$f\in C^1(\mathbb R),\ c\in\mathbb R.$ If $\forall x\geq a:|f'(x)|\leq c$ & $\int_a^\infty|f(x)|dx$ is convergent, then $\lim_{x\to\infty}f(x)=0$.

Question

Let $f:[a,\infty]\rightarrow\mathbb R$ be a continuously differentiable function, and let $c\in\mathbb R$ be a constant, such that $|f'(x)|\leq c$ for every $x\geq a$.

Prove that if:

$$\int_a^\infty|f(x)|dx$$

is convergent, then:

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

It seems that I need to do something with $\int_a^\infty f(x)f'(x)dx$ and then conclude something else about $f(x)^2$, but I couldn't figure out what to do.

Thanks!

Answer 1

If $c=0$ replace $c$ by $1$. So we can assume $c>0.$

Suppose by contradiction that $\neg (\lim_{x\to \infty}f(x)=0).$ Then for some $r>0$ the set $S=\{x\ge a: |f(x)|>r\}$ has no upper bound.

$(I).$ When $x\in S,$ for any $y\in (0,r/2c)$ there exists $z\in (x,x+y)$ such that $$|f(x+y)-f(x)|=y\cdot \left|\frac {f(x+y)-f(x)}{y}\right|=y\cdot|f'(z)|\le yc<(r/2c)c=r/2.$$ Now $(\,|f(x)|>r\land |f(x+y)-f(x)|<r/2\,)\implies |f(x+y)|>r/2.$ So for every $x\in S$ we have $$\int_x^{x+ r/2c}|f(t)|dt>\int_x^{x+r/2c}(r/2)dt=r^2/4c.$$

$(II). $ Let for any $x$ let $G(x)=\int_a^x |f(t)|dt.$ And let $L=\lim_{x\to \infty}G(x).$

There exists $x_0\ge a$ such that $\forall x>x_0 \,(|G(x)-L|<r^2/8c).$

But there exists $x\in S$ with $x>x_0,$ so for this $x$ we have $$r^2/4c <\int_x^{x+r/2c}|f(t)| dt=|G(x+r/2c)-G(x)|\le$$ $$\le |G(x+r/2c)-L|\,+\,|L-G(x)|<$$ $$<r^2/8c+r^2/8c=r^2/4c$$ implying $r^2/4c<r^2/4c,$ an absurdity.

Answer 2

Suppose $x_n \to \infty$ and $f(x_n) \geq \epsilon$ for all $n$. Then $|x-x_n| <\epsilon /(2c)$ implies $f(x) \geq f(x_n)-|f(x)-f(x_n)| \geq \epsilon - \epsilon /2$ because $|f(x)-f(x_n)| \leq c|x-x_n|$ by MVT. Hence $f(x) \geq \epsilon /2$ on the interval $(x_ n, x_n+ \epsilon /(2c)$ Can you see from this that $\int |f| =\infty$?

What I have proved is that $\lim \sup_{x \to \infty} f(x) $ cannot be positive. Now apply the same argument to $-f$ and conclude that $f(x)$ must tend to $0$ as $x \to \infty$.