Uniform convergence of a sine series

• member 428835
In summary, the conversation discussed the use of the Weierstrass M-test to prove the uniform convergence of a series of functions. The individual had made progress by showing that the series in question was uniformly Cauchy, but still needed to show that it was uniformly bounded. This can be done by using the fact that all Fourier sine coefficients have a finite sum, which implies that the series is uniformly bounded. This allows the Weierstrass M-test to be applied, ultimately leading to the conclusion that the series is uniformly convergent.
member 428835
Homework 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.

Delta2

Hello,

Thank you for sharing your progress so far. It seems like you are on the right track. The hint given was to use the Weierstrass M-test, which is a useful tool for proving the uniform convergence of a series of functions.

In your work, you have shown that the series ##\sum_{k=1}^\infty b_k \sin(k x)## is uniformly Cauchy, which means that it is a Cauchy sequence in the uniform norm. This is a good start, but you still need to show that the series is uniformly bounded in order to apply the M-test.

To do this, you can use the fact that ##\sum_{k=1}^\infty |b_k| < \infty##, which is true for all Fourier sine coefficients. This implies that the series is uniformly bounded by ##\sum_{k=1}^\infty |b_k|##, which is a finite number. Thus, you can conclude that the series is uniformly bounded, and therefore, the Weierstrass M-test can be applied.

I hope this helps. Keep up the good work!

1. What is uniform convergence of a sine series?

Uniform convergence of a sine series refers to a mathematical concept in which a series of sine functions, when summed together, approaches a specific function uniformly across its entire domain. This means that the difference between the sum of the series and the actual function becomes smaller and smaller as the number of terms in the series increases.

2. How is uniform convergence of a sine series different from pointwise convergence?

Pointwise convergence of a sine series only requires that the difference between the sum of the series and the actual function becomes smaller at each individual point in the domain. In contrast, uniform convergence requires that the difference becomes smaller across the entire domain, not just at individual points.

3. What is the significance of uniform convergence of a sine series?

Uniform convergence is significant because it allows for the approximation of complicated functions using simpler sine functions. This makes it a useful tool in mathematical analysis and can help in solving many real-world problems.

4. How is the uniform convergence of a sine series tested?

The uniform convergence of a sine series can be tested using various mathematical techniques, such as the Weierstrass M-test, the Cauchy criterion, and the Dini's test. These tests check for the existence of a uniform bound on the difference between the sum of the series and the actual function.

5. Can a sine series converge pointwise but not uniformly?

Yes, it is possible for a sine series to converge pointwise but not uniformly. This occurs when the series approaches the actual function at each individual point, but the difference between the sum of the series and the actual function does not become uniformly smaller across the entire domain.

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