# When convergence in mean implies uniform convergence?

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?

I'm not sure this answers your question, but here are some remarks that I think are pertinent:

Bessel's inequality holds more generally for any orthonormal sequence. If that sequence is complete, then the inequality becomes an equality and is called Parseval's identity.

If $f$ is square integrable, that is, if $\int_{-\pi}^\pi f^2$ is finite, then the Fourier series for $f$ automatically converges to $f$ in the $L_2$-norm. Thus the system of functions defining the Fourier Coefficients are complete in $L_2$, and and Parseval's equality holds.

This tells us nothing apriori about pointwise convergence though (however, there is a famous result known as Carleson's Theorem).

Concerning uniform and pointwise convergence:

If $f$ is $2\pi$-periodic, continuous, and $f'$ is piecewise continuous, then the Fourier series for $f$ converges uniformly to $f$.

Generally, if $f$ and $f'$ are piecewise continuous and $f$ is periodic, then the Fourier series of $f$ converges pointwise to $(f(x_+)+f(x_-))/2$ (the average of the right and left hand limits). In this case, one can not conclude that the Fourier series converges uniformly to $f$ on all of $[-\pi,\pi]$ (in particular, the limit may not be a continuous function, which implies that the convergence cannot be uniform).

Some helpful links towards illustrating the claims made in the previous two paragraphs are here and here.

 ...the requirement of the convergence in mean may imply the uniform
convergence, am I right?


I cannot tell if the question is, "Does $L^2$ convergence imply uniform convergence?" or "$L^2$ convergence plus what implies uniform convergence?"

If you mean the former, note that $f(x)=x^{-1/3}\in L^2[0,1]$, so its Fourier series converges in the $L^2$ sense on $[0,1]$, but it does not converge uniformly on $[0,1]$ since the uniform limit of continuous functions is again continuous.