Fourier Series on $L^1\left(\left[0,1\right)\right)\cap C\left(\left[0,1\right)\right)$

Question

My professor claimed that

if $f\in L^1\left(\left[0,1\right)\right)\cap C\left(\left[0,1\right)\right)$, then $$f\left(t\right)=\sum_{k\in\mathbb{Z}}c\left(k\right)\exp\left(2\pi ikt\right).$$

I think that this is false.

Should it be that $f\in L^2\left(\left[0,1\right)\right)$ instead?

One note: if $c(k)$ is the $k$th fourier coefficient of $f$, then it should be $f(t) = \sum_k c(k)exp(2\pi ikt)$ (i.e. no minus sign in the exponential). — mathworker21, Jan 20 '18 at 08:22

score 1 · Answer 1 · answered Jan 20 '18 at 16:12

Yes, if he means convergence for every $x$ then this is false. Of course it's true for convergence in the $L62$ norm, and (although this is a very difficult theorem) it's true for almost everywhere convergence.

It's well known that there exists a continuous (hence integrable) function on the circle with a Fourier series that diverges at at least one point. This is trivial from the Uniform Boundedness Principle.

score 0 · Accepted Answer · answered Jan 20 '18 at 08:06

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By $[0,1)$, they mean the Torus, which is compact. So $f \in C(\mathbb{T})$ implies it's in $L^\infty(\mathbb{T})$ and, in particular, in $L^1$ and $L^2$.

answered Jan 20 '18 at 08:06

mathworker21

34,399

In other words, we only need that $f\in C\left(\left[0,1\right)\right)$, right? – wjmolina Jan 20 '18 at 08:09
By the way, is it standard to denote $\mathbb R/\mathbb Z$ by $\left[0,1\right)$? – wjmolina Jan 20 '18 at 08:19
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Yes; Yes, I think – mathworker21 Jan 20 '18 at 08:21
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To complement mathworker21's answer. For the Fourier series to be defined it is enough that $f\in L^1$. Convergence of the series to the original function requieres further conditions on $f$. – Julián Aguirre Jan 20 '18 at 10:47

Fourier Series on $L^1\left(\left[0,1\right)\right)\cap C\left(\left[0,1\right)\right)$

2 Answers2