Bounds on dimension of a subspace

Question

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,

Answer 1

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.