Let $(a_n)_{n∈N}$ be a sequence of real numbers. Show that the series $∑_{n=1} ^{\infty} |a_n− a_{n+1}|$ converges if and only if the series $∑_{n=1}^{\infty} a_nb_n$ converges for every convergent series $∑_{n=1}^{\infty}b_n$ in $\mathbb{R}$.
It is an application of uniform bounded principle.
Can anyone give me some hint how to approach this problem.
The space $S$ of all summable sequences is isomorphic to $c$ (convergent sequences) via the map $$ x\in c \mapsto (x_0, x_1-x_0, x_2-x_1,\ldots)\in S, $$ while the space $D$ of all sequences $(a_n:n\in\mathbb{N})$ such that $\sum_n |a_n-a_{n+1}|$ converges is isomorphic to $\mathbb R\oplus\ell^1$ under the map $$ x\in D \mapsto (x_0, x_1-x_0, x_2-x_1,\ldots)\in \mathbb R\oplus\ell^1. $$
The result sought is just the standard application of the UBP for the duality $c'=\mathbb R\oplus\ell^1$, disguised by these isomorphisms.
This is to give more details to the clever solution by @Ruy.
$S=\{\mathbf{y}\in\mathbb{C}^{\mathbb{N}}:\sum_n\mathbf{x}(n)\,\text{converges}\}$ is a linear space over $\mathbb{C}$ and $$\|\mathbf{y}\|_S:=\sup_n\Big|\sum^n_{j=1}\mathbf{y}(n)\Big|$$ defines a norm on $S$. Let $\mathcal{c}=\{\mathbf{x}\in\ell_\infty(\mathbb{N}): \lim_n\mathbf{x}(n)\,\text{exists}\}$. This is a well known Banach space under the $\|\;\|_\infty$ norm. Furthermore, $\mathcal{c}^*\simeq\mathbb{R}\times\ell_1(\mathbb{N})$.
Consider the linear map $R:\mathcal{c}\rightarrow S$ given by $R\mathbf{x}(1)=\mathbf{x}(1)$ and $R\mathbf{x}(n)=\mathbf{x}(n)-\mathbf{x}(n-1)$ for $n\geq2$. Clearly $R$ is invertible and $R^{-1}\mathbf{y}(n)=\sum^n_{j=1}\mathbf{y}(j)$ for all $n\in\mathbb{N}$
It is clear that $$\begin{align} \|R\mathbf{x}\|_S=\sup_n|\mathbf{x}(n)|=\|\mathbf{x}\|_\infty\tag{0}\label{zero} \end{align}$$ Thus, $S$ and $\mathcal{c}$ are isometric and so, $(S,\|\;\|_S)$ is a Banach space.
If $\lambda\in S^*$ then $\lambda=(\lambda\circ R)\in \mathcal{c}^*$ and so, there is a unique $[\beta_{\lambda},-\mathbf{b}_\lambda]\in\mathbb{R}\times\ell_1(\mathbb{N})$ such that $$\begin{align} \lambda(\mathbf{x})=\beta_\lambda\sum_m\mathbf{x}(m) - \left(\sum_n\mathbf{b}_\lambda(n)\Big(\sum^n_{j=1}\mathbf{x}(j)\Big)\right)\tag{1}\label{one} \end{align}$$ As $S$ and $\mathcal{c}$ are isometric, so are $S^*$ and $\mathcal{c}^*$ ($R^\dagger:S^*\rightarrow \mathcal{c}^*$ given by $\lambda\mapsto\lambda\circ R$ is an isometry), and $$\begin{align} \|\lambda\|_{S^*}=\|R^*\lambda\|_{\mathcal{c}^*}=|\beta_\lambda|+\|\mathbf{b}_\lambda\|_{\ell_1}\tag{2}\label{two}\end{align} $$
$D=\{\mathbf{x}\in\mathbb{C}^{\mathbb{N}}: \sum^\infty_{n=1}|\mathbf{x}(n+1)-\mathbf{x}(n)|<\infty\}$ is a linear space on $\mathbb{C}$, and $$ \newcommand\vertiii[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}} \vertiii{\mathbf{x}}=|\mathbf{x}(1)|+\sum_{n\geq1}|\mathbf{x}(n+1)-\mathbf{x}(n)| $$ defines a norm on $D$.
Consider the linear map $ T:(D,\newcommand\vertiii[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}} \vertiii{\,})\rightarrow\mathbb{R}\times\ell_1(\mathbb{N}) $ given by $(T\mathbf{x})_1=\mathbf{x}(1)$ and $(T\mathbf{x})_2(n)=\mathbf{x}(n+1)-\mathbf{x}(n)$ for $n\geq1$. Clearly, $T$ is invertible and $T^{-1}[r,\mathbf{y}](1)=r$ and $T^{-1}[r,\mathbf{y}](n)=r+\sum^{n-1}_{j=1}\mathbf{y}(j)$ for $n\geq2$. Then $$\begin{align} \|T\mathbf{x}\|_{\mathbf{R}\otimes\ell_1}=|(T\mathbf{x})_1|+ \|(T\mathbf{x})_2\|_{\ell_1}=\newcommand\vertiii[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}} \vertiii{\mathbf{x}}\tag{3}\label{three} \end{align}$$ This implies that $D$ and $\mathbb{R}\times\ell_1(\mathbb{N})$ are isometric. In particular, $(D,\newcommand\vertiii[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}} \vertiii{\,})$ is a Banach space.
We have all the ingredients to provide a proof for sufficiency in the statement in the OP:
For each $n\in\mathbb{N}$, $$\lambda_n(\mathbf{y}):=\sum^n_{k=1}a_n\mathbf{y}(n)=a_n\sum^n_{j=1}\mathbf{y}(k) - \left(\sum^{n-1}_{k=1}(a_{k+1}-a_k)\Big(\sum^k_{j=1}\mathbf{y}(j)\Big)\right) $$ which corresponds to the element $[\beta_n,-\mathbf{b}_n]\in S^*$ with $\beta=0$, $\mathbf{b}_n(k)=a_{k+1}-a_k$, for $1\leq k<n$, $\mathbf{b}_n(n)=a_n$, and $\mathbf{b}_n(k)=0$ for $k>n$. Since for each $\mathbf{y}\in S$ the series $\sum^na_n\mathbf{y}(n)$ converges, the conditions of the uniform boundedness principle are satisfied. Hence $$\begin{align} \sup_n\big(|a_n|+\sum^{n-1}_{k=1}|a_{k+1}-a_k|\big)<\infty \tag{4}\label{four} \end{align} $$ and the desired conclusion follows.
