With Fourier series, I'm confused about Bessel's inequality and Parseval's identity.
I understand that Bessel's inequality becomes Parseval's equality if and only if both integrals $ \int_{-\pi}^\pi f(x) dx\ $ and $\int_{-\pi}^\pi f^2(x) dx $ converge and the Fourier series converges in mean, because :
$ \int_{-\pi}^{\pi} (f(x)-F_n(x))^2 dx =\cdots = \int_{-\pi}^\pi f^2(x) dx - \pi\left(\frac{a_0^2}{2} + \sum_{k=1}^n a_k^2 + b_k^2\right ) $ having $F_n(x) = \sum_{k=1}^n a_k\cos(kx) + b_k\sin(kx) $ the first term yields $0$ for $n \to \infty$ because of the convergence in mean, if I understand correctly.
On the other hand, if the Fourier series converges uniformly one can integrate $ f^2(x) = \frac{a_0}{2}f(x) + \sum_{n=1}^\infty a_n\cos(nx)f(x) + b_n\sin(nx)f(x) $
term-by-term which also gives Parseval's equality. So from the look of it, the requirement of the convergence in mean may imply the uniform convergence, am I right? Or is there any connection between these two types of convergence? Is the requirement of uniform convergence sufficient for the Parseval's equality to hold?