Set $\phi (x) = u(x)+iv(x)$, $x=(x_1,\ldots,x_N)$, a $T$-periodic function in $H^1_\text{loc}(\mathbb{R}^N)$, that is $\phi (x) = \phi (x_1 + T ,\ldots, X_N + T)$ for all $x$ and where $u = \operatorname{Re} \phi$ and $v = \operatorname{Im} \phi$. I wonder if $$ \int_{[0,T]^N}\frac{u_{x_1}^2}{2} + \frac{v_{x_1}}{\sqrt{2}}(u-1)^2\ dx > 0$$ for functions satisfying $$ \int_{[0,T]^N} v_{x_1}(u-1)\ dx = \varepsilon > 0$$ for some small and fixed $\varepsilon > 0$. Any idea or help to show this would be appreciate! I am stucked. Thank you in advance.

EDIT: maybe proving $$ \int_{[0,T]^N}\frac{|\nabla u|^2}{2} + \frac{v_{x_1}}{\sqrt{2}}(u-1)^2\ dx > 0$$ under the same constraint is easier?


1 Answer 1


Take $N=1$, $T=1$, and let $u=1/2$. Let $v(x) =1/2 - x$ for $x\in \mathbb{Z}+[h,1-h]$ for some small positive number $h$. By taking $h$ small enough, you can obviously make sure

$$ \int_{[0,1]} v'(u-1) = \int_h^{1-h} (-1)(1/2-1) = \dfrac{1}{2}(1-2h) $$

is positive. However,

$$ \int_{[0,1]} v'(u-1)^2 = \int_h^{1-h}(-1)(1/2-1)^2 = -\dfrac{1}{4}(1-2h) $$

is then negative (and equal to $\sqrt{2}$ times your first integral).

You definitely seem to need additional hypotheses on $u$ and $v$ to have chance of anything like what you propose being true (perhaps you missed some?)


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