# Prove norm of continuously differentiable function - repost from "urgent"

• Nov 18th 2008, 03:30 PM
Cere Kong
Prove norm of continuously differentiable function - repost from "urgent"
Hello everyone, I might be breaking all rules by reposting my question here as well but the "urgent homework" forum did not quite seem to be the right place.

Can someone prove that for any continuously differentable function http://www.mathhelpforum.com/math-he...1929cce7-1.gif on http://www.mathhelpforum.com/math-he...1891d927-1.gif:
http://www.mathhelpforum.com/math-he...a940b1ca-1.gif

Yes I know I am supposed to do it and I assume Cauchy-Schwarz is somehow involved but I cannot make out the details (Crying)

Best regards and thanks,
Thomas
• Nov 19th 2008, 12:08 AM
Opalg
Cauchy–Schwarz in $\displaystyle \mathbb{R}^2$ tells you that

\displaystyle \begin{aligned}|f(t)\cos t-f'(t)\sin t| &= |(f(t),f'(t))\mathbf{\cdot}(\cos t,-\sin t)|\\ &\leqslant \left(|f(t)|^2+|f'(t)|^2\right)^{1/2}\left(\cos^2t+\sin^2t\right)^{1/2}. \end{aligned}