Show that $f:(0,\infty) \to (0,\infty)$ differentiable and such that $f'(x) \le \frac{1}{2f(x)}$ implies $f(x) \le \sqrt{x}$

Solution 1:

Once you have $f(x)^2 \le x+f(a)^2-a$ for $0 < a < x$ you can indeed take the limit for $a \to 0^+$ and conclude that $f(x)^2 \le x$. That part is fine.

But there is still a problem with your approach: You are assuming that $f(x)f'(x)$ is integrable, which is not given. What you are using is the fundamental theorem of calculus for $g(x) = f(x)^2$: $$ g(x) - g(a) = \int_a^x g'(t) \, dt \, . $$ This is true if $g'(x) = 2f(x) f'(x)$ is Riemann integrable, or more generally, if $g'$ is Lebesgue integrable (which is equivalent to $g$ being absolutely continuous on compact intervals).

A sufficient condition is (for example) that $f'$ is continuous.

This additional condition is avoided if you argue instead that $f(x)^2-x$ is decreasing because its derivative is $\le 0$ (which might be what your textbook solution does).

Roughly speaking: $g' \le 0$ implies that $g$ is decreasing because of the mean-value theorem, and that requires only the differentiability of $g$. Using the fundamental theorem of calculus for the same conclusion requires stronger conditions on the given function.