Is the lower incomplete gamma function convex in terms of $x$, for $x>0$ and $s>0$? $$\gamma(s, x) = x^s \, \Gamma(s) \, e^{-x}\sum_{k=0}^\infty\frac{x^k}{\Gamma(s+k+1)}$$
My answer: It is convex for $x>0$ since it is product of two convex functions $e^{-x}$ and $\Gamma(s) \sum_{k=0}^\infty\frac{x^k}{\Gamma(s+k+1)}$. Is this argument right?
No, your argument is not correct. The product of two convex functions on an interval is not necessarily convex; e.g., $f(x) = e^{-x}$ is convex, and so is $g(x) = x^2$, but their product $f(x) g(x) = x^2 e^{-x}$ is not convex, having a local maximum at $x = 2$ and global minimum at $x = 0$.
In fact, since $$\gamma(s,x) = \int_{t=0}^x t^{s-1} e^{-t} \, dt,$$ we have $$\frac{d\gamma}{dx} = x^{s-1} e^{-x}, \quad \frac{d^2\gamma}{dx} = (s-x-1)x^{s-2} e^{-x},$$ and since convexity for a smooth function requires that the second derivative is nonnegative, we see that this condition is violated if $x > s-1$.