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

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.

• I always thought that the point of a hint was to make a problem easier, not harder – shalop May 19 '19 at 0:40
• @Shalop: Clearly not in this case. – 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, $$\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow\infty}\lim_{n\rightarrow\infty}F_n(x)=\lim_{n\rightarrow\infty}\lim_{x\rightarrow\infty}F_n(x)=0.$$

• 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. – RRL May 19 '19 at 0:28
• @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. – Thorgott May 19 '19 at 0:38
• OK - thanks. It's a very nice answer (+1) from me. – 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.

• How does this use the sequence $x\mapsto \int_x^{x+\frac1n}\int f(t)\ \mathsf dt$? – Math1000 May 18 '19 at 23:59
• I had a very similar idea, but as mentioned that does not use the sequence. – איתן לוי May 19 '19 at 0:09
• 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. – RRL May 19 '19 at 0:13
• @Math1000: It doesn't. – RRL May 19 '19 at 0:33