Let $f:[a,\infty)\rightarrow \mathbb{R}$ be a uniformly continuous function in that range. $\int_{a}^{\infty} f$ converges. Prove that $\lim_{x\to\infty} f(x)=0$

Hint: Use the sequence $F_n(x)=n\int_{x}^{x+\frac{1}{n}} f$.

Honestly I have been trying to solve this one for some time but the hint really confuses me.

I have tried to mess around with $F_n(x)$ a bit, for example by using the fundamental theorem but it still seems like such a random choice and I can't make anything out of it.

Any guidance/explanations will be appreciated.

Please use the hint in the question.

  • $\begingroup$ I always thought that the point of a hint was to make a problem easier, not harder $\endgroup$ – Shalop May 19 '19 at 0:40
  • $\begingroup$ @Shalop: Clearly not in this case. $\endgroup$ – RRL May 19 '19 at 0:40

First of all, note that for all $n\in\mathbb{N}$ $$\lim_{x\rightarrow\infty}F_n(x)=n\left[\lim_{x\rightarrow\infty}\left(\int_a^{x+\frac{1}{n}}f(t)\mathrm{d}t- \int_a^xf(t)\mathrm{d}t\right)\right]=n\left[\int_a^{\infty}f(t)\mathrm{d}t-\int_a^{\infty}f(t)\mathrm{d}t\right]=0.$$ Now let $\varepsilon>0$ be arbitrary. By uniform continuity, there is a $\delta>0$ such that for all $t,x\in\left[a,\infty\right)$, we have $|f(t)-f(x)|<\varepsilon$ whenever $|t-x|<\delta$. Pick $N\in\mathbb{N}$ such that $\frac{1}{n}<\delta$ for all $n\ge N$. Then for all $x\in\left[a,\infty\right)$, we have $$\left|n\int_x^{x+\frac{1}{n}}f(t)\mathrm{d}t-f(x)\right|=\left|n\int_x^{x+\frac{1}{n}}f(t)-f(x)\mathrm{d}t\right|\le n\int_x^{x+\frac{1}{n}}|f(t)-f(x)|\mathrm{d}t\le\varepsilon.$$ Since $\varepsilon$ and $x$ were arbitrary, we conclude that $\lVert F_n-f\rVert\rightarrow0$ as $n\rightarrow\infty$, i.e. $F_n\rightarrow f$ uniformly. Thus, $$\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow\infty}\lim_{n\rightarrow\infty}F_n(x)=\lim_{n\rightarrow\infty}\lim_{x\rightarrow\infty}F_n(x)=0.$$

  • $\begingroup$ I'm still puzzling over the interchange of limits, but at least for all $n > N$ we have $0 \leqslant |f(x)| \leqslant |f(x) - F_n(x)| + |F_n(x)| < \epsilon + |F_n(x)|$ and $0 \leqslant \liminf_{x \to \infty}|f(x)| \leqslant \limsup_{x \to \infty} |f(x)| < \epsilon$ for any $\epsilon > 0$ which gives the result. $\endgroup$ – RRL May 19 '19 at 0:28
  • $\begingroup$ @RRL This follows from a generalized version of the uniform limit theorem, as given in 7.11 of Baby Rudin for example. This particular case is easier though: suppose $f_n\colon\left[a,\infty\right)\rightarrow\mathbb{R}$ converge uniformly to $f$ and $f_n(x)\rightarrow L$ for $x\rightarrow\infty$ and all $n$. Then $|f(x)-L|\le|f(x)-f_n(x)|+|f_n(x)-L|$. Pick $n$ s.t. $||f_n-f||<\varepsilon/2$. Pick $X$ s.t. $|f_n(x)-L|<\varepsilon/2$ for $x>X$. Then $|f(x)-L|<\varepsilon$ for $x>X$. The statement follows. $\endgroup$ – Thorgott May 19 '19 at 0:38
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    $\begingroup$ OK - thanks. It's a very nice answer (+1) from me. $\endgroup$ – RRL May 19 '19 at 0:39

Assume that $\lim_{x \to \infty}f(x) =0$ does not hold and arrive at contradiction with the fact that the integral of $f$ is convergent.

If $\lim_{x \to \infty} f(x) = 0$ does not hold then there exists $\epsilon_0 > 0$ and a sequence $x_n \to \infty$ such that $|f(x_n)| \geqslant \epsilon_0$ for all $n$.

Assume WLOG that $f(x_n) \geqslant \epsilon_0$.

There exists by uniform continuity of $f$ a $\delta > 0$ such that $|f(t) - f(x_n)| < \epsilon_0/2$ for all $t \in [x_n - \delta,x_n + \delta].$

This implies $f(t) > \epsilon_0/2$ and

$$ \int_{x_n - \delta}^{x_n + \delta} f(t) \, dt > \epsilon_0\delta$$

This violates the Cauchy criterion for convergence of the improper integral since $x_n$ can be arbitrarily large.

  • $\begingroup$ How does this use the sequence $x\mapsto \int_x^{x+\frac1n}\int f(t)\ \mathsf dt$? $\endgroup$ – Math1000 May 18 '19 at 23:59
  • $\begingroup$ I had a very similar idea, but as mentioned that does not use the sequence. $\endgroup$ – איתן לוי May 19 '19 at 0:09
  • $\begingroup$ There's a common thread here but I thought to show you an indirect proof. If following the hint is essential, then I would definitely accept the other answer. $\endgroup$ – RRL May 19 '19 at 0:13
  • $\begingroup$ @Math1000: It doesn't. $\endgroup$ – RRL May 19 '19 at 0:33

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