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I've been trying to prove that there does not exist $k\in L^1(\mathbb{T})$ such that $k*f=f$ for every continuous function $f:\mathbb{T} \rightarrow \mathbb{C}$. The exercise says to use the Riemann-Lebesgue lemma.

So what I've tried is this:

Since $k$ is defined as above, we know from the Riemann-Lebesgue lemma that the Fourier coefficients $\widehat{k}(n)\rightarrow 0$ as $|n|\rightarrow \infty$. If $k*f=f$ for every $f$, then I assume that their Fourier coefficients should also be the same such that $\widehat{k*f}(n)=\hat{f}(n)$ for all $n\in\mathbb{Z}$. Now, a lemma in my notes states that $\widehat{k*f}(n)=\hat{k}(n)\widehat{f}(n)$ for all $n\in\mathbb{Z}$ (Fourier coefficients of a convolution equal the product of Fourier coefficients).

From here I'm kind of stuck. If I were to use the very first point stated in the proof, I don't feel like I've proven it for every $f$. Any help would be appreciated!

PLY
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Since $k*f=f$, we also have that $\widehat{f}(n)=\widehat{k*f}(n)$ for all $n\in\mathbb{Z}$, so $$\widehat{f}(n)=\widehat{k*f}(n)=\hat{k}(n)\widehat{f}(n).$$ Since there are a lot of continuous functions with $\widehat{f}(n)\neq 0$ for all $n\in\mathbb{Z}$, for example the $2\pi$-periodic extension of $f(x)=x^2$ on $(-\pi,\pi]$, we obtain $1=\lim\limits_{|n|\to\infty}\hat{k}(n)=0$ by the Riemann-Lebesgue lemma, so we obtained a contradiction.

Remark 1: You actually only need $\widehat{f}(n)\neq 0$ for infinitely many $n\in\mathbb{Z}$.

Remark 2: Even easier, as noted by David C. Ullrich, you only need to find for each $n\in\mathbb{Z}$ (or just infinitely many $n$) a continuous function $f_n$ on $\mathbb{T}$ such that $\widehat{f_n}(n)\ne0$. So you may choose $f_n(t)=e^{int}$.

The Phenotype
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    Actually all you need is that for every $n$ there exists $f$ with $\hat f(n)\ne0$; this implies that $\hat k(n)=0$ for every $n$. (Seems worth mentioning because that's so trivial: $f=e^{int}$.) – David C. Ullrich Apr 30 '18 at 13:59
  • @DavidC.Ullrich You're very right! I'll add it in the remark. – The Phenotype Apr 30 '18 at 14:02
  • @ThePhenotype Thank you! The remarks made it much easier indeed. I am still curious about one part of the original answer though: how was the relation of $1=\lim_{|n|\rightarrow\infty}\hat{k}(n)$ exactly deduced? The use of Riemann-Lebesgue is clear to me. – PLY Apr 30 '18 at 14:20
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    @Kirigiri For every $\widehat{f}(n)\neq 0$, you can divide $\widehat{f}(n)=\hat{k}(n)\widehat{f}(n)$ by $\widehat{f}(n)$ and get $\hat{k}(n)=1$ for these $n$. If this holds for infinitely many $n$, the limit cannot be equal to $0$. – The Phenotype Apr 30 '18 at 14:23
  • Of course, very clear! Thanks again. – PLY Apr 30 '18 at 14:28