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.
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 :)