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?

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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.

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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.)

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Is the inequality with the $\leq$ replaced with $\geq$ always true? – soup Nov 17 '12 at 16:03
But, it is true with $\le$, assuming $y\ne 0$. See the 2nd part of my answer. – Berci Nov 17 '12 at 16:07
It depends on exactly what you are trying to figure out. Any sequence that converges is bounded, so if the limit is non zero, you can always find a constant so that the equation holds for a particular sequence. This is a property of any sequence converging to a non-zero limit. But if you have a prescribed method for choosing the $a_{im}$ as a function of $y$ (such as Gram-Schmidt, or Galerkin discretization) then you need to say something about how the $a_{im}$ are computed (or at least, some properties of the mapping) if you want to conclude $\|y_m\|\leq C \|y\|$ in general. – copper.hat Nov 17 '12 at 17:06
Maybe this explains what I am getting at: In general you are choosing $y_m \in W_m =\text{sp}\{w_1,...,w_m\}$. For any finite set of indices I can choose $y_m$ arbitrarily in $W_m$ without affecting convergence. So, given any $C$, I can choose $y_1$ so that $\|y_1\|$ is arbitrarily large and violates the inequality (if the limit is non-zero, and $C$ is independent of $y$). – copper.hat Nov 17 '12 at 17:12
One last gasp: The Galerkin discretization has conditions on a bilinear form that correspond, roughly, to saying that the $y_m$ is 'close' to an orthogonal projection on to the $\text{sp}\{w_1,...,w_m\}$. This is what allows one to conclude the existence of such a $C$ in this case. – copper.hat Nov 17 '12 at 17:16