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Let f be a real-valued function over $ \Omega=(0,+\infty)$. Firstly, I assume that f is twice differentiable $(*)$.

Under the condition $(*)$ it's easy to see the following proposition

$\textbf{Proposition}$ If f is convex on $\Omega$ then the function $t \mapsto tf(\frac{1}{t})$ (let's say h) is also convex on $\Omega$ (by using the second derivative.)

My question is whether above proposition is still true if we eliminate the condition $(*)$. If it's wrong, please give a counterexample. Thanks everyone.

No name
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1 Answers1

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Hint: Take $t_{0}$ and $t_{1}$ in $\Omega$, and take $\alpha \in \left] 0,1 \right[$. You may write

$$h((1-\alpha)t_{0} + \alpha t_{1}) = ((1-\alpha)t_{0} + \alpha t_{1})\,f\Bigl(\frac{(1-\alpha)t_{0}}{(1-\alpha)t_{0} + \alpha t_{1}}\frac{1}{t_{0}} + \frac{\alpha t_{1}}{(1-\alpha)t_{0} + \alpha t_{1}} \frac{1}{t_{1}} \Bigr),$$ and then use convexity of $f$. Note that

$$\frac{(1-\alpha)t_{0}}{(1-\alpha)t_{0} + \alpha t_{1}} + \frac{\alpha t_{1}}{(1-\alpha)t_{0} + \alpha t_{1}} =1.$$ No fancy derivative test is required :)

weirdo
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