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Let $H=L^2(\Omega)$ and $V=H^1(\Omega)$.

Suppose that $\{v_j\}$ is a basis for $H$ and $V$ (not necessarily orthogonal).

Let $V_m = \text{span}(v_1, ..., v_m)$.

Define a projection operator $P_m:H \to V_m$ satisfying $$(P_m h - h, v_m) = 0 \qquad\text{for all $v_m \in V_m$}.$$

Since $v_j$ is a basis, we can write $h = \sum_{j=1}^\infty a_jv_j$ where $a_j$ are coefficients. Now if $v_j$ were an orthonormal basis of $V$ and an orthogonal basis of $H$, then we simply have $$P_m(h) = \sum_{j=1}^m a_jv_j.$$

Is there any such expression when $v_j$ is not orthogonal?

I am asking because this is the set up used in a Galerkin method. Is there a different way to define projection operators onto the finite dimensional subspace when we have no orthogonal basis?

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1 Answer 1

No matter what, the projection is still orthogonal, so it still takes the form $$\tilde h = P_m(h) = \sum_{j=1}^m a_jv_j$$ i.e. it is in the linear span of the $v_j$'s. If the $v_j$'s are orthonormal, then the coefficients are "easy" to calculate: $$ a_j = \left<h,v_j\right> $$

In general, the coefficients have to satisfy a system of equations. In particular, if you write

$$ \tilde h = \sum_{j=1}^m a_j v_j $$

then you can still solve for the coefficients using the inner product (you just can't do it directly any more):

$$ \left<\tilde h,v_i \right> = \sum_{j=1}^m a_j \left<v_j,v_i\right> $$

which is a system of equations for $a_j$, i.e.

$$ V\begin{pmatrix}a_1 \\ a_2 \\ \vdots \\ a_m \end{pmatrix} = \begin{pmatrix}h_1 \\ h_2 \\ \vdots \\ h_m \end{pmatrix} $$

where $(V)_{ij} = \left<v_j,v_i\right>$ and $h_i = \left<\tilde h,v_i \right>$.

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Thanks for answering. So you say if $h := \sum_{j=1}^\infty a_jv_j$ then $P_m(h) = \sum_{j=1}^m a_jv_j$ no matter what the basis is.. But this has to satisfy the identity $(P_mh-h, w_m) = 0$ but the LHS is $(\sum_{j=m+1}^\infty a_jv_j, \sum_{j=1}^m b_jv_j)$ which I can't show is zero ($w_m \in V_m$) –  weasd Feb 13 at 21:28
    
Yeah, maybe I'm being too hasty. If the $v_j, j>m$ aren't orthogonal to $V_m$, we have a problem. In your setting, is it possible to switch to an orthonormal basis for $V_m^\perp$? I.e. keep $V_m$ with its non-orthogonal basis, but "gram-schmidt" the rest? –  BaronVT Feb 13 at 22:40
    
Hmm I'm afraid probably not. We want to take limits $m \to \infty$ so we need to consider $V_m$ for each $m$ which probably constrains it to be defined as I did above. –  weasd Feb 14 at 9:04
1  
ok, I'm going to take a think on this, and I'll get back to you –  BaronVT Feb 14 at 17:01

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