I have written a proof which seems to imply that a function is $L^4$-integrable over $[0, 2\pi]$ if its Fourier coefficients are $l^1$ and $l^2$-summable over $\mathbb{Z}$. I don't think this is correct. Where am I going wrong in my argument?
Suppose $f : \mathbb{R} \rightarrow \mathbb{R}$ belongs to both $L^{1}(\mathbb{R})$ and $L^{2}(\mathbb{R})$, and that $\hat{f}(0) = 0$. Using Parseval's identity, we have:
$$\displaystyle \|f\|_{4}^{4} = \int_{0}^{2\pi} |f(x)|^{4} \ \mathrm{d}x = \int_{0}^{2\pi}|f(x)^2|^2 \ \mathrm{d}x = \sum_{n = -\infty}^{\infty}|\widehat{f(n)^2}|^2 = \sum_{n = 1}^{\infty}|\widehat{f(n)^2}|^2 + \sum_{n = 1}^{\infty}|\widehat{f(-n)^2}|^2 \\ = \|\widehat{f(n)^2}\|_{l^2}^2 + \|\widehat{f(-n)^2}\|_{l^2}^2$$
(There is a factor of $\sqrt{2\pi}$ in there also, but I left that out for ease of notation.) Now, the convolution theorem says that $\widehat{f\cdot g} = \hat{f} \ast \hat{g},$ where $\ast$ denotes the discrete convolution. Further, Young's inequality says that $\|x \ast y\|_{l^r} \leqslant \|x\|_{l^p} \cdot \|y\|_{l^q},$ where $x \in l^p$ and $y \in l^q$ with $1/p + 1/q = 1/r + 1$, $p,q,r \in [1,\infty].$ Therefore, setting $p = 1, q = 2, r = 2$, we have:
$$\displaystyle \|\widehat{f(n)^2}\|_{l^2}^2 + \|\widehat{f(-n)^2}\|_{l_2}^2 = \|(\hat{f} \ast \hat{f})(n)\|_{l^2}^2 + \|(\hat{f} \ast \hat{f})(-n)\|_{l^2}^2 \\ \leqslant \|\hat{f}(n)\|_{l^1}^2 \cdot \|\hat{f}(n)\|_{l^2}^2 + \|\hat{f}(-n)\|_{l^1}^2 \cdot \|\hat{f}(-n)\|_{l^2}^2 < \infty.$$
I am sure I must be doing something wrong. Could someone help me see what it is?