# Can it be proved that $f$ is integrable?

Let $x$‎ be a differentiable function on $\mathbb{R}$. I want to prove that for any time $t \geq t_0$‎ ‎$$‎\frac{1}{2} D^{\alpha} x^2(t) \leq x(t) D^{\alpha} x(t)‎, ‎\ \ \forall \alpha \in (0‎, ‎1), ‎$$‎ where $D^\alpha$ is the Caputo derivative. This is equivalent to showing that

‎$x(t) D^{\alpha} x(t)‎ - ‎\frac{1}{2} D^{\alpha} x^2(t) \geq 0‎, ‎\ \ \forall \alpha \in (0‎, ‎1)$.

After simplifying we have: $$\label{11.67}‎ ‎\biggl[\frac{[x(t) - x(t_0)]^2}{2\Gamma (1‎- ‎\alpha) (t‎- ‎t_0)^\alpha}\biggr]‎+ ‎\frac{\alpha}{2\Gamma (1‎- ‎\alpha)} \int_{t_0}^{t} \frac{[x(t)-x(\tau)]^2}{(t-\tau)^{\alpha+1}} \,d\tau \geq 0‎. ‎$$‎

If we prove that the function $f(\tau)=\frac{[x(t)-x(\tau)]^2}{(t-\tau)^{\alpha+1}}$ is integrable, then the theorem is proved. Can this result be obtained?

• For a differentiable function $|x(t)-x(\tau)|^2=O(|t-\t|^2)$, $\tau\to t$, so yes function $f$ is integrable. – Andrew Jan 19 '18 at 5:43

We have that $\tau \mapsto (t-\tau)^{-\varepsilon-\alpha}[x(t) - x(\tau)]^2$ is bounded on $(t_0, t)$ for some $\varepsilon > 0$. Then, up to a constant $C > 0$
$$\frac{[x(t) - x(\tau)]^2}{(t-\tau)^\alpha} \leq C (t-\tau)^{\varepsilon} \to 0 \text{ for } \tau \to t,$$
$$\frac{[x(t) - x(\tau)]^2}{(t-\tau)^{\alpha+1}} \leq C \frac{1}{(t-\tau)^{1-\varepsilon}}$$
which gives integrability of $f$ at $t$.