Equivalent norm in Sobolev space 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.
 A: 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 $1/\sqrt{x}$. 
A: 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 $\|\rho\|_{W^{1,2}}=\|\rho\|_{H^1}$.
