# Bounded sequence has Cauchy subsequence w.r.t. $(x|y)_0:=\sum^\infty_{n=1} 2^{-n}\phi_n(x)\phi_n(y).$

Let $X$ be a separable reflexive real Banach space and let $(\phi_n)$ be a dense sequence in $$\{ \phi\in X'\,|\,\|\phi\|\leq 1 \}.$$ Consider in $X$ the scalar product $( \cdot| \cdot)_0$ defined by $$(x|y)_0:=\sum^\infty_{n=1} 2^{-n}\phi_n(x)\phi_n(y).$$

Prove that:

(a) if $\|\cdot\|_0$ is the norm induced by such scalar product, then $\|x\|_0\leq \|x\|$ for all $x \in X$,

(b) every bounded sequence in $X$ admits a Cauchy subsequence with respect to the norm $\|\cdot\|_0$.

I did the first request while I had problems with b).

My attempts and questions:

a) $\|x\|^2_0:=\sum^\infty_{n=1} 2^{-n}(\phi_n(x))^2\leq\sum^\infty_{n=1} 2^{-n}\|\phi_n\|^2\|x\|^2\leq \sum^\infty_{n=1} 2^{-n}\|x\|^2=\|x\|^2$.

b) This request gave me some problems. My attempts: let $x_n$ be a bounded sequence. Since $X$ is reflexive there exists weakly convergent subsequence $x_{n_k}$... my idea is to prove that this sub-sequence is Cauchy with respect to $\|\cdot\|_0$ but I do not know how to prove it. Since $x_{n_k} \rightharpoonup x$ we have in particular that $\{\phi_n(x_{n_k}) \}_k$ is Cauchy in $\mathbb{R}$ for every $n$ fixed. Could we try to use this fact? What do you think of my idea? Is this the right way to solve the exercise?
Where it is necessary to use that $X$ is separable and that $(\phi_n)$ is dense in the ball?

I would like to have a hint for concluding point b).

UPDATE: my solution of point b) after the hints of @Kavi Rama Murthy.

Let $x_n$ be a bounded sequence. Since $X$ it is reflexive there exists a subsequence $x_{n_k}$ weakly convergent to $x\in X$. I claim that it is a Cauchy subsequence. For simplicity in the notation we denote this subsequence by $x_k$. Let $\varepsilon>0$. Since $\sum^\infty_{n=1} 2^{-n}$ is convergent there exists $N\in \mathbb{N}$ s.t. $$\|x_k-x_j\|^2_0=\sum^\infty_{n=1} 2^{-n}(\phi_n(x_k-x_j))^2=\sum^N_{n=1} 2^{-n}(\phi_n(x_k-x_j))^2+\sum^\infty_{n=N+1} 2^{-n}(\phi_n(x_k-x_j))^2$$ $$< \sum^N_{n=1} 2^{-n}(\phi_n(x_k-x_j))^2+\varepsilon^2/2\qquad\,\,\,\,\,\, (1)$$

By definition of weak convergence we have $\phi_n(x_k)\to\phi_n(x)$ for every $n$ fixed, in particular $\{\phi_n(x_k)\}_k$ is a Cauchy sequence for every $n$ fixed, that is there exists $k_0(n,\varepsilon)\in \mathbb{N}$ s.t. $$|\varphi_n(x_k-x_j)|<\dfrac{\varepsilon^2}{2\sum^N_{n=1} 2^{-n}}\qquad \forall \,i,j>k_0(n,\varepsilon)$$ Let $\overline k_0(\varepsilon):=\max\{k_0(1,\varepsilon),k_0(2,\varepsilon),...,k_0(N,\varepsilon)\}$, then for all $i,j>\overline k_0(\varepsilon)$ we have $$\sum^N_{n=1} 2^{-n}(\phi_n(x_k-x_j))^2 < \varepsilon^2/2$$ and by $(1)$ the proof is complete.

It is all correct? I'm also interested in proof different from mine.

Given $\epsilon >0$ choose $N$ such that $\sum_{n=N+1}^\infty \frac 1 {2^{n}} <\epsilon$. Then $||x-y||_0^{2}=\sum 2^{-n}(\phi_n(x-y))^{2}<\epsilon ||x-y||^{2}+\sum_{n=1}^{N}2^{-n}(\phi_n(x-y))^{2}$. Use a diagonal argument to get one subsequence $\{n_k\}$ such that $\phi_n (x_{n_{k}} )$ converges for each $n$. Replace $x$ by $x_{n_{k}}$, $y$ by $x_{n_{j}}$ and take the limit as $j,k \to \infty$.
• "Use a diagonal argument to get one subsequence $\{n_k\}$ such that $\phi (x_{n_{k}} )$ converges for each $n$" Is there a misprint? You wanted to write $\phi_n (x_{n_{k}} )$? Anyway, were we use the fact that $X$ is separable and that $(\phi_n)$ is dense in the ball? – Ef_Ci Jul 2 '18 at 7:01
• @Ef_Ci I have changed $\phi$ to $\phi_n$. The fact that $\{\phi_n\}$ is dense is only required to show that $||.||_0$ is a norm. it is not required to prove existence of the subsequence. – Kavi Rama Murthy Jul 2 '18 at 7:22