1
$\begingroup$

Wirtinger's inequality for one-dimensional functions states that if $f(x)$, $f'(x) = \frac{df(x)}{dx}$ $\in$ $\mathcal{L}^2(a,b)$ and either $f(a) = 0$ or $f(b) = 0$ then \begin{equation} \int_{a}^{b} |f(x)|^2dx \leq \frac{4}{\pi^2}(b-a)^2\int_{a}^{b}|f'(x)|^2dx, \end{equation} where $\mathcal{L}^2(a,b)$ is the space of square integrable functions defined on the closed interval $[a,b]$.

I am wondering if there is an equivalent theorem for multivariate functions, i.e., if we have a square integrable function $f(\mathbf{x})$ in some compact domain $Q$, can we say that \begin{equation} \int_{Q} |f(\mathbf{x})|^2d\mathbf{x} \leq C\int_{Q}\|\nabla{f(\mathbf{x})}\|^2d\mathbf{x}, \end{equation} where $\|.\|$ indicates $\ell_2$ norm of a vector. For this to hold true may be the function $f(\mathbf{x})$ has to be equal to zero on the boundary of Q? Can we also obtain C?

$\endgroup$
0
$\begingroup$

This generalization is called the Poincare inequality, and is a fundamental result in PDE. There is a more general result called Sobolev inequality (Theorem 7.10 in Gilbarg and Trudinger), which states that if $f$ is $C^1$ of compact support, then

$$(*) \bigg( \int_Q |f(x)|^{p^*} dx \bigg)^{1/p^*} \leq C \bigg( \int_Q |\nabla f(x)|^p dx\bigg)^{1/p}$$

for all $p<n$ and $p^* = pn/(n-p)$. The proof is not difficult, just by applying fundamental theorem of Calculus and some manipulation of the integral. Using this, if $n\geq 3$, set $p=2$, then $2^*>2$ and by Holder inequality, write $q$ such that $1/p + 1/q = 1$,

$$\sqrt{\int_Q |f(x)|^2 dx} \leq \bigg( \int_Q |f(x)|^{2^*} dx\bigg)^{1/2^*} \bigg(\int 1 dx\bigg)^{1/q} \leq Vol(Q)^{1/q} C \sqrt{\int_Q |\nabla f(x)|^2 dx}$$

Squaring both sides would obtain Poincare inequality. The constant $C$ in (*) is called the Sobolev constant and are closely related to $Q$. You might search the term "Isoperimetric inequality" to get more information.

On the other hand, the $C$ in Poincare inequality can be viewed as the first eigenvalue of the problem

$$ -\Delta f = \lambda f,\ \ f|_{\partial Q} = 0\ .$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.