$\ell^1$ bounds for a particular sequence I have a research problem which has been reduced to what seems like it should be elementary but I can't seem to prove it. 
I have a sequence $a_n \in \ell^1$ and fixed constants $0 < \delta < 1$ and $0 \leq \epsilon < 1$ which also satisfy $\delta + \epsilon \leq 1$. Define a new sequence by $$c_n = \delta^n \sum\limits_{m=n}^{\infty} a_m \binom{m}{n}\epsilon^{m-n}.$$
Then I can prove that $c_n \in \ell^1$ also but I would like more. I am interested in computing $K$ such that $$||c_n ||_{\ell^1} \leq K ||a_n ||_{\ell^1}$$
and in particular I would like to find the sharpest (or near) such $K$. 
In fact, I believe there exists a $K$ which is uniform in $\delta,\epsilon$ (satisfying the above conditions) which would be nice but not entirely necessary. In any case, this seems like a problem which is probably an exercise in a book somewhere but I can't seem to get anywhere. 
 A: Consider
$$f(z) := \sum_{n = 0}^{\infty} a_n z^n\quad\text{and}\quad \tilde{f}(z) = \sum_{n = 0}^{\infty} \lvert a_n\rvert z^n.$$
Since $(a_n)$ is in $\ell^1$, the series converge uniformly on the closed unit disk, and $f$ and $\tilde{f}$ are of course holomorphic on the open unit disk. By the maximum modulus principle, we therefore have
$$\lvert f(z)\rvert \leqslant \sup\, \{ \lvert f(\zeta)\rvert : \lvert \zeta\rvert = 1\} \leqslant \sum_{n = 0}^{\infty} \lvert a_n\rvert = \lVert a\rVert_{\ell^1}$$
for all $z$ in the open unit disk, and the first inequality is strict unless $f$ is constant. The same holds for $\tilde{f}$.
Now, since $\delta + \varepsilon \leqslant 1$, we can compute
\begin{align}
f(\delta + \varepsilon) &= \sum_{m = 0}^{\infty} a_m (\delta +\varepsilon)^m \\
&= \sum_{m = 0}^{\infty} a_m \sum_{n = 0}^m \binom{m}{n}\delta^{n}\varepsilon^{m-n} \\
&= \sum_{n = 0}^{\infty} \sum_{m = n}^{\infty} a_m \binom{m}{n}\delta^n\varepsilon^{m-n} \\
&= \sum_{n = 0}^{\infty} c_n.
\end{align}
Repeating the computation with the sequence $(a_n)$ replaced by $(\lvert a_n\rvert)$ shows that
$$\lVert c\rVert_{\ell^1} = \sum_{n = 0}^{\infty} \lvert c_n\rvert \leqslant \sum_{n = 0}^{\infty} \sum_{m = n}^{\infty} \lvert a_n\rvert\binom{m}{n}\delta^n\varepsilon^{m-n} = \tilde{f}(\delta + \varepsilon) \leqslant \lVert a\rVert_{\ell^1}.$$
Thus $K = 1$ works for all $\delta,\varepsilon \in [0,1)$ with $\delta + \varepsilon \leqslant 1$, and choosing a sequence with $a_n = 0$ for $n > 0$ shows that $K = 1$ is the optimal constant.
A: WLOG, $(a_m)$ is a nonnegative sequence. Then
$$\sum_{n=0}^{\infty}c_n =  \sum_{n=0}^{\infty}\sum\limits_{m=n}^{\infty} \delta^n a_m \binom{m}{n}\epsilon^{m-n}$$ $$= \sum_{m=0}^{\infty}\sum_{n=0}^{m}\delta^n a_m \binom{m}{n}\epsilon^{m-n}=\sum_{m=0}^{\infty}a_m\epsilon^m\sum_{n=0}^{m}\binom {m}{n}\left (\frac{\delta}{\epsilon}\right )^n$$ $$ = \sum_{m=0}^{\infty}a_m\epsilon^m \left (1+\frac{\delta}{\epsilon} \right )^m = \sum_{m=0}^{\infty}a_m (\epsilon+\delta)^m \le \sum_{m=0}^{\infty}a_m.$$
So $K=1$ works, and as Daniel Fischer said, it is sharp.
