Do we have such kind of estimates? Let $0<a_0\leq a(x)$ be a smooth function on $\mathbb{T=[0,2\pi]}$, and $a(0)=a(2\pi)$, then whether it holds that
$$
\int_{\mathbb{T}}a(x)|\partial_x\phi|^2 dx\geq \int_{\mathbb{T}}|\partial_xa|^2|\phi|^2 dx
$$
for all $\phi\in H_{per}^1(\mathbb{T})$ ?   More precisely,
$$
\phi(0)=\phi(2\pi)=0,\quad \int_{\mathbb{T}}\phi dx=0
$$
Thanks.

Controlling $|\phi|$ by $|\phi'|$ sounds reasonable. But we can't control $|a'|$ by $|a|$. Example: let $a(x) = 2+\sin nx$, where $n$  is large. As $n\to \infty$, 
$$\int_{\mathbb{T}}a(x)|\partial_x\phi|^2 dx$$  stays bounded but $$\int_{\mathbb{T}}|\partial_xa|^2|\phi|^2 dx$$ blows up (unless $\phi\equiv 0$).

You may want to consider 
$$\int_{\mathbb{T}}a(x)|\partial_x\phi|^2 dx\geq  C\int_{\mathbb{T}}|\partial_xa|^2|\phi|^2 dx$$ with $C$ depending on $a$.
