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given $a \in \mathbb{R}$ and $f$, a continuously differentiable function with a bounded derivative on $[a, \infty]$, such that $\int_0^{\infty}|f(x)|dx$ converges, prove that $\lim_{x\rightarrow\infty}f(x)=0$

hint: observe $\int f(x)f'(x)$

I was thinking maybe $\int f(x)f'(x)dx = \frac{f^2(x)}{2}$ but I don't see how it helps. more hints are welcome

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user11555739
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2 Answers2

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We can in fact prove a stronger statement :

If $f$ is uniformly continuous and $\int_0^\infty |f(x)|dx < \infty$ then $\lim_{x \to \infty} f(x) = 0$.

To see that this is stronger, show that a continuously differentiable function with bounded derivative is uniformly continuous.

Proof : Suppose not. Then there is an $\epsilon > 0$, such that there exists a sequence $r_i \to \infty$ such that $|f(r_i)| > \epsilon$ for all $i$. By uniform continuity, there exists a $\delta > 0$ such that on each of the intervals $[r_i -\delta,r_i+\delta]$, the value of $|f|$ stays above $|\frac{\epsilon}{2}|$.

We note : $$ \int_{0}^\infty |f| \geq \int_{\cup [r_i-\delta,r_i+\delta]}\frac{|\epsilon|}2 dx $$

and now conclude that $\int_0^\infty |f|dx = \infty$ because that union has infinite measure (why? Actually, we first choose the $\delta$ then the $r_i$ so that the intervals are disjoint, of course their total measure must be infinite).

This contradiction shows that the limit must exist and be zero.

Sarvesh Ravichandran Iyer
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  • thanks, by the way, is it necessary for f to be continuously differentiable? I remember showing every function with a bounded derivative is uniformly continuous, so we can make a stronger case can't we? – user11555739 May 07 '20 at 16:35
  • Sure, I mean, you can introduce kinks using the $|x|$ function : this lemma will apply to those functions, but those are not even differentiable, let along continuously differentiable. – Sarvesh Ravichandran Iyer May 07 '20 at 16:36
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If you want to use the hint. Then here is my idea.

$f$ has bounded derivative then there exists $M >0$ such that $\mid f'(x) \mid <M$ for all $x \geq a$. Choose any $\epsilon >0$ we prove that there exists $N>0$ such that $f^2(x) <\epsilon$ for all $x >N$.

Since $\int_0^{\infty}|f(x)|dx$ bounded, for $\frac{\epsilon}{4M}$ there exists $K>a$ such that $\int_K^{\infty}|f(x)|dx<\frac{\epsilon}{4M}$.

Using Mean Value Theorem, there exists $N \in (K,K+1)$ such that $\mid f(N)\mid = \int_K^{K+1}|f(x)|dx<\frac{\epsilon}{4M}$

We have for all $x > N$ $$\mid \int_N^{x} f(t)f'(t)dt\mid \leq \int_N^{x} \mid f(t)\mid \mid f'(t)\mid dt \leq M\int_N^{x} \mid f(x)\mid dx \leq \frac{\epsilon}{4}$$ So $$\mid f^2(x)-f^2(N) \mid \leq \frac{\epsilon}{2}$$ which implies $f^2(x) \leq \frac{\epsilon}{2}+\frac{\epsilon^2}{16M^2}<\epsilon$ for all $x>N$.

We conclude $\lim_{x \to \infty}f(x)=0.$

N.Quy
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