Is a summable sequence 'uniformly bounded'? Let $(\rho_n)_{n\in\mathbb{N}}$ be a sequence of (non-negative) functions $\rho_n : [0, \epsilon_0)\rightarrow\mathbb{R}_+$, some $\epsilon_0>0$, such that
$$\tag{1}
\sum_{n=0}^\infty 2^{-n}\rho_n(\epsilon) \ \leq \ \epsilon, \quad \forall\, \epsilon\in [0, \epsilon_0).
$$
Can we conclude that there is a (right-continuous at zero) function $\varphi : [0,\epsilon_0)\rightarrow[0,\epsilon_0)$ with $\varphi(0)=0$ and such that
$$
\sup_{n\in\mathbb{N}_0}\rho_n(\epsilon) \ \leq \ \varphi(\epsilon) \quad \text{for each } \ 0\leq\epsilon<\epsilon_0?
$$
Edit: A counterexample to the original question was provided by Ian below. What happens if $(\rho_n)_n$ is a bounded function of $n$?
 A: A family of counterexamples to the original claim without the uniform boundedness assumption are given by $\rho_n(\epsilon)=Cn\epsilon$ where $C \in (0,1/2]$ (noting that $\sum_{n=0}^\infty n 2^{-n}=2$).
If instead $\sup_n \rho_n(\epsilon)<\infty$ for each $\epsilon$, then you can consider $\varphi(\epsilon)=\sup_n \rho_n(\epsilon)$. You will have $\varphi(0)=0$ since the given inequality implies $\rho_n(0) \equiv 0$.
As for right continuity of $\varphi$ at zero together with having $\varphi(0)=0$, you can't necessarily have both. The problem is that $\lim_{\epsilon \to 0^+} \sup_n \rho_n(\epsilon)$ can be positive, since the bound only constraints $\rho_n$ for large $n$ just a little bit due to the really small coefficient.
For example $\rho_n(\epsilon)=\begin{cases} n \epsilon/2 & \epsilon \in [0,1/n] \\ 1/2 & \text{otherwise} \end{cases}$ is uniformly bounded and satisfies your bound. But $\sup_n \rho_n(\epsilon)=\begin{cases} 1/2 & \epsilon>0 \\ 0 & \epsilon=0 \end{cases}$ and there is no way to regularize that while keeping it zero at zero.
