Equivalent norm in Sobolev space

Question

Let $\rho\in H^{1}(0,\pi)$ be a function, and consider the functional $$ I(\rho)=\bigg(\int_{0}^{\pi}{\sqrt{\rho^2(t)+\dot\rho^2(t)}\,dt}\bigg)^2. $$ I'm asking if it is equivalent to the norm $$ \lVert \rho \rVert_{H^1}=\lVert \rho \rVert_{L^2}+\lVert \dot\rho \rVert_{L^2} $$ on $H^{1}(0,\pi)$. Obviously $I(\rho)\leq \lVert \rho \rVert_{H^1}^2$, i'm asking if the other inequalities holds.

Answer 1

To be precise, you are asking if $\sqrt{I(\rho)}$ is equivalent to $\|\rho\|_{H^1}$. As you noted, $\sqrt{I(\rho)}$ is dominated by $\|\rho\|_{H^1}$. However, the converse fails.

Consider $\rho(x)=\sqrt{x+\epsilon}$. Since $\rho'(x) = \dfrac{1}{2\sqrt{x+\epsilon}}$, we have $\|\rho\|_{H^1}\to\infty$ as $\epsilon \to 0$.

On the other hand, $\sqrt{I(\rho)}$ stays bounded as $\epsilon\to 0$: $$\int_{0}^{\pi}{\sqrt{\rho^2(t)+\dot\rho^2(t)}\,dt} \le \int_{0}^{\pi}{\sqrt{\pi+\epsilon + \frac{1}{ 4(x+\epsilon)} }\,dt} = O(1) $$ since the singularity at $x=0$ is like $1/\sqrt{x}$.

Answer 2

The functional $\sqrt{I(\rho)}$ cannot be equivalent to the norm $\|\rho\|_{H^1}$ becuase the latter cannot be equivalent to the norm $$ \|\rho\|_{W^{1,1}}=\|\rho\|_{L^1}+\|\dot{\rho}\|_{L^1}\,. $$ More precisely, due to an obvious fact $$ \frac{1}{\sqrt{2}}(|\rho|+|\dot{\rho}|)\leqslant\sqrt{|\rho|^2+|\dot{\rho}|^2} \leqslant(|\rho|+|\dot{\rho}|) $$ holds the following inequality $$ \frac{1}{\sqrt{2}}\|\rho\|_{W^{1,1}}\leqslant\sqrt{I(\rho)} \leqslant\|\rho\|_{W^{1,1}}\,, $$ implying the equivalence of $\sqrt{I(\rho)}$ to the norm $\|\rho\|_{W^{1,1}}$ which is apparently weaker than the norm $\|\rho\|_{W^{1,2}}=\|\rho\|_{H^1}$.