Basis, dense subset and an inequality Let $V \subset H$, where $V$ is separable in the Hilbert space $H$.
So there is a basis $w_i$ in $V$ such that, for each $m$, $w_1, ..., w_m$ are linearly independent and the finite linear combinations are dense in $V$.
Let $y \in H$, and define $y_m = \sum_{i=1}^m a_{im}w_i$ such that $y_m \to y$ in $H$ as $m \to \infty$.
Then, why is it true that $\lVert y_m \rVert_H \leq C\lVert y \rVert_H$?
I think if the $w_i$ were orthonormal this is true, but they're not. So how to prove this statement?
 A: It is not true. Choose $y=0$ and $a_{1,m} = \frac{1}{m}$. Then $y_m \to y$, but it is never the case that $\|y_m\| \leq C \|y\|$.
Elaboration:
This is because $\|y_m\| = \frac{1}{m} \|w_1\|$, the $w_i$ are linearly independent, hence non-zero. Hence $\|y_m\| = \frac{1}{m} \|w_1\| > 0$ for all $m$. There is no choice of $C$ that will satisfy the equation $\|y_m\| \leq C \|y\|$. 
The above is true even if $w_i$ are orthonormal. 
(I think you need to be more explicit about your choice of $a_{im}$. A 'nice' choice would be to let $y_m$ be the closest point to $y$ in $\text{sp}\{w_i\}_{i=1}^m$. This is what the first part of the answer by Berci does below.)
A: For simplifying calculations, we can assume that $w_m$ is an orthonormal basis, by the following argument:
By the Gram-Shmidt orthogonalization process we can define $v_m$ out of the $w_1,..,w_m$'s, such that $v_1,..v_m$ is an orthonormal basis of $\langle w_1,..,w_m\rangle$. Then rewriting the coordinates accordingly, we have
$$y_m=\sum_{i\le m} b_{i(m)} v_i $$
Then, $||y_m||^2=\sum_{i\le m}(b_{i(m)})^2$, and, using the continuity of $x\mapsto \langle x,v_i\rangle$, we see that $b_{i(m)}$ tends to $\langle y,v_i\rangle=:b_i$ as $m\to\infty$, and that
$$||y||^2=\sum_{i=1}^\infty {b_i}^2.$$
But maybe all these are not needed for your inequality, all is needed that the scalar product is continuous, and hence so is the norm, so
$$||y_m||\to ||y||$$
From which, if $y\ne 0$, the sequence $\displaystyle\frac{||y_m||}{||y||}$ convergences (to 1), hence bounded.
Well, if $y$ happens to be $0$, then the claim can also be false.
