# If $f$ is a convex increasing function, then $\lim_{x\rightarrow -\infty} \frac{1}{x}\int_{x}^{x_0}f(t)dt=-\lim_{x\rightarrow -\infty}f(x)$

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?

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

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.