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This is my first post here.

I have some troubles with this property of the Fourier coefficients. Indeed, let $f(x)$ be a continuous real function, with compact support $[a,b] \subset (0,2\pi)$, and $$ b_k := \frac{1}{\pi}\int_0^{2\pi}f(x)\sin kx \; \mathrm{d}x $$

My question is: does the series $$ \sum_{k=1}^\infty \frac{b_k}{k}$$ converge? If the answer is yes, is its sum equal to $\int_0^{2\pi}\frac{\pi-x}{2}f(x) \; \mathrm{d}x$?

How can I prove/disprove this? What do you suggest? Thanks a lot.

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As a hint on the first part, these Fourier coefficients are square summable (why?) and $\frac{1}{k}$ is square summable. Apply Cauchy Schwartz. –  Chris Janjigian Apr 10 '12 at 12:29
    
Hint on the second part: This is a convolution, and convolutions in real space correspond to multiplications in Fourier space. –  joriki Apr 10 '12 at 12:31
    
@Chris It's confusing that there is Schwarz and Schwartz. But on second thought it's not actually that confusing: one, Herr Hermann Schwarz, was German and lived around 1900, the other, Monsieur Laurent Schwartz, was French (hence probably the misspelling of "schwarz" which means "black" in German) lived around 1950. Latter doing a lot of cool stuff with distributions hence the name Schwartz space. –  Matt N. Apr 10 '12 at 14:13
    
Maybe you can memorise it by remembering that the one doing more elaborate stuff has one more letter in his name than the other. : ) –  Matt N. Apr 10 '12 at 14:14
    
@MattN. In hindsight I think the Russians don't get enough credit. Bunyakovsky it is :) Thanks for the heads up though. –  Chris Janjigian Apr 10 '12 at 22:25
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1 Answer

up vote 2 down vote accepted

For arbitrary continuous function $\varphi\in C([0,2\pi])$ denote $$ a_0(\varphi)=\frac{1}{\pi}\int\limits_{0}^{2\pi}\varphi(x)dx $$ $$ a_k(\varphi)=\int\limits_{0}^{2\pi}\varphi(x)\cos(kx)dx\qquad b_k(\varphi)=\int\limits_{0}^{2\pi}\varphi(x)\sin(kx)dx $$ Consider function $$ F(x)=\int\limits_{0}^x f(t)dt-\frac{a_0(f)x}{2} $$ Since $f\in C([0,2\pi])$ then $F\in C^1([0,2\pi])$. Note that $$ F(x+2\pi)-F(x)=\int\limits_{0}^{2\pi}f(t)dt-a_0(f)\pi=0. $$ So $F$ is $2\pi$ periodic. Since $F$ is $2\pi$ periodic continuously differentiable function its Fourier series uniformly converges to $F$. Thus we have for all $x\in[0,2\pi]$ the following equality $$ F(x)=\frac{a_0(F)}{2}+\sum\limits_{k=1}^\infty\left(a_k(F)\cos(kx)+b_k(F)\sin(kx)\right) $$ Integration by parts gives us $$ a_0(F)=\frac{1}{\pi}\int\limits_{0}^{2\pi}F(x)dx= \frac{1}{\pi}\left(xF(x)|_0^{2\pi}-\int\limits_{0}^{2\pi}x\left(f(x)-\frac{a_0(f)}{2}\right)dx\right)= \frac{1}{\pi}\left(-\int\limits_{0}^{2\pi}xf(x)dx+\int\limits_{0}^{2\pi}\pi f(x)dx\right)= \int\limits_{0}^{2\pi}\frac{\pi-x}{\pi}f(x)dx $$ $$ a_k(F)=\int\limits_{0}^{2\pi}F(x)\cos(kx)dx= \left(F(x)\frac{\sin(kx)}{k}\right)_0^{2\pi}- \int\limits_{0}^{2\pi}\left(f(x)-\frac{a_0(f)}{2}\right)\frac{\sin(kx)}{k}dx= $$ $$ -\frac{1}{k}\int\limits_{0}^{2\pi}f(x)\sin(kx)dx=-\frac{1}{k}b_k(f) $$ $$ b_k(F)=\int\limits_{0}^{2\pi}F(x)\sin(kx)dx= \left(F(x)\frac{-\cos(kx)}{k}\right)_0^{2\pi}- \int\limits_{0}^{2\pi}\left(f(x)-\frac{a_0(f)}{2}\right)\frac{-\cos(kx)}{k}dx= $$ $$ \frac{1}{k}\int\limits_{0}^{2\pi}f(x)\cos(kx)dx=\frac{1}{k}a_k(f) $$ Hence $$ F(x)=\frac{a_0(F)}{2}+ \sum\limits_{k=1}^\infty\left(-\frac{b_k(f)}{k}\cos(kx)+\frac{a_k(f)}{k}\sin(kx)\right) $$ Substitute $x=0$, then $$ 0=\frac{a_0(F)}{2}-\sum\limits_{k=1}^\infty\frac{b_k(f)}{k} $$ i.e. $$ \sum\limits_{k=1}^\infty\frac{b_k(f)}{k}= \frac{a_0(F)}{2}= \int\limits_{0}^{2\pi}\frac{\pi-x}{2\pi}f(x)dx $$

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Great! Very clear proof. Thank you very much for your kind reply. –  Romeo Apr 10 '12 at 18:53
    
Not at all, Romeo :) –  Norbert Apr 10 '12 at 19:25
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