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Let $f(x)$ be an increasing, strictly concave function with $f(0)=0$. I have to show that given $x<y$, $f(y+\varepsilon) - f(x+\varepsilon) < f(y) - f(x)$, where $\varepsilon$ is a small, positive number. The idea is that a concave function has a decreasing slope (i.e. $f''<0$ if $f$ is differentiable). Any help would be appreciated!

Alessandro
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2 Answers2

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$f''<0$ implies $$f'(a+ε) < f'(a) $$ $$\implies \int_x^y f'(a+\epsilon) da < \int_x^y f'(a) da$$ $$\implies f(y+\epsilon) -f(x+\epsilon) < f(y) - f(x)$$

Madhusudana
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Let $ \alpha = \frac{ \epsilon}{y - x}$, which is between 0 and 1 for small positive $\epsilon$.

$$\begin{array} { l l l }f(y) &= &f\left( (1-\alpha)(y+\epsilon) + \alpha (x+\epsilon) \right) &> ( 1 - \alpha)f(y+\epsilon) + \alpha f(x+ \epsilon) \\ f(x+\epsilon) &= &f\left( (1-\alpha)(x) + \alpha y \right) &> ( 1 - \alpha)f(x) + \alpha f(y). \end{array} $$

Sum up these 2 inequalities, combine terms, and divide by $ 1 - \alpha$, we get

$$f(y) - f(x) > f(y + \epsilon) - f(x + \epsilon)$$

As a follow up, try the case where $ \epsilon > y -x$ in a similar manner.
Hence, the inquality is true for all $ \epsilon > 0$.

Calvin Lin
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  • Coming back to this question after a long time.. In both proofs (with and without f twice differentiable) the assumption that f is increasing is never used, right? So the statement is true for all strictly concave functions, regardless of whether they are increasing or not, correct? – Alessandro Apr 21 '23 at 15:29
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    @Alessandro Correct. – Calvin Lin Apr 21 '23 at 15:35