Does Non-Uniform Convergence of Fourier Series Imply Discontinuity?

[torquerotates]

I don't understand what my professor told me. He said that if a fourieer series of a function f is not continuous then it doesn't converge uniformly to f.

But the given theorem only states that, "if a sequence of functions is continuous and converges uniformly to f, then f is continuous."

I'm given that f is not continuous. and that the fourieer series of f is not continuous. How does this mean that the fourieer series doesn't converge uniformly to f?

This is just a application of logic here

A&B=> C

not C=> (not A) or (not B)

knowing (not A) doesn't really give me (not B)

 

[hamster143]

Even if the function f is not continuous, each individual term and all partial sums of its Fourier series are still continuous (all terms are just sines and cosines, the sum of any finite number of sines and cosines is continuous everywhere).

The sequence of partial sums is continuous, but the limit is not - therefore, convergence is not uniform.

A & not C => not B

 

[torquerotates]

Oh i see that makes much more sense. So the sequence of functions are continuous even though the fourieer series is not?

Heres what my professor said, "If the FS series converges uniformly, then the FS must have been continuous (since the uniform limit of a sequence of continuous functions is continuous). Since the FS is not continuous , we conclude that the converge is not uniform."

I don't quite understand this statement.

 

[rock.freak667]

Oh i see that makes much more sense. So the sequence of functions are continuous even though the fourieer series is not?

Heres what my professor said, "If the FS series converges uniformly, then the FS must have been continuous (since the uniform limit of a sequence of continuous functions is continuous). Since the FS is not continuous , we conclude that the converge is not uniform."

I don't quite understand this statement.

If f(x) has Fourier series

$$f(x)=\frac{a_0}{2}+\sum_{n=1} ^{\infty}a_n cos(\frac{n \pi x}{L})+b_n sin(\frac{n \pi x}{L})$$

f(x) is continuous everywhere for all x except at points x₀.

So for all x≠x₀, the Fourier series converges to f(x).

at x=x₀, the series converges to the mean of the left and right limits i.e.

$$f(x)=\frac{1}{2}(\lim_{x \rightarrow x_0^-} f(x) +\lim_{x \rightarrow x_0^+} f(x))$$

This usually occurs when you have piecewise functions.

 

[LCKurtz]

Oh i see that makes much more sense. So the sequence of functions are continuous even though the fourieer series is not?

Heres what my professor said, "If the FS series converges uniformly, then the FS must have been continuous (since the uniform limit of a sequence of continuous functions is continuous). Since the FS is not continuous , we conclude that the converge is not uniform."

I don't quite understand this statement.

I think that when your professor talks about "the FS" he means "the sum of the FS". If the convergence of the FS had been uniform, the sum would be a continuous function. Since the sum isn't, the convergence couldn't have been uniform. Is that where your confusion lies?