when dealing with the linear heat equation \begin{align*} \partial_t u &= \partial_{xx} u, \quad 0<x<l, 0<t<T\\ \partial_x u(0,t) &= \partial_x u(l,t) = 0, \quad 0<t<T\\ u(x,0) &= g(x), \quad 0<x<l \end{align*}
one can define the corresponding solution operator $T:L^2(0,l) \rightarrow L^2(0,T;H^1(0,l)), g(x) \mapsto u(x,t)$. How can I show that this operator is bounded, i.e. \begin{align*} \int_0^T \lVert u\rVert_{H^1(0,l)} dt \leq C \int_0^l \lvert g(x) \rvert^2 dx \end{align*} I tried to reformulate the left side \begin{align*} \int_0^T \lVert u\rVert_{H^1(0,l)} dt = \int_0^T \lVert u \rVert_{L^2(0,l)}+ \lVert \partial_x u \rVert_{L^2(0,l)} dt \end{align*} but I am stuck immediately. Can somebody give me a hint?
It holds $$ \partial_t \int_0^l |u|^2 = \int_0^l 2\,u \,\partial_tu = 2 \int_0^l u \,\partial_{xx}u $$ hence, integrating by parts and using the fact that $\partial_x u = 0$ at $x=0$ and $x=l$, $$\tag{1}\label{1} \partial_t \int_0^l |u|^2 = -2 \int_0^l |\partial_{x}u|^2. $$ In particular, $\|u\|_{L^2}$ is decreasing in time so $$\tag{2}\label{2} \int_0^T \|u(t,\cdot)\|_{L^2}^2\,\mathrm d t \leq \int_0^T \|g\|_{L^2}^2\,\mathrm d t \leq T\,\|g\|_{L^2}^2. $$ Taking the integral in time of Inequality \eqref{1} yields $$ \int_0^l |u(T,x)|^2\,\mathrm d x - \int_0^l |u(0,x)|^2\,\mathrm d x = -2 \int_0^T\!\!\!\int_0^l |\partial_{x}u(t,x)|^2\mathrm d x\,\mathrm d t, $$ which implies $$ 2 \int_0^T \|\partial_{x}u(t,\cdot)\|_{L^2(0,l)}^2\,\mathrm d t \leq \|g\|_{L^2(0,l)}^2. $$ Combining this inequality with Equation \eqref{2}, leads to $$ \|u(t,\cdot)\|_{L^2((0,T),H^1(0,l))}^2 \leq \left(T + \tfrac{1}{2}\right) \|g\|_{L^2(0,l)}^2. $$