When the series of orthogonal functions converges absolutely?

Question

It is given a family of functions $\phi_n(t)$ orthogonal in $[0, T]$, $0 < T < \infty$. What conditions must $\phi_n$ satisfy in order to have $$ \sum_{n = 1}^\infty |\phi_n(t)| \leq c < \infty? $$

Any hint or reference would be appreciated.

Answer 1

If your family of functions belongs to a RKHS $\mathcal{H}$ of functions from $[0,T]$ to $\mathbb{C}$ with the scalar product defined as: $$\langle f,g\rangle:=\int_0^Tf(t)\overline{g(t)}\operatorname{dt}$$ then the pointwise absolute convergence is easily achieved, because point evaluations are continuous.

Under the hypothesis that $$\sum_{n=1}^{+\infty}\|\varphi_n\|_2^2<+\infty$$ we have (by orthogonality) that $(\sum_{n=1}^{N}\varphi_n)_{N\in\mathbb{N}}$ is a Cauchy sequence in $\mathcal{H}$, and so (by completeness) it converges in norm to a (unique) $\varphi\in\mathcal{H}$. So, if $k_t\in\mathcal{H}$ is the element in $\mathcal{H}$ that represents the point evaluation in $t$ via the scalar product of $\mathcal{H}$ (it exists and it is unique thanks to Riesz representation theorem for Hilbert spaces) then:

$$\varphi(t)=\langle\varphi,k_t\rangle=\langle \lim_{N\rightarrow\infty}\sum_{n=1}^N\varphi_n,k_t\rangle = \lim_{N\rightarrow\infty}\langle \sum_{n=1}^N\varphi_n,k_t\rangle \\ = \lim_{N\rightarrow\infty} \sum_{n=1}^N\langle \varphi_n,k_t\rangle = \lim_{N\rightarrow\infty} \sum_{n=1}^N\varphi_n(t)=\sum_{n=1}^{+\infty}\varphi_n(t).$$

On the other hand, we get the same conclusion if $n :\mathbb{N}\rightarrow \mathbb{N}$ is any bijection because, thanks to orthogonality, the series $(\sum_{n=1}^{N}\varphi_n)_{N\in\mathbb{N}}$ is unconditionally convergent to $\varphi$ and so with the very same proof we get: $$\sum_{k=1}^{+\infty}\varphi_{n_k}(t)=\varphi(t).$$ So the series $$\sum_{n=1}^{+\infty}\varphi_{n}(t)$$ is unconditionally convergent to $\varphi(t)\in\mathbb{C}$. But a series of complex numbers converges if and only if its real and imaginary part converge, so we get that $$\sum_{n=1}^{+\infty}\varphi_{n}(t)$$ has real and imaginary part unconditionally convergent. Then, thanks to Riemann theorem we get that both real and imaginary part of $$\sum_{n=1}^{+\infty}\varphi_{n}(t)$$ are absolutely convergent and then, from: $$\sum_{n=1}^{+\infty}|\varphi_{n}(t)|\le \sum_{n=1}^{+\infty}|\operatorname{Re}(\varphi_{n}(t))| + \sum_{n=1}^{+\infty}|\operatorname{Im}(\varphi_{n}(t))| < +\infty,$$ we get the conclusion.