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Please suggest something as I am completely stumped here.

$F:[0,\pi] \to \mathbb R$ is smooth function with $F(0)=0=F(\pi)$.We need to show that: $$ \int_0^\pi {(F'(t))}^2dt \ge \int_0^\pi {(F(t))}^2 dt$$

Thanks,any hint also would be appreciated.

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2 Answers 2

up vote 1 down vote accepted

Perhaps see Wirtinger's inequality.

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Link given by sourisse was quite enough for me but I found an elementary proof for this question so I wanted to share.

Put $G(t)=\frac {F(t)}{\sin t}$,then,we have $$\int_0^\pi {(F'(t))}^2 dt = \int_0^\pi {(G'(t)\sin t +G(t)\cos t )}^2dt$$ $$ = \int_0^\pi {(G'(t))}^2 \sin^2t dt + \int_0^\pi2G'(t)G(t)\sin t \cos t dt +\int_0^\pi{(G(t))}^2\cos^2 t dt$$

Now,$$\int_0^\pi 2G'G\sin t \cos t dt= {[G^2\sin t \cos t]}_0^\pi - \int_0^\pi G^2(\cos^2 t - \sin^2 t)dt $$ $$=\int_0^\pi G^2(\sin^2 t-\cos^2 t)dt$$

So,$$\int_0^\pi(F')^2dt=\int_0^\pi {(G')}^2 \sin^2t dt +\int_0^\pi G^2(\sin^2 t-\cos^2 t)+\int_0^\pi G^2\cos^2 t dt$$ $$=\int_0^\pi({(G')}^2+G^2)\sin^2 tdt=\int_0^\pi F^2dt+\int_0^\pi{(G')}^2\sin^2tdt$$ Hence,it follows that, $$\int_0^\pi{(F'(t))}^2dt - \int_0^\pi {(F(t))}^2dt \ge 0$$

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