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A proof for deducing Lypaunov's inequality seems to be centered on showing that if $f$ is convex on $0< a \le x \le b$, then $f(x)/x$ must be non-decreasing on the interval. I am not aware that $f$ needs to be differentiable.

Starting from a definition of convexity, I can write $$ f(x) \le f(a) + \frac{(x-a)[f(b) - f(a)]}{b-a} \implies \frac{f(x)}{x} \le \frac{f(a)}{a} + \frac{(x-a)[f(b) - f(a)]}{x(b-a)}. $$ We can define $g(x) = f(x)/x$ to obtain $$ \frac{g(x) - g(a)}{x-a} \le \frac{f(b)-f(a)}{x(b-a)}. $$ None of this seems to be heading in the right direction, what am I doing wrong? Is there another requirement on the function $f$?

Gregory
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1 Answers1

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Without extra assumption, it is false. Let $f:(1,3)\rightarrow \mathbb{R}$ be defined as $f(x) = |x-2|$. Clearly, $f$ is convex. However, $x\mapsto f(x)/x$ is not non-decreasing. Denote $g(x) = f(x)/x$. Obviously $g(1) = 1$ but $g(2) =0$.

Danny Pak-Keung Chan
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  • This at least is a relief as I was unable to prove it haha – Gregory Feb 17 '21 at 05:42
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    Assuming differentiability would not help. Imagine that you can smooth out the corner at $x=2$ (this is possible by convoluting $f$ with a $C^\infty_c$ function $h$ with support contained in $(-\delta, \delta)$, $h\geq 1$ and $\int h=1$). The situation will not improve. – Danny Pak-Keung Chan Feb 17 '21 at 05:44