I'm trying to show that \begin{align} \left|\int_{-\pi}^\pi [f(t)\cos t - f'(t) \sin t]\, dt\right| \le \sqrt{2\pi}\left(\int_{-\pi}^\pi [|f(t)|^2 + |f'(t)|^2]\, dt\right)^{1/2}. \end{align}

However I do not see how to do this. If I use the triangle inequality I get \begin{align} \left|\int_{-\pi}^\pi [f(t)\cos t - f'(t) \sin t]\, dt\right| = \left| \int_{-\pi}^\pi f(t)\cos t\, dt - \int_{-\pi}^\pi f'(t)\sin t\, dt\right| \le \left| \int_{-\pi}^\pi f(t)\cos t\, dt\right| + \left|\int_{-\pi}^\pi f'(t)\sin t\, dt\right|. \end{align}

If I then use Schwarz inequality I get that $$ \left| \int_{-\pi}^\pi f(t)\cos t\, dt\right| \le \sqrt{\pi} \left(\int_{-\pi}^\pi |f(t)|^2dt\right)^{1/2}, $$ so that \begin{align} \left|\int_{-\pi}^\pi [f(t)\cos t - f'(t) \sin t]\, dt\right| \le \sqrt{\pi} \left(\int_{-\pi}^\pi |f(t)|^2dt\right)^{1/2} + \sqrt{\pi} \left(\int_{-\pi}^\pi |f'(t)|^2dt\right)^{1/2}. \end{align}

I feel close to the expected result, but don't know how to improve. Any help will be appreciated. Cheers

  • 1
    $\begingroup$ It definitely isn't as strong as your bound, but you have $$|f(t)\cos(t)-f'(t)\sin(t)|\leq2\max\{|f(t)|,|f'(t)|\}.$$ Applying Cauchy-Schwarz and the fact that $(\max\{x,y\})^2=\max\{x^2,y^2\}\leq x^2+y^2$ for $x,y\geq0$ gives $$2\sqrt{2\pi}\left(\int_{-\pi}^\pi f(t)^2+f'(t)^2\,dt\right)^{1/2}.$$ $\endgroup$ – Clayton May 2 '17 at 0:09
  • $\begingroup$ @Clayton Thanks, I'm still working on it. If I make any progress I shall let you know $\endgroup$ – Vladimir Vargas May 2 '17 at 0:12

By Cauchy's inequality $$\vert f(t)\cos t-f'(t)\sin t \vert =\vert (f(t),f'(t))\cdot (\cos t,-\sin t) \vert \le \sqrt{(f(t))^2+(f'(t))^2} \sqrt{(\cos t)^2+(\sin t)^2} =\sqrt{(f(t))^2+(f'(t))^2}.$$ Then apply Holder's

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