If $f$ is a convex increasing function, then I need to show that $$ \lim_{x\rightarrow -\infty} \frac{1}{x}\int_x^{x_0}f(y)dy=-\lim_{x\rightarrow -\infty}f(x).$$

I can prove this equality using L'Hopital's rule and the Fundamental Theorem of Caluclus. However, is there a way of proving this without use L'Hopital's rule?

  • 1
    $\begingroup$ Yes, you are right. I fixed the satement. $\endgroup$
    – MEG
    Jul 5, 2022 at 20:58
  • $\begingroup$ The only aspect of convexity here that is relevant is the fact that $\lim_{x \to -\infty} f(x)$ exists (may be $-\infty$). $\endgroup$
    – copper.hat
    Jul 5, 2022 at 22:31

2 Answers 2


L'Hopital's rule shows that the identity holds whenever the limit on the right exists. In your special case ($f$ increasing and convex) there is a simpler short proof:

First note that $A = \lim_{x\to -\infty}f(x)$ exists (as a real number or $-\infty$) because $f$ is increasing.

For $x < x_0$ is $$ (x_0-x) f(x) \le \int_x^{x_0} f(y) \, dy \le (x_0-x) f(\frac{x+x_0}{2}) \, , $$ where the left inequality holds because $f$ is increasing, and the right inequality holds because $f$ is convex.

Dividing by $x$ and squeezing shows that $$ \lim_{x \to -\infty} \frac 1x\int_x^{x_0} f(y) \, dy = -A \, . $$


I think the limit on the right should have a negative sign in front of it. I didn't need the convexity assumption for this proof. Let $L = \lim_{x \to -\infty}f(x)$ and $g(x) = f(x) - L$. Suppose $L$ is finite. Then $g$ is also increasing and its limit at $-\infty$ is $0$. Now we show that the limit on the left with $f$ replaced by $g$ is also $0$. Obviously $$ \frac{1}{x} \int_x^{x_0}g(y)\ dy \le 0 $$ for all $x < \textrm{min}\{x_0, 0\}$. Notice that the limit of this expression does not depend on $x_0$. Hence we can choose $x_0 < 0$ so that $g(x) < \varepsilon$ for all $x < x_0$ and some fixed $\varepsilon > 0$. Then $$ \frac{1}{x}\int_x^{x_0}g(y)\ dy > \frac{\varepsilon(x_0 - x)}{x}\ .$$ Therefore $$ \liminf_{x \to -\infty} \frac{1}{x}\int_x^{x_0}g(y)\ dy \ge -\varepsilon\ .$$ Since $\varepsilon$ was arbitrary, we see that the limit exists and is equal to $0$. It follows that $$0 = \lim_{x \to -\infty} \frac{1}{x}\int_x^{x_0}f(y) - L\ dy = \lim_{x \to -\infty} \frac{1}{x}\int_x^{x_0}f(y) \ dy - \lim_{x\to -\infty}\frac{L(x_0 - x)}{x} \\= \lim_{x \to -\infty} \frac{1}{x}\int_x^{x_0}f(y)\ dy + L\ .$$ This concludes the proof. The case where $L = -\infty$ is similar and I leave it to you.


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