't seem to follow for all of $L^1$ (as far as I can tell).

I'll put my solution (without directly using the DCT) in the answers because it's pretty long.

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    $\begingroup$ I think the trick here is to fix $\epsilon>0$ and consider $\int_0^{\epsilon} nf(x)e^{-nx}dx+ \int_{\epsilon}^{\infty}nf(x)e^{-nx}dx$. Then you can say, restricting attention to the interval $x \geq \epsilon$, that $|f(x)ne^{-nx}|$ is bounded by the function $c|f(x)|$ for some suitable $c>0$ and use Lebesgue Dominated Convergence theorem (LDC) for $\lim_{n\rightarrow\infty}\int_{\epsilon}^{\infty}nf(x)e^{-nx}dx$. Then take care of the first integral $\int_0^{\epsilon}$ separately (without LDC). $\endgroup$
    – Michael
    Sep 12 '18 at 23:10
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    $\begingroup$ This is great, I got it now! Would you care to write up a solution so that I can accept it? $\endgroup$
    – Greg K
    Sep 12 '18 at 23:27
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Here is detail on my comment, as requested:

We assume $c=\int_0^{\infty} |f(x)|dx<\infty$. Fix $\epsilon>0$ and notice that $$\left| \int_{\epsilon}^{\infty} f(x)ne^{-nx}dx \right| \leq ne^{-n\epsilon}\int_{\epsilon}^{\infty} |f(x)|dx \leq ne^{-n\epsilon}c \rightarrow 0$$ where the limit is as $n\rightarrow\infty$. (You could also use LDC to claim this, as in my first comment). So then $$ \int_0^{\infty} = \int_0^{\epsilon} + \int_{\epsilon}^{\infty} $$ and $\lim_{n\rightarrow\infty} \int_0^{\epsilon}$ can be bounded (in terms of $\epsilon$) again by direct "sandwich" type methods.

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Extend $f$ to $\mathbb{R}$ so that $f(-x) = f(x)$ for all $x$. Define $g_n(x) = \frac{1}{2}ne^{-n|x|}$ for all $x \in \mathbb{R}$. Then, notice that

\begin{align*} (f*g_n)(0) &= \int_\mathbb{R} f(x) g_n(-x) \, dx\\ &= \frac{1}{2} \int_\mathbb{R} nf(x) e^{-n|x|} \, dx\\ &= \int_0^\infty n f(x) e^{-nx} \, dx \end{align*}

So what we really want to show is that $\lim_{n\rightarrow \infty} (f*g_n)(0) = \lim_{x\rightarrow 0} f(x) = 2016$.

Note that because $\lVert g_n\rVert_1 = 1$ for all $n$, then for any $y \in \mathbb{R}$, we have

\begin{align*} |(f*g_n)(y) - f(y)| &= \left|\int_\mathbb{R} (f(y-x) - f(y))g_n(x) \, dx\right|\\ &\leq \int_\mathbb{R} |f(y-x) - f(y)|g_n(x) \, dx \end{align*}

Denoting $f_x(y) = f(y-x)$, we can integrate both sides of the above equation with respect to $y$ and apply Fubini's theorem (since the integrand is nonnegative) to find that

\begin{align*} \lVert f*g_n - f\rVert_1 &\leq \int_\mathbb{R} \lVert f_x - f \rVert_1 g_n(x) \, dx\\ &= \int_\mathbb{R} \lVert f_x - f \rVert_1 ng_1(nx) \, dx\\ &= \int_\mathbb{R} \lVert f_{x/n} - f \rVert_1 g_1(x) \, dx \end{align*}

But since the integrand is bounded by $2\lVert f\rVert g_1 \in L^1(\mathbb{R})$, we can now apply the dominated convergence theorem to find that $f*g_n$ converges to $f$ in the $L^1$ norm. In particular, we will have $f*g_n(0) \rightarrow 2016$, which is what we wanted to prove.

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