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- 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.

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.