• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Uniform convergence of a sine series

1,682
50
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.
 

Want to reply to this thread?

"Uniform convergence of a sine series" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top