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. $\endgroup$ Commented May 24, 2014 at 18:56
  • 2
    $\begingroup$ counterexample: $\phi(x)=\cos(x)$ $a(x)= 2+\sin(Nx)$ gives $$ 2\pi > N^2\frac{\pi}{2}.$$ $\endgroup$
    – username
    Commented 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 Ming
    Commented 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. $\endgroup$
    – username
    Commented May 26, 2014 at 20:53

1 Answer 1


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 right-hand side should be positive. Take for instance $\varphi=\cos x$ and $a=1+b$ with $b$ non-zero (and big) only where $\varphi_x^2+2\varphi\varphi_{xx}=\sin^2x-2\cos^2x$ 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 cut-off 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. $\endgroup$
    – Wang Ming
    Commented May 26, 2014 at 0:20

Not the answer you're looking for? Browse other questions tagged or ask your own question.