Exercise :
Assume that $T$ satisfies the equation $T_t(t,z) = aT_{zz}(t,z)$ for $t>0, z \in (0,1)$ and $a > 0$ a constant. Moreover, suppose that $T(0,z) = T_0(z)$ for $z \in [0,1]$, where $T_0 : [0,1] \to \mathbb R$ and that $T(t,0) = T(t,1) = 0$.
Show that : $$\int_0^1 \left(\frac{\partial T}{\partial z}(t,z)\right)^2\mathrm{d}z \geq 2 \int_0^1 T^2(t,z)\mathrm{d}z$$
Attempt-thoughts :
Since we have the boundary conditions $T(t,0) = T(t,1) = 0$ for the problem and our integrations are over the interval $[0,1]$, we can "see" the given inequality as a norm-2 inequality and use the Wirtinger inequality which gives us a weaker lower bound than the one desired though, as it would be : $$\text{Wirtinger :} \;\|T_z\|_2^2 \geq \pi^2\|T\|_2^2 \Leftrightarrow \int_0^1 \left(\frac{\partial T}{\partial z}(t,z)\right)^2\mathrm{d}z \geq \pi^2 \int_0^1 T^2(t,z)\mathrm{d}z$$
So, that intuition falls short.
Important : Another thought, since its often carried out in such cases (and also a hint given by our professor) is that the Cauchy-Schwarz inequality shall be used. I cannot see how though.
Finally, I have previously proven via substitution from the PDE and integration by parts, that :
$$\frac{\mathrm{d}}{\mathrm{d}t} \int_0^1 T^2(t,z)\mathrm{d}z = -2a\int_0^1 \left(\frac{\partial T}{\partial z}(t,z)\right)^2\mathrm{d}z $$
I don't know if that can be of any help.
Any hints or elaborations will be greatly appreciated.
Would this work for $f=T(t,\cdot)$ or am I completely mistaken?
Let $f: [0,1] \rightarrow \mathbb{R}$ be any $\mathscr{C}^1$ function st $f(0)=f(1)=0$. For $0 \leq x \leq 0.5$, $f^2(x)=\left(\int_0^x{f’}\right)^2 \leq x\int_0^{1/2}{f’^2}$ by Cauchy-Schwarz.
Similarly, if $0.5 \leq x \leq 1$, $f^2(x) \leq (1-x)\int_{0.5}^1{f’^2}$.
Thus $\int_0^1{f^2} \leq \int_0^1{f’^2}\int_0^{1/2}{xdx}=\frac{1}{8}\int_0^1{f’^2} \leq \frac{1}{2}\int_0^1{f’^2}$.