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Let F : [ 0 , π ] → R {\displaystyle F:[0,\pi ]\to \mathbb {R} } be a smooth function such that F ( 0 ) = F ( π ) = 0 {\displaystyle F(0)=F(\pi )=0} . Then:
∫ 0 π ( d F d t ) 2 d t ≥ ∫ 0 π F ( t ) 2 d t {\displaystyle \int _{0}^{\pi }\left({\frac {dF}{dt}}\right)^{2}dt\geq \int _{0}^{\pi }F(t)^{2}dt}
![{\displaystyle F:[0,\pi ]\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/1eba3386d846b846ec55ce108c359496b5ec61a2)
be a smooth function such that 
. Then:
