For $j\in\mathbb{N}$, consider continuous functions $f\_j:[0,1]\to\mathbb{\mathbb{R}^+}$ such that
$$\sup\_{t\in[0,1]}\sum\_jf\_j(t)<+\infty,$$
namely $f\_j(t)\in L\_t^{\infty}((0,1),l\_j^1(\mathbb{N}))$. I would like to understand whether the quantity
$$S\_f:=\sum\_{j,k\in\mathbb{N}}\int\_0^1f\_j(t)f\_k(t)dt$$
is finite. Some observations:


* If for every $j$ the function $f\_j$ is increasing in $t$, then $S\_f$ is finite. Indeed,


$$S\_f\leq \sum\_{j\in\mathbb{N}}f\_j(T)\sum\_{k\in\mathbb{N}}\int\_0^1 f\_k(t)dt\leq \sum\_{j\in\mathbb{N}}f\_j(T)\int\_0^1\sup\_{t\in[0,1]}\sum\_{j\in\mathbb{N}}f\_k(t)dt<+\infty$$


Similarly, if the functions $f\_j$ attains their maxima on a finite set of points in $[0,1]$, then $S\_f$ is finite.


* More generally, $S\_f$ is finite if $f\_j(t)\in L\_j^1L\_t^{\infty}$, but of course this is not always the case. For example, one can take (a smooth modification of) $f\_j(t):=\delta\_{t=j^{-1}}$. Nevertheless, also in this case $S\_f$ is finite.
* I tried to consider the case when the $f\_j$ have plenty of oscillations, but I was unable to find a counterexample.


So my question is the following: **is it true that $S\_f$ is always finite?**


In the context I'm interested, there is an additional (weak) control of the derivatives of $f\_j$, of the form $2^{-j}f\_j'\in L^1(0,1)$, so one could also try to prove the finiteness of $S\_f$ under this additional assumption.


Thank you for any suggestion.



Finiteness of a bilinear combination
The answer is yes. Indeed,
$$M:=\sup\_{t\in[0,1]}\sum\_jf\_j(t)<\infty,$$
and hence
$$\begin{aligned}S\_f&=\sum\_{j,k\in\mathbb{N}}\int\_0^1f\_j(t)f\_k(t)\,dt \\ 
&=\int\_0^1\sum\_{j,k\in\mathbb{N}}f\_j(t)f\_k(t)\,dt \\
&=\int\_0^1\Big(\sum\_{j\in\mathbb{N}}f\_j(t)\Big)^2\,dt \\ 
&\le\int\_0^1 M^2\,dt=M^2<\infty,
\end{aligned}$$
as desired.


