Is the $L_\infty$ norm of the $\mathcal{l}_2$ norm of this sequence of functions finite? I am interested in proving or disproving the following claim and am stuck.
We define a series of functions with the following properties.
For each $i\in \mathbb{N}$ let $f_i\colon \mathbb{R}^+ \to \mathbb{R}$  be such that for all $t\in\mathbb{R}^+$ we have $f_i(t)\in \mathcal{l}_2$, (we can also assume that each $f_i$ is continuous but I would rather not do so).  Furthermore assume that $\sup_{t\in\mathbb{R}^+}   \sum_{i=1}^\infty (f_i(t))^2 = M < \infty$.  So the $L_\infty$ norm of the $\mathcal{l}_2$ norm is finite.  Let $\{\lambda_i\}_{i=1}^\infty\subset \mathbb{R}^+$ be a positive, increasing and unbounded sequence of real numbers. We also know that $1/\lambda_i\in \mathcal{l}_2$, Finally we define a new sequence of function $u_i\colon \mathbb{R}^+\to \mathbb{R}$ as
$$u_i(t) = \int_{\tau = 0}^t \lambda_i \exp(-\lambda_i(t-\tau))f_i(\tau) \mathrm{d} \tau $$ 
Then the $L_\infty$ norm of the $\mathcal{l}_2$ norm of $u_i(t)$ is also finite.
The following are a couple seemingly dead ends I have pursued.  I include even the trivial steps so hopefully it is clear what manipulations are being done.  I never use that $1/\lambda_i\in \mathcal{l}_2$ 
Dead end #1
$$
\begin{align*}\sup_{t\in \mathbb{R}^+} \sum_{i=1}^\infty (\, f_i(t)\, )^2
&= \sup_{t\in\mathbb{R}^+} \sum_{i=1}^\infty \left( \int_{\tau=0}^t \lambda_i \exp(-\lambda_i (t-\tau) )f_i(\tau)  \mathrm{d} \tau \right)^2\\
&\le \sup_{t\in \mathbb{R}^+} \sum_{i=1}^\infty \lambda_i^2 \left(\sup_{\tau\in (0,t)} f_i(\tau)^2 \right) \left(\int_{\tau=0}^t \exp(-\lambda_i( t-\tau)  )  \mathrm{d} \tau \right)^2 \\
&=\sup_{t\in \mathbb{R}^+} \sum_{i=1}^\infty \lambda_i^2 \left(\sup_{\tau\in (0,t)} f_i(\tau)^2 \right) \left(\int_{\tau=0}^t \exp(-\lambda_i(\tau)  )  \mathrm{d} \tau \right)^2 \\
&\le \sup_{t\in \mathbb{R}^+} \sum_{i=1}^\infty \lambda_i^2 \left(\sup_{\tau\in (0,t)} f_i(\tau)^2 \right) \left(\int_{\tau=0}^\infty \exp(-\lambda_i(\tau)  )  \mathrm{d} \tau \right)^2 \\
&\le \sup_{t\in \mathbb{R}^+} \sum_{i=1}^\infty \left(\sup_{\tau\in (0,t)} f_i(\tau)^2 \right)  \\
\end{align*}
$$
I would like to go from this last line to 
$$
\sup_{t\in \mathbb{R}^+}\sum_{i=1}^\infty f_i(t)^2, 
$$ the desired inequality, but that does not follow.  As the supremums of each $f_i$ might not be in $\mathcal{l}_2$ even though at every $t$, $f_i(t)\in \mathcal{l}_2$. 
