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a) $f(x)>0$ and $f(x)\in C[a,b]$

Prove $$\left(\int_a^bf(x)\sin x\,dx\right)^2 +\left(\int_a^bf(x)\cos x\,dx\right)^2 \le \left(\int_a^bf(x)\,dx\right)^2$$

I have tried Cauchy-Schwarz inequality but failed to prove.

b) $f(x)$ is differentiable in $[0,1]$

Prove $$|f(0)|\le \int_0^1|f(x)|\,dx+\int_0^1|f'(x)|dx$$

Any Helps or Tips,Thanks

Shuhao Cao
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Xiaolang
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  • a) One thing that came to mind was use $(a+b)^2 = a^2+b^2+2ab$ and argue by cases. One case (when $ab>0$) gives the result very quickly. Just an idea to start with. My guess is when $ab<0$ either there is a contradiction. – toypajme Jun 26 '13 at 21:33
  • @toypajme I use this method :$(\int^a_bf(x)\sin xdx)^2 +(\int^a_bf(x)\cos xdx)^2$=$(\int^a_b \sqrt{f(x)} \sqrt{f(x)} \sin xdx)^2 +(\int^a_b \sqrt{f(x)} \sqrt{f(x)} \cos xdx)^2$ and use the Schwartz inequality – Xiaolang Jun 26 '13 at 21:39
  • I see, that way gives you an annoying $\sqrt 2$ – toypajme Jun 26 '13 at 21:44
  • Can you give me some idea about (b)? – Xiaolang Jun 26 '13 at 21:45
  • and one more question if $f(x)$ is second-order differentiable and $f(0)=f(1)=f'(1)=0$ $f'(1)=1$ Prove $\int^0_1 (f''(x))^2dx) \ge 4$ I can prove $\int^0_1 (f''(x))^2dx) \ge 1$ – Xiaolang Jun 26 '13 at 21:50
  • You should ask that in a separate post @Xiaolang – Pedro Jun 27 '13 at 00:35

2 Answers2

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Hint: For part a), use Jensen's inequality with weighted measure $f(x)\,\mathrm{d}x$. Since $f(x)>0$, Jensen says that for a convex function $\phi$ $$ \phi\left(\frac1{\int_Xf(x)\mathrm{d}x}\int_Xg(x)\,f(x)\mathrm{d}x\right) \le\frac1{\int_Xf(x)\mathrm{d}x}\int_X\phi(g(x))\,f(x)\mathrm{d}x $$

Hint: For part b), note that for $x\in[0,1]$, $$ f(0)-f(x)\le\int_0^1|f'(t)|\,\mathrm{d}t $$ and integrate over $[0,1]$.

robjohn
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I think for case a) double integral should help you.

$$ \left(\int_a^bf(x)\sin x\,dx\right)^2 +\left(\int_a^bf(x)\cos x\,dx\right)^2 = \int_a^b\int_a^bf(x)f(y)sinxsinydxdy+\int_a^b\int_a^bf(x)f(y)cosxcosydxdy =\int_a^b\int_a^bf(x)f(y)(sinxsiny+cosxcosy)dxdy=\int_a^b\int_a^bf(x)f(y)cos(x+y)dxdy\le \int_a^b\int_a^bf(x)f(y)dxdy=\left(\int_a^bf(x)\,dx\right)^2 $$

pointer
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