I need to prove the following:
If $(\Phi_n)_{n\ge0}$ is an orthonormal system of integrable functions defined on some interval $[a,b]$, and $(c_n)_n$ is a sequence of reals such that $\sum c_n \Phi_n$ converges uniformly to an integrable function $f$ on $[a,b]$, then:
$$c_m = \int_a^b f(x) \Phi_m(x) dx, \ \ \forall \ m \in \mathbb N$$
If it can be guaranteed that for a given $m \in \mathbb N$, $\sum_{n} c_n \Phi_n \Phi_m$ converges uniformly as well, then we can do:
$$\int_a^b f(x) \Phi_m(x) dx = \int_a^b \left( \sum_{n=0}^{\infty} c_n \Phi_n(x) \right) \Phi_m(x) dx = \int_a^b \sum_{n= 0}^{\infty} c_n \Phi_n(x) \Phi_m(x) dx = \sum_{n=0, n \neq m}^{\infty} c_n \int_a^b \Phi_n(x) \Phi_m(x) dx + c_m \int_a^b (\Phi_m(x))^2 dx = 0 + c_m \times 1 = c_m$$
The question is: why need the series $\sum_{n} c_n \Phi_n \Phi_m$ converge uniformly as well for a given $m \in \mathbb N$? For this to be achieved, $\Phi_m$ should be bounded, that's, $\exists$ $M>0$, such that $||\Phi_m||_{\infty} \le M$. Why need this occur?
Thanks.
Let us start with a general claim:
Let $(g_n),\quad n=1,2,\ldots$, be a sequence of functions which converges uniformly to the function $g$. Let $h$ be a square-integrable function. Then the sequence $(g_nh)$ converges by integral to $gh$.
To prove the claim, we may assume $g=0$. By the Cauchy-Schwarz inequality we have $$\left(\int g_nhdx\right)^2\leq\int g_n^2dx\cdot\int h^2dx,$$and the claim follows since uniform convergence yields square integral convergence.
We return to the posted question. If we may assume that $\Phi_m$ is square integrable, we deduce from the above claim that the sum in question converges by integral to the desired function. This is actually enough, as we all agree.