My problem : Let $r<-1$ be a fixed real number and define the function $f:[1, +\infty)\rightarrow\mathbb{R}$ by $f(x) = -x^r$. Hence, $f$ is concave and increasing on $[1, +\infty)$. My question is:
Is the function $\zeta(-r, x+1)-\zeta(-r) + \frac{1}{2}x^r + \frac{x^{r+1}}{r+1}$ convex on $[1, +\infty)$?
Where $\zeta(-r)$ and $\zeta(-r, x+1)$ are Riemann zeta function and Hurwitz zeta function, respectivly. Can you help me?
I know that if $f^{\prime\prime}(x)$ exists on $(a, b)$, then $f$ is convex if and only if $f^{\prime\prime}(x)\geq 0$.
My attempt: Since $f$ is concave, so $\frac{1}{2}x^r + \frac{x^{r+1}}{r+1}$ is convex. On the other hand, the second derivative $\zeta(-r, x+1)$ is $r(r-1)\zeta(2-r, x+1)$, so $f^{\prime\prime}\geq 0$, and so $\zeta(-r, x+1)-\zeta(-r) + \frac{1}{2}x^r + \frac{x^{r+1}}{r+1}$ is convex. is it right?