I am studying Carother's Real Analysis for my qualifying exams. In the book I am to prove that if $f : [0, \infty) \rightarrow [0, \infty)$ is an increasing function, $f(0) = 0$, and $f(x) > 0$ for all $x > 0$, then $\frac{f(x)}{x}$ being decreasing for $x > 0$ implies that $f$ is subadditive, or that $f(x + y) \leq f(x) + f(y)$.

So far I have tried:

$\frac{f(x+y)}{x+y} \leq \frac{f(y)}{y}$ and $\frac{f(x+y)}{x+y} \leq \frac{f(x)}{x}$ implies that $2\frac{f(x+y)}{x+y} \leq \frac{f(x)}{x} +\frac{f(y)}{y}$, so $\frac{f(x+y)}{x+y} \leq 2\frac{f(x+y)}{x+y} \leq \frac{f(x)}{x} +\frac{f(y)}{y} \leq f(x) + f(y)$

I think that I am close but I can't get rid of the $x + y$ in the denominator. Any help would be appreciated.


1 Answer 1


For $x \gt 0$ and $y \gt 0$, you get

$$\frac{f(x)}{x} \geq \frac{f(x+y)}{x+y} \; \implies f(x) \geq \frac{xf(x+y)}{x+y} \tag{1}\label{eq1}$$


$$\frac{f(y)}{y} \geq \frac{f(x+y)}{x+y} \; \implies f(y) \geq \frac{yf(x+y)}{x+y} \tag{2}\label{eq2}$$

Adding \eqref{eq1} and \eqref{eq2} gives

$$f(x) + f(y) \geq \frac{xf(x+y)}{x+y} + \frac{yf(x+y)}{x+y} = \frac{(x+y)f(x+y)}{x+y} = f(x+y) \tag{3}\label{eq3}$$

For $x$ and/or $y$ being $0$, since $f(0) = 0$, then \eqref{eq3} still holds. Overall, this is what you were asked to prove, i.e., that $f$ is subadditive.

Note this didn't use that $f$ is an increasing function. Just having $f(0) = 0$ and $\frac{f(x)}{x}$ being a decreasing function is sufficient.

  • 1
    $\begingroup$ @Aphyd - I am literally doing the SAME problem right now and made the same mistake, no worries my friend. $\endgroup$ Commented Aug 23, 2020 at 0:58

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