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Let $\phi \in \mathcal{C}^2(\mathbb{R})$ such that $\phi'(z),\phi''(z) > 0$ for all $z \in \mathbb{R}$, i.e. $\phi$ is a strictly increasing and strictly convex function.

I have the strong believe that it holds that \begin{align} z \phi''(z) \to 0 \quad \text{for} \quad z \to -\infty. \end{align}

However, I couldn't show this statement so far. Any ideas?

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I think this is not the case. Consider, for example, $$ \phi''(x) := \sum_{n=1}^\infty n\, h_{[-n, -n + 1/n^3]}(x), $$ where $h_{[a,b]}$ denotes the "triangular" profile $$ h_{[a,b]}(x) := \begin{cases} 0, &\text{if}\ x \leq a\ \text{or}\ x \geq b,\\ 1- \frac{2}{b-a} \left| x - \frac{a+b}{2}\right|, & \text{if}\ a < x < b. \end{cases} $$ (If you prefer a strictly positive function, you can add for example $e^x$ to $\phi''$.)

Then let $$ \phi'(x) := \int_{-\infty}^x \phi''(t)\, dt, \qquad \phi(x) := \int_0^x \phi'(t)\, dt. $$ Your $\phi$ satisfies all the assumptions, but $\phi''$ is not bounded from above in $(-\infty, a)$ for any $a$.

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  • $\begingroup$ Hi, thanks for the quick answer and this counterexample. Is there a quick way to see that the function $\phi$ is well-defined and finite? By the construction of the triangles which have an area of $\frac{1}{2n^2}$, it is quite straight-forward to see that $\phi'$ is finite. However, I'm not sure how to do the second integration in order to see that also the function $\phi$ is finite. $\endgroup$ – Timo Dimi May 9 '17 at 19:58
  • $\begingroup$ The second integration is on the interval $[0,x]$. The function $\phi$ is simply the primitive of $\phi'$ vanishing at $0$. $\endgroup$ – Rigel May 10 '17 at 6:25

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