This is my question:
How do I should that, for $f \in C[0,\pi]$ with $f(0) = f(\pi) = 0$, the Fourier sine series $$\tilde f_n = \sum_{r=0}^n b_r \sin(r s)$$ converges uniformly to $f$ on $[0,\pi]$?
I would be able to do it given a specific $f$ (I would hope!), say by the Weierstrass M-Test, but I don't know how to in general. Also note that there is no differentiability condition on $f$.
This is part of a longer question, of which the first part is the same but for the cosine (without $f(\pi) = 0$) - the section is about polynomial approximation, and it asks me to use Chebyshev polynomials (of the first kind) and the Weierstrass approximation theorem. I just don't know how to adapt this to the sine case. Could I perhaps write $$F(x) = \int_0^x f(y) dy,$$ since $f$ is continuous and so integrable, and thus $F$ is continuously differentiable. I could then get a Fourier cosine series approximation for $F$, then differentiate term-by-term (as uniform convergence allows) to give the required answer. Would this be legitimate? (The issue is trying to differentiate through limits/inequalities...)
I have since gone through my above suggestion, and I believe it does work. My question is now the following:
Is there a more direct method of doing this?
That is, it seems a slightly unusual way to go about answering the question, so is there a direct proof?
Thanks in advance for any help!
(PS - Sorry if this is similar to a duplicate question, although it's certainly not an exact duplicate!)