Let $f,g$ are the functions of $S^{1}$. Are there $f, g$ such that $$\int_{S^{1}} |f'|^{2}+|g'|^{2}d\theta-2\int_{S^{1}}f'g<0,$$ where $f'=\frac{\partial f}{\partial \theta}$?
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$\begingroup$ $f(\theta)=\theta$ is not well defined on $S^{1}.$ $\endgroup$– lidingCommented Jul 31, 2020 at 7:24
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$\begingroup$ OK, comment deleted. $\endgroup$– abxCommented Jul 31, 2020 at 9:21
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1 Answer
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No, there are no such functions. Indeed, if $h$ is periodic and has mean zero, then $\|h\|_2 \le \|h'\|_2$. Take $f,g$ as above and write $g=g_1+c$ with $g_1$ having mean zero. Then $$ \|f'\|_2^2+\|g'\|_2^2<2\int_{S^1}f'g=2 \int_{S^1} f'g_1 \le 2\|f'\|_2\|g_1'\|_2=2\|f'\|_2\|g'\|_2 \le \|f'\|_2^2+\|g'\|_2^2. $$
answered Jul 31, 2020 at 9:43
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$\begingroup$ Thank you! For general constant $C$, are there functions $f,g$ such that $$\int_{S^{1}} |f'|^{2}+|g'|^{2}d\theta-C\int_{S^{1}}f'g<0$$? $\endgroup$– lidingCommented Jul 31, 2020 at 17:03
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$\begingroup$ Of course, yes. Take $f(t)=-\cos t, g(t)=\sin t$, then you get equality with $C=2$. $\endgroup$ Commented Jul 31, 2020 at 18:20