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Let $f_k$ be a real numbers such that $\sum_{k=1}^\infty f_k < \infty$. For each $R > 0$, define the convergent sum $$v(R) = \sum_{k=1}^\infty f_k(b_k(R)e^{-ky} - c_k(R)e^{ky})$$ where $0 \leq y \leq R$ and where $b_k$ and $c_k$ are functions of $k$ and $R$ and satisfy $$b_k(R) = 1+ \frac{1}{e^{2kR}-1}\quad \text{and}\quad c_k(R) = b_k(R)-1 = \frac{1}{e^{2kR}-1}$$

Apparently this sum uniformly converges (see this) in $n$ as function of $R$. How do I show this? I.e. if $v_n(R) = \sum_{k=1}^n f_k(b_k(R)e^{-ky} - c_k(R)e^{ky})$ how do I show that $v_n \to v$ uniformly as function of $R$?

I didn't understand how the explanation is given in that thread and I think my question is too basic to be asked there. Both the coefficients have look "like" $e^{-2kR}$ so it seems like it should be uniform but I don't know the argument.

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  • $\begingroup$ Is $R>\delta >0$? Or can $R$ be any real number? $\endgroup$ – Mark Viola Apr 21 '15 at 14:23
  • $\begingroup$ @Dr.MV Sorry $R > 0$, and $0 \leq y \leq R$. $\endgroup$ – RealMax Apr 21 '15 at 14:24
  • $\begingroup$ Yes, but is it restricted to be greater than a pre-assigned number? $\endgroup$ – Mark Viola Apr 21 '15 at 14:25
  • $\begingroup$ No, just that it is a positive number. $\endgroup$ – RealMax Apr 21 '15 at 14:26
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    $\begingroup$ You're welcome. My pleasure. And note, that if $f_k$ is the $k$th term of a Generalized Fourier series, then we certainly have that it converges in $\mathscr{L}^2$ (i.e., $\sum_k |f_k|^2$ converges). $\endgroup$ – Mark Viola Apr 21 '15 at 16:17
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I'm just fumbling here, but if you compute a bit you get $$ b_k(R) e^{-ky} - c_k(R) e^{ky} = \frac 1 {e^{2kR}-1} \left( e^{2kR} e^{-ky} - e^{ky} \right). $$ For $R$ large enough, you can easily bound this by some constant times $$ e^{-ky} - e^{k(y-2R)}. $$ Since $R \ge y$, the second term will be $\le e^{-ky}$. So for large $R$, you can bound your summand by $$ | f_k (b_k(R) e^{-ky} - c_k(R) e^{ky}) | \le C e^{-ky} |f_k| \le C |f_k|. $$ Since the $\sum f_k$ converges, so does this new sum (assuming $f_k\ge 0$), and since the bound doesn't depend on $R$, it converges uniformly in $R$ (again, for large $R$).

Please double check all of this, I'm not sure if I messed up along the way...

EDIT: Ouch, Dr. MV had the more elegant idea obviously.

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Assume that $\sum_k f_k$ converges absolutely.

Note that

$$\begin{align} |b_k(R)e^{-ky}-c_k(R)e^{ky}|&=\left|\frac{e^{2kR}e^{-ky}-e^{ky}}{e^{2kR}-1}\right|\\\\ &=\left|\frac{\sinh k(R-y)}{\sinh kR}\right|\\\\ &<1 \end{align}$$

Thus

$$\begin{align} \left|v(R) \right|&=\left|\sum_{k=1}^{\infty}f_k\left(b_k(R)e^{-ky}-c_k(R)e^{ky}\right) \right|\\\\ &\le \sum_{k=1}^{\infty}|f_k|\,\left|\left(b_k(R)e^{-ky}-c_k(R)e^{ky}\right)\right| \\\\ &\le \sum_{k=1}^{\infty}|f_k| \\\\ \end{align}$$

which converges by hypothesis!


Assume that $\sum_k |f_k|^2$ converges.

Then,

$$\begin{align} \left|v(R) \right|^2&=\left|\sum_{k=1}^{\infty}f_k\left(b_k(R)e^{-ky}-c_k(R)e^{ky}\right) \right|^2\\\\ &\le \sum_{k=1}^{\infty}|f_k|^2\,\,\sum_{k=1}^{\infty}\left|\left(b_k(R)e^{-ky}-c_k(R)e^{ky}\right)\right|^2 \\\\ &=\sum_{k=1}^{\infty}|f_k|^2\,\,\sum_{k=1}^{\infty}\left|\frac{\sinh k(R-y)}{\sinh kR}\right|^2 \end{align}$$

The first term converges by hypothesis. For the second term, note that

$$\left|\frac{\sinh k(R-y)}{\sinh kR}\right|^2\le e^{-2ky}$$

and $\sum_{k=1}^{\infty} e^{-2ky}$ converges for any fixed $y>0$.

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