# Uniform convergence of a sine series

#### joshmccraney

Problem Statement
Given a smooth function $f\in C^1[0,\pi], f(0)=f(\pi)=0$, prove that its Fourier series with respect to the basis $\sin(k x)$ uniformly converges to $f$.
Relevant Equations
Hint: extend $f$ to an odd function on $[−\pi,\pi]$. Use the fact that the standard Fourier series of a smooth $2\pi$ periodic function $g$ converges to $g$.
I'm not too sure how to use the hint here. What I had so far was this: an odd extension of $f$ implies $f = \sum_{k=1}^\infty b_k \sin(k x)$. Notice for $m>n$ $$\left|\sum_{k=1}^m b_k\sin(k x) - \sum_{k=1}^n b_k\sin(k x)\right| = \left| \sum_{k=n+1}^m b_k\sin(k x)\right| \leq \sum_{k=n+1}^m |b_k\sin(k x)| \leq \sum_{k=n+1}^m |b_k|.$$

Since $\sum_{k=1}^\infty |b_k| < \infty$ (isn't this true of all Fourier sine coefficients if we let $x \to \pi/2$?) we can make the right hand side of the above as small as we wish provided $n$ is sufficiently large. Thus, we can make the left hand side as small as we wish, independently of $x$, as long as $n$ is sufficiently large.

It follows that $\sum_{k=1}^\infty b_k \sin(k x)$ is uniformly Cauchy, and, thus, uniformly convergent.

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