Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is integrable on $L^{1}([0,2\pi])$, with Fourier coefficients $\hat{f} \in l^1(\mathbb{Z})$. Is the following manipulation valid?
$$\displaystyle \int_{0}^{2\pi}|f(x)|^4 \ \mathrm{d}x = \int_{0}^{2\pi}\left| \sum_{n = -\infty}^{\infty} \hat{f}(n)e^{inx}\right|^4 \ \mathrm{d}x \leqslant \int_{0}^{2\pi} \left( \sum_{n = -\infty}^{\infty} | \hat{f}(n) | \right)^4 \ \mathrm{d}x = 2\pi \left( \sum_{n = -\infty}^{\infty}|\hat{f}(n)| \right)^4$$
Is this valid, or is it also necessary that $f \in L^{4}([0,2\pi])$? Are there additional conditions that $f$ must satisfy for this to hold that I haven't mentioned? I feel as though I am missing something obvious here.