Dead end #2.  Here I assume continuity of the $f_i$ 's and use the positivity of the exponential to apply the mean value theorem for integrals, to produce a $\widehat{\tau}_i$.  
$$
 \begin{align}
 \sup_{t\in \mathbb{R}^+} \sum_{i=1}^\infty \left( f_i(t) \right)^2
&= \sup_{t\in \mathbb{R}^+} \sum_{i=1}^\infty  \lambda_i^2 \left( \int_{\tau=0}^t  \exp(-\lambda_i (t-\tau) f_i(\tau) \mathrm{d} \tau \right)^2\\
&= \sup_{t\in \mathbb{R}^+} \sum_{i=1}^\infty  \lambda_i^2 \left( \int_{\tau=0}^t  \exp(-\lambda_i (\tau) f_i(t-\tau) \mathrm{d} \tau \right)^2\\
&= \sup_{t\in \mathbb{R}^+} \sum_{i=1}^\infty \lambda_i^2 f_i(\widehat{\tau}_i)^2 \left(\int_{\tau=0}^t \exp(-\lambda_i \tau)\, \mathrm{d} \tau   \right)^2\\
&\le \sup_{t\in \mathbb{R}^+} \sum_{i=1}^\infty f_i(\widehat{\tau}_i)^2 \lambda_i^2 \left(\int_{\tau=0}^\infty \exp(-\lambda_i \tau) \, \mathrm{d}\tau \right)^2\\
&= \sup_{t\in \mathbb{R}^+} \sum_{i=1}^\infty f_i(\widehat{\tau}_i)^2 
 \end{align}
$$
Now we have pretty much the same problem as before.  We can't conclude the desired inequality because we don't know that $f_i(\tau_i)\in \mathcal{l}_2$.
I appreciate any help.  It seems that I need some sort of mean value theorem for weighted sums.  I mean, if we could conclude that
$$
\sum_{i=1}^\infty \lambda_i^2 \left( \int_{\tau = 0}^t \exp(-\lambda \tau) f_i( t-\tau)\, \mathrm{d} \tau\right)^2 = \sum_{i=1}^\infty \lambda_i^2 f_i(\widehat{\tau})^2 \left(\int_{\tau=0}^t \exp(-\lambda_i \tau) \, \mathrm{d} \tau\right)^2
$$
for some $\widehat{\tau}$ (independent of $i$) the desired result  would follow.  Is there such a mean value theorem?  Any help and ideas are appreciated.
Edited: TeX corrections
@ David Ullrich To be honest I am not familiar with the idea of $L_2$ modulus of continuity.  Could you tell me a definition?  I appreciate it.
@ David Ullrich:  Thank you.  I want to accept your answer but I don't have a green box showing up.  I really appreciate it.  I will figure out how to accept the answer.  I really liked your counterexample.  Thanks again.
 A: It's false. The counterexample is simpler than I expected at first; the condition $\sum f_n^2\in L^\infty$ is just too fragile, those convolutions smear the mass of $f_n$ around a bit and it's destroyed.
As a general notational convention we're going to say $$\delta=1/\lambda.$$Let $e_\lambda$ be your exponential kernel, so that $$u_n=e_{\lambda_n}*f_n.$$
Note first that continuity and compactness show that $$e_1*\chi_{(0,1)}\ge c\chi_{[1,2]}.$$Since $e_1\ge0$ this shows that if $0\le a<b$ and $b-a\ge 1$ then $$e_1*\chi_{(a,b)}\ge c\chi_{[b,b+1]}.$$Now a change of variables gives
Lemma If $0\le a<b$ and $b-a\ge\delta$ then $$e_\lambda*\chi_{(a,b)}\ge c\chi_{[b,b+\delta]}.$$
We will show that if $\lambda_n\to\infty$ then there exist $f_n$ giving a counterexample. To this end we construct an increasing sequence $(n_j)$ of integers and a sequence $(a_j)$ of reals increasing to $1$ as follows: Let $a_0=0$ and $n_0=0$.
Supposing we've chosen $n_j$ and $a_j$, choose $n_{j+1}>n_j$ large enough that $$\delta_{n_{j+1}}<(1-a_j)/2.$$ Let $a_{j+1}=1-\delta_{n_{j+1}}$.
Note that $a_j-a_{j-1}>\delta_{n_j}$. Now we set $f_n=0$ if $n\ne n_j$ and $$f_{n_j}=\chi_{(a_{j-1},a_j)}.$$So $\sum f_n^2\le 1$, while the lemma shows that $$u_{n_j}\ge c\chi_{[a_j,a_j+\delta_{n_j}]}=c\chi_{[1-\delta_{n_j},1]};$$hence $\sum u_n^2$ is not bounded.
