Let $I=(0,1)$ and let $C>1$ be a constant. Let $L^2(I)$ and $H^1(I)$ be the standard Sobolev spaces on $I$. Suppose that $U$ is a **subspace** of $H^1(I)$ with the additional property that:
$$ \| u\|\_{H^1(I)} \leq C \|u\|\_{L^2(I)}\quad \forall\, u\in U.$$
By compact embedding of $H^1(I)\subset L^2(I)$, it is trivial to see that $U$ must be finite dimensional (note here that the fact that $U$ is a subspace is vital).


My question is whether one can put an explicit bound on the dimension of $U$ in terms of the constant $C$.


Thanks,



Bounds on dimension of a subspace
Just use the Wirtinger's inequality that says that if $u=0$ at the center of an interval $I\subset \mathbb R$, then $\int\_I|u|^2\le (|I|/\pi)^2\int\_I|u'|^2$. Thus, if the dimension is $>n$, we can create a function that vanishes at the centers of $n$ subintervals of the interval $(0,1)$ of length $1/n$ and have $\|u\|\_{L^2}\le \frac 1{\pi n}\|u'\|\_{L^2}$, so we must have $\pi n\le C$. On the other hand, we can take trigonometric polynomials $f(x)=\sum\_{|k|\le N}c\_ke^{2\pi i k x}$ and observe that for them $\|f'\|\_{L^2}\le 2\pi N\|f\|\_{L^2}$ and the dimension is $2N+1$, so the bound is tight if we use the norm of the derivative alone. Including the norm of the function into the definition of $H^1$ norm changes the answer by $O(1)$ (and brings it down), so the inequality $\rm{dim\,}U\le \frac{C}\pi+1$ still holds and is nearly the best possible one. That is, of course, in a perfect alignment with Giorgio's comment, but this case is simple enough to do it without formally invoking the spectral theory for the Laplacian.


