# A question regarding a problem in Folland

Let $f_n(x) = ae^{-nax}-be^{-nbx}$ where $0<a<b$.

Prove that $$\sum_1^\infty f_n \in L^1([0,\infty),m)\quad\text{and}\quad\int_0^\infty\sum_1^\infty f_n(x)\,dx=\log(b/a).$$

This is the problem 27 of chapter 2 of Folland Real Analysis. Could anyone give me some helps about how to attack this problem? I can't think of any way...

• Can you do $b=2,a=1$? – GEdgar Jan 27 '15 at 18:11
• Was the list of problems preceded by something that says "Prove the following assertions"? If so, why don't you tell us that? – Michael Hardy Jan 27 '15 at 18:16
• This is routine: $\sum_{n \geq 1} f_n(x)$ is a sum of two geometric series. Just compute it and then integrate from $0$ to $\infty$. – user203787 Jan 27 '15 at 18:20
• Because they are too trivial.. – Keith Jan 27 '15 at 18:20

Hint : $$\sum_{n=1}^{\infty}f_n(x)=\sum_{n=1}^{\infty}ae^{-nax}-\sum_{n=1}^{\infty}be^{-nbx}=a(\frac{e^{-ax}}{1-e^{-ax}})-b(\frac{e^{-bx}}{1-e^{-bx}})\in L^1([0, \infty),m)$$
Consider part (2) first, then for every $$m > 0$$, $$ae^{-anx},be^{-bnx}$$ are bounded on $$[0,m]$$ and we have $$\int_{0}^{m} ae^{-anx} = \frac{1}{n}(1 - e^{-nam})$$ for any $$n$$, by monotone convergence: $$\int_{0}^{\infty} f = \lim_{m} \frac{1}{n} (e^{-nbm} - e^{-nam}) = 0$$ and hence $$\sum^{\infty} \int_{0}^{\infty} f = 0$$.
Then consider part (1). For any $$n$$ and $$x \geq \chi := \frac{\log (b/a)}{(b-a)n} > 0$$, $$ae^{-anx} \geq be^{-bnx}$$ so $$\int_{[0,\infty]} f \geq \int_{[\chi,\infty]} (ae^{-anx} - be^{-bnx}) = \frac{1}{n} (e^{-na\chi} - e^{-nb\chi}) = \frac{1}{n} C$$ for a constant $$C = \frac{b}{a}(e^{-\frac{a}{b-a}} - e^{-\frac{b}{b-a}}) > 0$$. Because the series $$\sum_{n=1}^{\infty} \frac{1}{n} C$$ diverges, by comparison, $$\sum_{n=1}^{\infty} \int f$$ diverges as well.
Then consider part (3). $$\sum_{n=1}^{\infty} f_{n}(x) = \frac{a e^{-ax}}{1 - e^{-ax}} - \frac{b e^{-bx}}{1 - e^{-bx}}$$, and $$\vert{\sum_{n=1}^{\infty} f_{n}}\vert = \sum_{n=1}^{\infty} f_{n}$$because $$\frac{a e^{-ax}}{1 - e^{-ax}} - \frac{b e^{-bx}}{1 - e^{-bx}} = \frac{e^{-bx}e^{-ax}}{(1-e^{-bx})(1-e^{-ax})}(ae^{bx} - be^{ax} - (a -b)) > 0$$ by Taylor expansion of exponentials. Then $$\int \vert{\sum_{n=1}^{\infty} f_{n}}\vert = \int \sum_{n=1}^{\infty} f_{n} = \lim_{m\to \infty} \log(\frac{1-e^{-b/m}}{1 - e^{-a/m}}) = \log \frac{b}{a}$$