Assume $f,g:[-\pi,\pi]\to\mathbb{R}$ are continuous functions. We define the $L^2$ norm of $f$, and the scalar product between $f$ and $g$ as $$\|f\|_{L^2}=\sqrt{\int_{-\pi}^{\pi}|f(x)|^2\,dx}$$ and $$\langle f,g \rangle=\int_{-\pi}^{\pi}f(x)g(x)\,dx.$$

Consider $u:[-\pi, \pi]\to\mathbb{R}$ piecewise continuous with finitely many singularities, and its corresponding partial Fourier series for $N\in\mathbb{N}$ $$S_{N}f(x)=\frac{a_0(f)}{2}+\sum_{k=1}^{N}a_k(f)\cos(kx)+b_k\sin(kx)$$ where the $a_k$'s and $b_k$'s are the Fourier coefficients of $f.$

I'm attempting to show that for any $j\in[0,N]$ one has $$\langle f-S_Nf, \cos(jx) \rangle=0$$

Here is my attempt:

Let $j\in[0,N].$ Observe that \begin{align*} \langle f-S_Nf, \cos(jx) \rangle &= \int_{-\pi}^{\pi} (f-S_Nf)(\cos jx)\,dx \\ &=\underbrace{\int_{-\pi}^{\pi}f\cos jx \,dx}_{:=A} - \underbrace{\int_{-\pi}^{\pi}S_Nf\cos jx\,dx}_{:=B} \end{align*}

We first deal with the non-trivial case, where $j\neq0.$ Notice that \begin{align*} B&=\int_{-\pi}^{\pi} \cos jx \left[ \frac{a_0(f)}{2}+\sum_{k=1}^{N} a_k(f) \cos kx + b_k (f) \sin kx \right]\,dx \\ &=\underbrace{\int_{-\pi}^{\pi} \frac{a_0(f)}{2}\cos jx \,dx}_{=0} +\underbrace{\int_{-\pi}^{\pi}\sum_{k=1}^{N}a_k(f)\cos kx \cos jx \,dx}_{=a_j(f)\pi} + \underbrace{\int_{-\pi}^{\pi}\sum_{k=1}^{N}b_k(f)\sin kx \cos jx\,dx}_{=0}.\\ &=a_j(f)\pi. \end{align*}

At this point, I'm stuck. I would like $A$ to be equal $a_j(f)\pi,$ as that will prove what I want to show. However, I don't see very much I can do with the quantity $A.$

Any thoughts?

  • $\begingroup$ For one thing, shouldn't the last term in $B$, where you have integral of sum of $cos(jx)sin(kx)$ be equal to zero? Since you are integrating unti-symmetric function on symmetric domain. $\endgroup$
    – them
    May 2 '19 at 2:36
  • $\begingroup$ @them you are correct! I just modified the post. $\endgroup$ May 2 '19 at 2:39
  • $\begingroup$ In the other term in $B$, the integral with $cos(kx)cos(jx)$, where did the corresponding $a_k(f)$ disappear, how did you get it to be equal to $\pi$? This seems incorrect. $\endgroup$
    – them
    May 2 '19 at 2:42
  • $\begingroup$ @them True. It would evaluate to $a_j(f)\pi$. Fixed it now. $\endgroup$ May 2 '19 at 2:46
  • $\begingroup$ Now use the definition of $a_j(f)$ (and it still seems like the $\pi$ should not be there... but I didn't check) $\endgroup$
    – them
    May 2 '19 at 2:53

By definition, $$a_j(f)\pi=\pi\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(jx)\,dx,$$ implying that $A-B=0,$ as desired.


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