I want to show that $$\frac{1}{\pi} \int_{-\pi}^{\pi} f(x)g(x)dx = \frac{a_0\alpha_0}{2} + \sum_{n=1}^{\infty} (a_n\alpha_n + b_n\beta_n)$$
where $f,g: [-\pi,\pi] \to \mathbb{R}$ are integral functions and that the Fourier series of $f$ and $g$ are uniformly convergent to $f,g$ respectively.
$a_0,a_n,b_n,\alpha_0,\alpha_n,\beta_n$ are the fourier coefficients of $f,g$ respectively.
I thought we could simply expand the LHS, simplify, and then show equality. I start by multiplying $f(x)g(x)$ via their respective Fourier series.
$$a_0\alpha_0 + a_0(\sum_{n=1}^{\infty} \alpha_n\cos(nx) + \beta_n\sin(nx))+\alpha_0(\sum_{n=1}^{\infty} a_n\cos(nx) + b_n\sin(nx)) + (\sum_{n=1}^{\infty} \alpha_n\cos(nx) + \beta_n\sin(nx))(\sum_{n=1}^{\infty} a_n\cos(nx) + b_n\sin(nx))$$
How do I simplify this? Because we have uniform convergence, we can switch the places of the Sigma and Integral symbols yeah? Even then though..
Best Answer
You know that
$$f(x) = \frac{a_0}{2} + \sum_{n = 1}^\infty (a_n \cos(nx) + b_n\sin(nx)), \quad x \in [-\pi,\pi].$$
So
$$\int_{-\pi}^\pi f(x)g(x)\, dx = \frac{a_0}{2}\int_{-\pi}^\pi g(x)\, dx + \sum_{n = 1}^\infty \left(a_n \int_{-\pi}^\pi \cos(nx)g(x)\, dx + b_n \int_{-\pi}^\pi \sin(nx)g(x)\, dx\right).$$
This step is justified by the uniform convergence assumptions given in the problem. By definition of the Fourier coefficients $\alpha_n$ and $\beta_n$, we have
$$\pi \alpha_0 = \int_{-\pi}^\pi g(x)\, dx,$$
$$\pi \alpha_n = \int_{-\pi}^\pi \cos(nx)g(x)\, dx,\quad n \ge 1$$
and
$$\pi \beta_n = \int_{-\pi}^\pi \sin(nx)g(x)\, dx,\quad n \ge 1.$$
Therefore
$$\int_{-\pi}^\pi f(x)g(x)\, dx = \frac{\pi a_0\alpha_0}{2} + \sum_{n = 1}^\infty (\pi a_n\alpha_n + \pi b_n \beta_n),$$
or
$$\int_{-\pi}^\pi f(x)g(x)\, dx = \frac{a_0 \alpha_0}{2} + \sum_{n = 1}^\infty (a_n \alpha_n + b_n \beta_n).$$