# Equivalence of norms on finite dimensional subset ($L^2$ and $H^1$ norms)

Let $b_i$ be a basis of $H^1(\Omega)$ and define $X=\text{span}(b_1, ..., b_n)$ for $n$ fixed.

Consider $X_1 = (X, \lVert \cdot \rVert_{H^1(\Omega)})$ and $X_1 = (X, \lVert \cdot \rVert_{L^2(\Omega)})$.

Both $X_1$ and $X_2$ are finite-dimensional sets equipped with different norms. So we know that by equivalence of norms, the $L^2$ norm and $H^1$ norm are equivalent (restricted to elements of $X$).

This seems a bit weird to me. I can put any norm on it and I will get a equivalence of norms result to any other norm. Is it correct to think like this?

First of all, there is no finite(!) basis of $H^1(\Omega)$. I assume that you take the $b_i$ to be just linearly independent.
In that case, what you say is correct, but the point here is that the constant of equivalence of the norms is dependent on the $b_i$ and in particular on the number of elements $n$.
If you let $n \rightarrow \infty$, your constants will deteriorate.
• Yeah I just meant $\{b_i\}_{i=1}^\infty$ is a basis. I am taking the first $n$ elements to make $X$. – studentX May 24 '14 at 16:48