Let $H$ be the Hilbert space generated by $$ \varepsilon =\left\{\sqrt{\frac{2}{T}} \, \cos\left( \frac{2\pi kt} T \right),\sqrt{\frac{2}{T}} \, \sin\left( \frac{2\pi kt} T \right) \right\} $$ using the scalar product:
$$(x,y)_1=\int_0^T(x'(t),y'(t))+(x(t),y(t)) \, dt.$$
Let $L_T^2$ be the Hilbert space generated by $\varepsilon$ using the scalar product:
$$(x,y)_2=\int_0^T(x(t),y(t))\,dt.$$
Prove that
$$\forall x\in H, \int_0^T|x(t)|^2 \, dt\le \frac{T^2}{4\pi} \int_0^T|x'(t)|^2 \, dt$$
What I did:
I used Parseval identity and an integration by parts on $x(t)$ then on $x'(t)$ to prove the inequality. But I used the fact that $x'(t)\in L^2_T$ to apply Parseval on $x'(t)$. Is this correct? Or should I prove that $x'(t)\in H$ to be able to apply the Parseval on $x'(t)$?
Is $L^2_T$ the same $L^2$ space of functions $f$, s.t.: $\int_0^T|f(t)|^2\, dt<\infty$ ?
Thank you for your help.
The space $H$ is the closure with respect to the norm induced by $(\cdot,\cdot)_1$ of the linear hull of $\epsilon$. Thus, for all elements in $H$, the induced norm is finite. This implies that $x\in L^2$ and $x'\in L^2$ for all $x\in H$.