Let $f:[0, \infty) \to \mathbb{R}$ be a continuous function such that $\lim _{x \to \infty} f(x)=0$
(a) prove that $f$ is bounded
(b) prove that $$\lim _{x \to \infty} \frac{1}{x}\int ^x_0 f(t)dt=0$$
My attempt:
How to prove (a)
for (b) if we know that $|f(t)| \le \epsilon$ fro all $t \ge S.$ then
$\int ^x_0f(t)dt=\int ^S_0 f(t) dt+\int ^x_0 f(t)dt$ ..how to processed from this
a) For some $M>0$, $|f(x)|<1$ for all $x>M$. Now $|f|$ is continuous on compact $[0,M]$, and hence $\max_{x\in[0,M]}|f(x)|$ exists, so $|f(x)|\leq\max\{1,\max_{x\in[0,M]}|f(x)|\}$ for all $x\in[0,\infty)$.
b) Given $\epsilon>0$, for some $M>0$, $|f(x)|<\epsilon$ for all $x\geq M$, then $\dfrac{1}{x}\displaystyle\int_{0}^{x}|f(u)|du\leq\dfrac{1}{x}\int_{0}^{M}|f(u)|du+\dfrac{1}{x}\int_{M}^{x}|f(u)|du\leq\dfrac{1}{x}\int_{0}^{M}|f(u)|du+\epsilon$ for all such $x$. Since $\displaystyle\int_{0}^{M}|f(u)|du$ is constant, now taking $\limsup$ as $x\rightarrow\infty$, we have $\limsup_{x\rightarrow\infty}\dfrac{1}{x}\displaystyle\int_{0}^{x}|f(u)|du\leq 0+\epsilon$.