$F_N(t)=\int_{-1/N}^0\overline{f}(t+h)dh$ implies $\|F_N\|_{L^2((0,T); X)}\leq\|f\|_{L^2((0,T); X)}$?

Question

Let $f\in L^2((0,T); X)$ where $X$ is a Banach space and $0<T<\infty$. Define $F_N: \mathbb{R}\to X$ by $$F_N(t)=\int_{-1/N}^0\overline{f}(t+h)dh$$

where $\overline{f}$ is the extension by $0$ of $f$ outside $(0,T)$. How can I prove that $$\|F_N\|_{L^2((0,T); X)}\leq\|f\|_{L^2((0,T); X)}$$

Remark: The integral above is the Bochner integral and $$\|u\|_{L^2((0,T); X)}=\left(\int_0^T\|u(t)\|_X^2\right)^{1/2}$$

Remark 1: By aaplying some simple inequalities, I was able to reach in the following inequality $$\int_0^T\|F_N(t)\|_X^2\leq N\int_0^T\int_{-1/N}^0\|\overline{f}(t+h)\|_X^2dhdt$$

Can I conclude something from it?

Thank You.

Answer 1

I think I got it, please verify if it is correct. Note that

\begin{eqnarray} \|F_N\|_{L^2((0,T),X)} &\leq& N\int_0^T\int_{-1/N}^0\|\overline{f}(t+h)\|^2_X dhdt \nonumber \\ &=& N\int_{-1/N}^0\left(\int_{0}^T\|\overline{f}(t+h)\|^2_X dt\right)dh \nonumber \\ &=& N\int_{-1/N}^0\|\overline{f}(\cdot+h)\|^2_{L^2((0,T),X)}dh \\ &\leq & \sup_{-1/N\leq h\leq 0}\|\overline{f}(\cdot+h)\|^2_{L^2((0,T),X)} \\ &\leq & \|f\|^2_{L^2((0,T),X)} \end{eqnarray}

In the last inequality, I have made a change of variables.