Convergence, Integration

Question

Assume that f is a non-negative real function, and let $a>0$ be a real number.

Define $I_a(f)$ to be

$I_a(f)$=$\frac{1}{a}\int_{0}^{a} f(x) dx$

We now assume that $\lim_{x\rightarrow \infty} f(x)=A$ exists.

Now I want to proof whether $\lim_{a\rightarrow \infty} I_a(f)=A$ is true or not. I have concluded that this is not always true.

My approach has been to construct the following counterexample $f(x)=A+\frac {1}{x}$, it is easily seen that $\lim_{x\rightarrow \infty} f(x)=A$.

By integrating the chosen function I get that

$\lim_{a\rightarrow \infty}I_a(f)=A+\lim_{a\rightarrow \infty}\frac{1}{a}\int_{0}^a \frac{1}{x}dx = A+\lim_{a\rightarrow \infty} [\frac{\log(a)}{a}-(-\infty)]\rightarrow \infty$.

Therefore I concluded that $\lim_{x\rightarrow \infty} f(x)=A$ does not in general imply that $\lim_{a\rightarrow \infty}I_a(f)=A$.

I am of course unsure whether my calculations are correct, because you could also write the $\log(0)$ as a limit of $\log(\epsilon)\rightarrow \log(0)$, an then by division by $a$, and letting $a\rightarrow \infty$ get that the whole expression with the integral of $\frac{1}{x}$ goes to zero, I this case, it might be true that $\lim_{x\rightarrow \infty} f(x)=A$ implies that $\lim_{a\rightarrow \infty}I_a(f)=A$.

Does anybody have an idea to this?

Answer 1

Your approach is fine, including the fact that you noticed that there's a problem with $0$. But that problem is easy to solve: define $f(x)=A+\frac1{x+k}$, for some $k>0$. Then the problem will go away.

Answer 2

Assume $f$ is continuous on $[0,\infty)$ and $\lim_{x\to \infty} f(x)=L.$ Then it is $\lim_a I_a(f)=L.$ Indeed, we have

$$\lim_{a\to \infty} \dfrac{\int_0^a f(t)dt}{a}=\lim_{a\to \infty} \dfrac{f(a)}{1}=\lim_{a\to \infty}f(a) =L,$$ where we have used the Fundamental theorem of calculus and L'Hôpital's rule.

Answer 3

Your counter-example works because f is not integrable on $(0,a)$ for some $a$. If $f$ is integrable on $(0,a)$ for every a, it's true.

WLOG, assume $\lim\limits_{x\to \infty} f(x) = 0$. Otherwise, since $f$ is integrable, we may show that $\frac{1}{a}\int_0^a (f(x)-A) dx \to 0$. For every $\epsilon >0$, there exists $M > 0$ s.d. $\forall x\geq M$, $|f(x)| < \epsilon$. Then for $a \geq M$,

$$\left|\frac{1}{a} \int_0^a f(x) dx \right| = \left| \frac{1}{a}\int_0^M f(x) dx + \frac{1}{a}\int_M^a f(x) dx \right| < \frac{1}{a}\left| \int_0^M f(x)dx \right| + \frac{a-M}{a}\epsilon $$

Take $a\to \infty$ first then $\epsilon \to 0$.