# Why of this bound $||u_{m0}|| \leq ||u_0||$, where $u_{m0}$ is a projection of $u_0$?

I'm dealing with Navier-Stokes equations, using the book of Teman. Theres a bound of a projection of a function that i didn't understood, so i will introduce the main concepts with i am dealing with.

Consider $$u_m=\sum_{i=1}^m g_{im}(t)w_i$$ an approximation of $$u$$ via galerkin method, where $$w_i$$ are the basis and $$g_{im}(t)$$ the coeficients, $$u_m(0)=u_{m0}$$ and $$u(0)=u_0$$.

The space $$V$$ is basecally the closure, in $$H^1_0(\Omega)$$, of the set of all function $$C^\infty$$ with support on a open set $$\Omega \subset \mathbb{R}^2$$, with it's divergent being null.

Let $$\{w_j\}$$ be a base of $$V$$ which we can take $$u_{m0}$$ as a projection in $$V$$ of $$u_0$$ spanned by $$w_1,w_2, \cdots,w_m$$.

So, why this bound is valid, $$||u_{m0}|| \leq ||u_0||$$?

$$||\cdot||$$ is the norm in $$H^1_0$$.

## 1 Answer

Usually, in the context of Galerkin method, "base" of a separable normed space $$X$$ means a linearly indpendent countable set whose span is dense in $$X$$ (the existence of this family follows from the separability - see Lemma 4.1, p. 83, in Le Dret). On the other hand, in the context of Hilber spaces, "base" usually means a maximal orthonormal set which is characterized as follows:

Theorem. Let $$\{e_j\}_{j\in J}$$ be an orthonormal family in a Hilbert space $$H$$. The following assertions are equivalent:

• $$\{e_j\}_{j\in J}$$ is maximal.
• $$x=\sum_{j\in J}(x,e_j)e_j$$ for all $$x\in \mathcal{H}$$.
• $$\|x\|^2=\sum_{j\in J}|(x,e_j)|^2$$ for all $$x\in \mathcal{H}$$.
• $$\overline{\operatorname{span}\{e_j\mid j\in J\}}=\mathcal{H}$$.

Proof: See Corollary 12.8, p. 532, in Knapp and Proposition 2.3, p. 479, in Taylor.

From this theorem, if $$X$$ is Hilbert separable, then a base in the "Hilbert space sense" is also a base in the "Galerking sense".

Therefore, assuming that $$(w_j)_{j\in\mathbb N}$$ is base of $$V$$ (which is Hilbert separable) in the "Hilbert space sense", using the above theorem we conclude that

\begin{aligned} \|u_{m0}\|^2&=(u_{m0},u_{m0})\\ &=\left(\sum_{i=1}^m(u_0,w_i)w_i,\sum_{j=1}^m(u_0,w_j)w_j\right)\\ &=\sum_{i=1}^m(u_0,w_i)\sum_{j=1}^m\overline{(u_0,w_j)}\left(w_i,w_j\right)\\ &=\sum_{i=1}^m|(u_0,w_i)|^2\left(w_i,w_i\right)\\ &=\sum_{i=1}^m|(u_0,w_i)|^2\\ &\leq \sum_{i=1}^\infty|(u_0,w_i)|^2\\ &=\|u_0\|^2 \end{aligned}