I have $\sigma(s) \in C^0[0,\infty)$, non-decreasing with $\sigma(0)=0$. Define for $t>0$, $$ \tau(t) = \frac{1}{t^2} \int_{0}^{2t} \int_{0}^{2r} \sigma(s)ds dr $$ with $\tau(0)=0$. Is $\tau(t)$ necessarily twice-differentiable and non-decreasing as well? How can I show it?
I know that it is second-order anti-derivative of $\sigma$. But, is the reverse true? Thanks!
By changing the order of integration, we have that $$\tau(t) = \frac{1}{t^2} \int_{r=0}^{2t} \int_{s=0}^{2r} \sigma(s)ds dr =\frac{1}{t^2} \int_{s=0}^{4t} \sigma(s)ds\int_{r=s/2}^{2t} dr =\frac{1}{2t^2} \int_{s=0}^{4t} (4t-s)\sigma(s)ds.$$ Hence for $t>0$, $$\tau'(t)=-\frac{2}{2t^3} \int_{s=0}^{4t} (4t-s)\sigma(s)ds+\frac{1}{2t^2} \int_{s=0}^{4t} 4\sigma(s)ds =\frac{1}{t^3} \int_{s=0}^{4t} (s-2t)\sigma(s)ds$$ and $\tau$ is differentiable. We may compute the second derivative as well: $$\tau''(t)=\frac{8\sigma(4t)}{t^2}-\frac{1}{t^4} \int_{s=0}^{4t} (3s-4t)\sigma(s)ds.$$ Moreover, by using Question related to Chebyshev Sum Inequality we have that $$\tau'(t)=\frac{1}{t^3} \int_{s=0}^{4t} (s-2t)\sigma(s)ds\geq \frac{1}{4t^4} \int_{s=0}^{4t} (s-2t)\cdot\int_{s=0}^{4t}\sigma(s)ds=0$$ and we may conclude that $\tau$ is non-decreasing.
P.S. $\tau$ may be not differentiable at $0$. Take for example $\sigma(x)=\sqrt{x}$. If $\sigma$ is differentiable at $0$, then $\tau'(0)=16\sigma'(0)/3$.
