Sum of tensor products is a bounded operator I came across this in a book:
Let $B$ be a Banach space with Schauder basis $\{\omega_j\}$, let $B'$ be its dual with $\{\nu_i\}$ being a corresponding biorthogonal system, i.e. $\langle \omega_j,\nu_i\rangle=\delta_{ji}$. Then it is well-known that the sequence of operators $\{F_n\}$ defined as $$F_n=\sum\limits_{j=1}^n\omega_j \otimes \nu_j$$
is bounded. I am not seeing this. So bounded means there exist some $M>0$ such that for all $n\in\mathbb{N}$ and some $v\in B$ $$\frac{\|F_n v\|}{\|v\|}=\frac{\|\sum\limits^{n}_{j=1}(\omega_j\otimes\nu_j)(v)\|}{\|v\|}\leq M$$
Why is that bounded when I choose an infinite vector and then taking the limit $\lim\limits_{n\to\infty}F_n$?
 A: Let's have a look at the definitions: A Schauder basis $\{\omega_j\}$ is a sequence so that for each $b\in B$ there exists a unique sequence of scalars $(c_j)$ such that $\lim_{n\to\infty}\sum_{j=1}^n c_j\omega_j=b$ (the limit means convergence in norm).
The biorthogonal system means that $v_j$ are the unique functionals $v_j:B\to\mathbb{C}$ such that $v_j(\omega_i)=\delta_{i,j}$ (the Kronecker delta). The wikipedia article on Schauder basis has a proof that these functionals are well-defined, bounded and also, in the unique expansion $b=\sum_{j=1}^\infty c_j\omega_j$, we have $v_j(b)=c_j$.
Edit (Added this Comment): This answer works only if we know that the operators $F_n$ are bounded, since we are applying the principle of uniform boundedness which only applies to collections of bounded operators. Note that knowing that the operators $F_n$ are bounded is actually equivalent to knowing that the functionals $v_j$ are bounded (why?), and, judging from OP's recent comment, this is not the case. I have added a 2nd answer that does not use this assumption.
Now to answer OP's question, we use the principle of uniform boundedness. If we show that for all $v\in B$ we have
$$\sup_{n\ge1}\|F_nv\|<\infty$$
then by the principle of uniform boundedness we are allowed to conclude that $\sup_{n\ge1}\|F_n\|<\infty$.
So let $v\in B$ be fixed. We have $\|F_nv\|=\|\sum_{j=1}^n\omega_j\otimes v_j(v)\|=\|\sum_{j=1}^nv_j(v)\omega_j\|$, so $$\lim_{n\to\infty}\|F_nv\|=\lim_{n\to\infty}\|\sum_{j=1}^nv_j(v)\omega_j\|=\|\lim_{n\to\infty}\sum_{j=1}^nv_j(v)\omega_j\|=\|v\|.$$
Since this limit exists and it is finite, the sequence $\{\|F_nv\|\}$ is bounded, which is what we wanted to show.
A: I am adding a 2nd answer which is more appropriate to the question. My other answer works under the assumption that the operators $F_n$ are bounded operators, so I am not deleting it as it was not known before the comments if OP knew that the operators $F_n$ are bounded. The following proof is due to Banach.
Let $F_n:B\to B$ be the operators defined by $F_n(b)=\sum_{j=1}^n v_j(b)\omega_j$, where $v_j(b)$ are the (unique) scalars appearing in the expansion $b=\sum_{j=1}^\infty v_j(b)\omega_j$ given by the Schauder basis. We do not know whether the linear operators $F_n$ are bounded or not. We are asked to show that there exist some constant $C>0$ s.t. $\|F_n\|\leq C$ for all $n$ (and this will also imply that all these operators are bounded).
We define $\|\cdot\|':B\to[0,\infty)$ by setting $\|b\|':=\sup_{n\ge1}\|F_n(b)\|$. Note that, by the definition of a Schauder basis, we do know that $b=\lim_{n\to\infty}\sum_{j=1}^nv_j(b)\omega_j=\lim_{n\to\infty}F_n(b)$, so the sequence $\|F_n(b)\|$ is bounded for all $b\in B$, i.e. $\|b\|'<\infty$ for all $b$. It is only a matter of calculations to verify that $\|\cdot\|'$ is indeed a norm on $B$.

Claim: $(B,\|\cdot\|')$ is complete.

After we show this claim, the proof is straight-forward: the identity operator $I:(B,\|\cdot\|')\to(B,\|\cdot\|)$ is bounded: indeed, $\|x\|=\lim_{n\to\infty}\|F_n(x)\|\leq\sup_{n\ge1}\|F_nx\|=\|x\|'$. By the open mapping theorem (which can be applied, since both the target and the domain are complete spaces), its inverse, i.e. the identity operator $(B,\|\cdot\|)\to(B,\|\cdot\|')$ is bounded, i.e. there exists $C>0$ so that $\|x\|'\leq C\|x\|$ for all $x\in X$. In particular, $\sup_{n\ge1}\|F_n(x)\|\leq C$ for all $x\in X$ with $\|x\|=1$, i.e. $\sup_{\|x\|=1}\sup_n\|F_n(x)\|\leq C$, i.e. $\sup_n\|F_n\|<\infty$.
The claim is proved by elementary theory, but the proof is very, very technical. I can recommend the interested reader to N.L. Carother's book, "A short course on Banach space theory", theorem 3.1. If OP is interested in this proof and has questions, I will gladly add these details.
