I want to check that a Riesz Basis is really a (Schauder) basis.
So let $\{Ue_k\}_{k=1}^\infty$ be a Riesz Basis for a Hilbert space $\mathcal{H}$, so $\{e_k\}_{k=1}^\infty$ is an orthonormal basis for $\mathcal{H}$ and $U:\mathcal{H}\to\mathcal{H}$ a bounded bijective operator.
Let $f\in \mathcal{H}$ arbitrary but fixed.
Since $\{e_k\}_{k=1}^\infty$ is a basis, there is a unique coefficient sequence $\{c_k(f)\}_{k=1}^\infty$ s.t. $f=\sum\limits_{k=1}^\infty c_k(f)e_k$, so $$|\sum\limits_{k=1}^nc_k(f)e_k - f|_\mathcal{H} \xrightarrow{n\to\infty}0.$$
$U$ is a bounded operator so $\left\lVert U\right\rVert = M <\infty$. Choose $\{\tilde c_k(f)\}_{k=1}^\infty$ s.t. $\tilde c_k(f) = \frac{1}{M}c_k(f)$. Then $$\begin{align}|\sum\limits_{k=1}^n\tilde c_k(f)Ue_k - f|_\mathcal{H} &= |U\sum\limits_{k=1}^n\tilde c_k(f)e_k -f|_\mathcal{H} \\&\leq |M\sum\limits_{k=1}^n\tilde c_k(f)e_k -f|_\mathcal{H} \\&= |M\sum\limits_{k=1}^n \frac{1}{M}c_k(f)e_k -f|_\mathcal{H} \\&= |\sum\limits_{k=1}^n c_k(f)e_k -f|_\mathcal{H}\xrightarrow{n\to\infty}0.\end{align}$$
Hence $\{\tilde c_k(f)\}_{k=1}^\infty$ is a fitting coefficient sequence for $f$ and the given Riesz Basis but what about the uniqueness (I mean for $M$ you could use any bound of the operator norm)? Is my construction right?
Let $X$ be a Banach space. A sequence of vectors $\left\{f_k\right\}_{k=1}^{\infty}$ belonging to $X$ is a (Schauder) basis for $X$ if, for each $f \in X$, there exist unique scalar coefficients $\left\{c_k(f)\right\}_{k=1}^{\infty}$ such that $$ f=\sum_{k=1}^{\infty} c_k(f) f_k $$ Now we have that, for a Riesz basis, $f_k=U e_k$ and we know that every $f\in \mathcal H$ has an expansion like this: $$f=\sum_k \langle f , e_k \rangle e_k$$ This holds also if we consider $U^{-1}f$ instead of $f$: $$U^{-1}f=\sum_k \langle U^{-1}f , e_k \rangle e_k=\sum_k \langle f , \left(U^{-1}\right)^*e_k \rangle e_k$$ and since $U$ is bijective: $$f=\sum_k \langle f , \left(U^{-1}\right)^*e_k \rangle Ue_k$$ But $\langle f , \left(U^{-1}\right)^*e_k \rangle$ is unique and thus the claim.