# 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?

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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.

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 ...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.

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