This is the exercise $7$ on page $347$ of Analysis I of Amann and Escher. The proof seems easy but Im not completely sure that all things are correct (almost sure but not totally sure).

Let $(f_n)$ a sequence of real valued functions on $X$ and $(g_n)$ a sequence of $\Bbb K$-valued functions on $X$ which satisfy the following conditions:

(a) For each $x\in X$, $(f_n(x))$ is decreasing.

(b) $(f_n)\to 0$ uniformly.

(c) $\sup_n\|\sum_{k=0}^ng_k\|_\infty<\infty$.

Show that $\sum f_n g_n$ converges uniformly.

Let define $$s_n(x):=\sum_{k=0}^n f_k(x)g_k(x),\quad s(x):=\sum_{k=0}^\infty f_k(x)g_k(x)$$

Then we want to show that

$$\|s_n-s\|_\infty=\sup_{x\in X}\left|\sum_{k=n+1}^\infty f_k(x)g_k(x)\right|\to 0\quad (n\to\infty)$$

Observe that if $(f_n)\to 0$ uniformly this imply that $\|f_n\|_\infty<\infty$ for all $n\in\Bbb N$, and because $(f_n(x))$ is decreasing for all $x\in X$ then $(f_n(x))\downarrow 0$ for all $x\in X$.

Now define $G_n(x):=\sum_{k=0}^n g_k(x)$, $\alpha_n:=\|G_n\|_\infty$ and $\alpha:=\sup_n\alpha_n$ what is well-defined by the (b) condition, hence $\|\sum_{k=0}^\infty g_n\|_\infty\le\alpha$.

Now if we use summation by parts in $\sum_{k=n+1}^\infty f_k(x)g_k(x)$ then

$$\begin{align}\sum_{k=n+1}^\infty f_k(x)g_k(x)&=f_k(x)G_{k-1}(x)\bigg|_{n+1}^\infty-\sum_{k=n+1}^\infty(f_{k+1}(x)-f_k(x))G_{k}(x)\\&=-f_{n+1}(x)G_n(x)-\sum_{k=n+1}^\infty\Delta f_k(x)G_k(x)\end{align}$$

provided that $\lim_{n\to\infty} f_n(x)=0$, and using the notation $\Delta f_k:=f_{k+1}-f_k$. Then

$$\begin{align}\sup_{x\in X}\left|\sum_{k=n+1}^\infty f_k(x)g_k(x)\right|&=\sup_{x\in X}\left|f_{n+1}(x)G_n(x)+\sum_{k=n+1}^\infty\Delta f_k(x) G_k(x)\right|\\&\le\sup_{x\in X}|f_{n+1}(x)G_n(x)|+\sup_{x\in X}\left|\sum_{k=n+1}^\infty\Delta f_k(x) G_k(x)\right|\\&\le\alpha\sup_{x\in X} f_{n+1}(x)+\alpha\sup_{x\in X}f_n(x)\end{align}$$

And because $\sup_{x\in X}f_{n}(x)\to 0$ as $n\to\infty$ we conclude that $\sum g_n f_n$ converges uniformly.$\Box$

  • $\begingroup$ why $\sup_{x\in X}\left|\sum_{k=n+1}^\infty f_k(x)g_k(x)\right|\le\sup_{x\in X}\left|\sum_{k=n+1}^\infty g_k(x)\right|\sup_{x\in X}|f_{n+1}(x)|$ ? Use the summation by parts for making $\sum_{k=0}^ng_k$ appear $\endgroup$ – reuns Jan 27 '17 at 22:52
  • $\begingroup$ Oh! I see.. if $g_n$ is not positive (or negative) I cant do that... mmm, let me check it. $\endgroup$ – Masacroso Jan 28 '17 at 1:00
  • $\begingroup$ You might know the Dirichlet test (based on the summation by parts) $\endgroup$ – reuns Jan 28 '17 at 1:20

As pointed out in @user1952009 comment, the estimate is not correct. You need to use Abel partial summation to get the estimate of Cauchy sum of $\sum_{n=1}^{\infty} f_n(x)g_n(x)$.

Let $A_n(x)=\sum_{k=m}^n g_k(x)$. Then $$ |A_n(x)|<\sup_n\left\|\sum_{k=0}^ng_k\right\|_\infty=M\tag1 $$ Now we apply Abel partial summation here. Since \begin{align} \sum_{k=m}^n f_k(x)g_k(x)&=\sum_{k=m}^n f_k(x)(A_k(x)-A_{k-1}(x)) \\ &=\sum_{k=m}^n f_k(x)A_k(x) -\sum_{k=m}^n f_k(x)A_{k-1}(x) \\ &=\sum_{k=m}^{n-1} A_k(x)(f_k(x)-f_{k+1}(x))+f_n(x)A_n(x)\tag{2} \end{align} Note that $A_{m−1}(x)=0$.

Since $f_n(x)\to0$ uniformly, given $\epsilon>0$, there is a $N$ such that for any $n>N$ and all $x\in X$ $$ |f_n(x)|<\epsilon\tag3 $$ And since $\:f_n(x) \downarrow$ on $X$, for all $k>0$ and $x\in X$ we have $$ f_k(x)-f_{k+1}(x)\geqslant0\tag4 $$ By $(1), (2), (3), (4)$, for all $m, n>N$ and $x\in X$, there is \begin{align} \left|\sum_{k=m}^n f_k(x)g_k(x)\right|&\leqslant\sum_{k=m}^{n-1} |A_k(x)||f_k(x)-f_{k+1}(x)|+|f_n(x)A_n(x)| \\ &<M\sum_{k=m}^{n-1} (f_k(x)-f_{k+1}(x))+M\:\epsilon \\ &=M(f_m(x)-f_{n}(x))+M\:\epsilon \\ &<3M\epsilon \end{align} So by Cauchy Criterion, $\sum_{k=1}^{\infty} f_k(x)g_k(x)$ converges uniformly in $X$.


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