How to show that a given sequence doesn't satisfy a given condition If $\{ f_k \}_{k=1}^{\infty}$ is a Riesz basis,  how to prove that $\{ f_k + f_{k+1} \}_{k=1}^{\infty}$ does not satisfy the frame condition.
Since  $\{ f_k \}_{k=1}^{\infty}$ is a frame for $\mathbb{C}^n,$ by definition there exists constants $A, B >0$ such that 
$$A \| f \|^2 \leq \sum_{k=1}^{\infty}|\left\langle f , f_k \rangle\right|^2 \leq B \| f \|^2.$$  To show that the given sequence is not a frame we should prove that either bound doesn't hold. So computing the sum
$$\sum_{k=1}^{\infty}|\left\langle f , f_k+f_{k+1} \rangle\right|^2 = \sum_{k=1}^{\infty}|\left\langle f , f_k \rangle + \langle f , f_{k+1} \rangle\right|^2.$$
I don't know how to continue from here onwards. Any help is much appreicated.
 A: Since $\{ f_k \}_{k=1}^{\infty}$ is a Riesz basis for a Hilbert space, $\mathcal{H},$ there exist constants $A, B > 0$ such that
$$A \| f \|^2 \leq \sum_{k=1}^{\infty}|\left\langle f , f_k \rangle\right|^2 \leq B \| f \|^2.~~(*)$$ 
Let $\{ g_k \}_{k \in \mathbb{Z}}$ be the biorthogonal basis associated with $\{  f_k \}_{k \in \mathbb{Z}}.$ Let 
$$h_j = \sum_{k=1}^{j} (-1)^{k} g_k$$
Claim:
(i) $\displaystyle{\sum_{k \in \mathbb{Z}} |\langle h_j < f_{k}+f_{k+1} \rangle|}^2 = 2.$ 
(ii) $\| h_j \|^2 \geq j/{B},$ where $B$ is an upper frame bound for $\{ f_k \}_{k \in \mathbb{Z}}.$ 
(i) follows from the following computation.
$$\sum_{k=1}^{j} |\langle h_j , f_{k}+f_{k+1} \rangle|^2  = \sum_{k=1}^{j} |\langle h_j , f_{k} \rangle + \langle h_j , f_{k+1}  \rangle|^2  
=\sum_{k=1}^{\infty} \langle \sum_{j=1}^{i} (-1)^i g_{i} , f_k  \rangle + \langle \sum_{j=1}^{i} (-1)^i g_{i} , f_{k+1}  \rangle|^2 
=\sum_{k=1}^{\infty} \left| \sum_{j=1}^{i} (-1)^i \langle  g_{i} , f_k  \rangle +\sum_{j=1}^{i} (-1)^i \langle g_{i} , f_{k+1}  \rangle \right|^2 
=\sum_{k=1}^{\infty} \left| \sum_{j=1}^{i} (-1)^i \cdot (\delta_{i,k} + \delta_{i,k+1}) \right|^2=2.$$
Now, consider the upper frame condition in $(*).$ Since it is true for every $f \in \mathcal{H},$ with $f=h_j$ we have
$$\sum_{k=1}^{\infty} |\langle h_j,f_k  \rangle|^2 \leq B \| h_j  \|^2.$$
Using the given expression for $h_j,$ we have
$$\sum_{k=1}^{\infty} \left| \sum_{i=1}^{j} (-1)^i \langle g_i,f_k  \rangle \right|^2 \leq B \| h_j  \|^2.$$
From this it follows that $ j \leq B \| h_j \|^2$ or $\| h_j \|^2 \geq j/{B},$ as desired.
Now, claim (i) states that the upper frame condition holds for $\{ f_k + f_{k+1} \}.$ Since claim (ii) holds for all $j,$ if follows that the lower frame condition doesn't hold for  $\{ f_k + f_{k+1} \}.$ Hence it is not a frame.
A: On page $66$ of the book you have linked you can find the following characterisation of a Riesz basis:

$\{f_k\}_k$ is a Riesz basis in $\mathcal H$ iff $\overline{\mathrm{span}\{f_k\}_k}=\mathcal H$ and $\exists A,B >0$ so that for all finite sequences $c_n$ you have:
  $$A\sum_k |c_k|^2 ≤\left\|\sum_k c_k f_k \right\|^2≤B\sum_k |c_k|^2\tag{1}$$

So if we want to show that $\{f_k+f_{k+1}\}$ cannot be a Riesz basis if $\{f_k\}$ is one, we should show one of the properties does not hold.
Making use of the inequality $\|a+b\|^2≤2\|a\|^2+2\|b\|^2$ which holds in Hilbert spaces:
$$\left\|\sum_k c_k (f_k+f_{k+1})\right\|^2=\left\| \sum_k c_k f_k + \sum_k c_{k+1}f_{k+1}\right\|^2≤2\left\| \sum_k c_k f_k\right\|^2 +2\left\| \sum_k c_{k+1}f_{k+1}\right\|^2$$
so from $f_k$ being Riesz it follows there exists a $B$ with:
$$\left\|\sum_k c_k (f_k+f_{k+1})\right\|^2≤4B\sum_k |c_k|^2$$
And it is the other part of $(1)$ that must fail. However here if we take $c_k=(-1)^k$ for $k\in\{1,..,n\}$ we find:
$$A\sum_k |c_k|^2=n A \overset{!}{≤}\left\|\sum_k (-1)^k (f_k+f_{k+1})\right\|^2=\|f_1-f_{n+1}\|^2≤2\|f_1\|^2+2\|f_{n+1}\|\tag{2}2$$
Now the result that the elements Riesz basis have bounded norm will give a contradiction, namely the right hand side of $(2)$ will be bounded independently of $n$ whereas the left and side will grow unboundedly.
How can we see that the $f_n$ must have bounded norm? Well the righthand equation of $(1)$ gives you with $c_n=\delta_{n,k}$:
$$\|f_n\|^2≤B$$
for all $n$.

I had also written down the case for an ONB before I looked into the book, I figured I might give it here also as it uses your definition:
If you choose $f=\sum_k a_k(f_k - f_{k+1})$ you have that
$$\langle f,f_k\rangle = a_k-a_{k-1}$$
so
$$|\langle f, f_k\rangle + \langle f,f_{k+1}\rangle |^2=|a_k-a_{k-1}+a_{k+1}-a_{k}|^2=|a_{k+1}-a_{k-1}|^2$$
Make all the $a_k$ zero for $k$ odd and you get
$$\|f\|^2=\sum_k |\langle f,f_k\rangle|^2=\sum_k |a_{2k}|^2\\
\sum_k|\langle f,f_k+f_{k+1}\rangle|^2=\sum_k |a_{2k}-a_{2k-2}|^2$$
If you take $a_{2k}=\frac1{x^{k}}$ with $|x|<1$ then the upper thing is $=\frac1{1-x^2}$ and the lower thing is
$$\sum_k \left|\frac{1}{x^{k}}-\frac1{x^{(k+1)}}\right|^2=\sum_k \left|\frac{x^{k+1}-x^k}{x^k x^{k+1}}\right|^2=\sum_k \left|\frac{x-1}{x}\frac1{x^k}\right|^2=\left|\frac{x-1}x\right|^2\cdot\left|\frac1{1-x^2}\right|$$
So the ratio is $\left|\frac{x-1}x\right|^2$ which for $x\to1$ can be made as large as you like.
