A lemma in the proof of Fourier's Main Theorem.

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?

• 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.
– them
May 2 '19 at 2:36
• @them you are correct! I just modified the post. May 2 '19 at 2:39
• 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.
– them
May 2 '19 at 2:42
• @them True. It would evaluate to $a_j(f)\pi$. Fixed it now. May 2 '19 at 2:46
• Now use the definition of $a_j(f)$ (and it still seems like the $\pi$ should not be there... but I didn't check)
– 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.