Let $0 < a_0 \leq a(x)$ be a smooth function on $\mathbb{T}=[0,2\pi]$, $a(0)=a(2\pi)$, whether it holds that $$ \int_\mathbb{T} a(x)\partial_x \varphi^2 \geq \int_\mathbb{T}\frac{\partial^2_x a }{2}\cdot \varphi^2dx $$ for all $\varphi \in \dot{H}^1_{per}(\mathbb{T})$? More precisely, $$ \varphi(0)=\varphi(2\pi), \quad \int_\mathbb{T} \varphi dx =0. $$ Prove it or give a counterexample please.

1$\begingroup$ This reads like homework. Voted to close. $\endgroup$– José FigueroaO'FarrillCommented May 24, 2014 at 18:56

2$\begingroup$ counterexample: $\phi(x)=\cos(x)$ $a(x)= 2+\sin(Nx)$ gives $$ 2\pi > N^2\frac{\pi}{2}.$$ $\endgroup$– usernameCommented May 24, 2014 at 22:16

$\begingroup$ Thanks for your answer. For your example, the right hand side is 0, which does not give a contradiction. $\endgroup$– Wang MingCommented May 26, 2014 at 0:14

$\begingroup$ It's a question of notations. If $\partial_x^2 a =(\partial_x a)^2$ my counter example works. If $\partial_x^2 a = \partial_{xx} a$ the answer below gives the counter example. $\endgroup$– usernameCommented May 26, 2014 at 20:53
1 Answer
The inequality does not hold.
Take $a,\varphi$ smooth, periodic, then integrations by parts yield $$ 0 = \int_\mathbb{T}(a\varphi^2)_{xx} = \int_\mathbb{T}(a_{xx}\varphi^2+a(\varphi^2)_{xx}+2a_x(\varphi^2)_x) = 2\int_\mathbb{T}a_{xx}\varphi  2\int_\mathbb{T}a(\varphi^2)_{xx} = 2\Bigl(\int_\mathbb{T}a_{xx}\varphi^2\int_\mathbb{T}a\varphi_x^2\Bigr)  2\int_\mathbb{T}a(\varphi_x^2+2\varphi\varphi_{xx}) $$ that is $$ \int_\mathbb{T}a\varphi_x^2  \int_\mathbb{T}a_{xx}\varphi^2 = \int_\mathbb{T}a(\varphi_x^2+2\varphi\varphi_{xx}) $$ and there is no reason the righthand side should be positive. Take for instance $\varphi=\cos x$ and $a=1+b$ with $b$ nonzero (and big) only where $\varphi_x^2+2\varphi\varphi_{xx}=\sin^2x2\cos^2x$ is positive.
[update] The inequality does not hold even in its general version. Say you want to prove that $\int a (\varphi_x)^2\geq\lambda\int a_{xx}\varphi^2$, then by integration by parts $\int a_{xx}\varphi^2=\int a(\varphi^2)_{xx}$ and the supposed inequality reads $\int a((\varphi_x)^2\lambda(\varphi^2)_{xx})\geq0$. By localization (with a smooth cutoff function), it is sufficient to find a $\varphi$ (not necessarily periodic, by the same localization argument) such that $(\varphi_x)^2\lambda(\varphi^2)_{xx}<0$ on an open set. This is easy, take $\varphi(x)=e^{a e^{bx}}$ and choose $a,b$ suitably.

$\begingroup$ Thanks for your nice answer. For your post, I make some correctons on previous problem. See above. Your example is not sufficient now. The question arised in the study whether the operator $a\partial_x^2$ is positive (under the above assumptions). But now it seems little is known. $\endgroup$ Commented May 26, 2014 at 0:20