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Background.I have been studying the proof of the unboundedness of the eigenvalues of a quadratic functional $$I[\phi]=\int_{\Omega}\left(p|\nabla\phi|^2+q\phi^2\right)d\boldsymbol{x}$$ subject to $$H[\phi]=\int_{\Omega}\rho\phi^2d\boldsymbol{x}=1$$ ($p,q,\rho>0$, $p\in C^1(\Omega)$) The crux of the argument by contradiction as set out in Courant & Hilbert and originally due to Rellich is that assuming the eigenvalues $\lambda_n=I[\phi_n]$ are bounded ($\phi_n$ being the corresponding eigenfunction) it is possible to extract a uniformly convergence subsequence of the eigenfunctions. Thus $$\lim_{m,n\to\infty}H[\phi_m-\phi_n]=0$$ whereas from orthonormality condition it follows that $$H[\phi_m-\phi_n]=2$$ for $n\ne m$. In this they rely on Arzela-Ascoli theorem calling it "accumulation principle". In one-dimensional case C&H prove uniform boundedness and equicontinuity directly, for example, using Schwarz inequality $$f(x_1)-f(x_2)=\int_{x_1}^{x_2}f'(x)dx \le\sqrt{\int_{x_1}^{x_2}f'^2(x)dx\cdot\int_{x_1}^{x_2}1^2dx}$$ Hence, if $\int_{x_1}^{x_2}f'^2(x)dx<M$ we have $$(f(x_1)-f(x_2))^2<M|x_2-x_1|$$ In case of 2 independent variables C&H refer to a lemma due to Rellich proved in the above mentioned article. This lemma asserts that if a for a set of functions $\phi(x,y)$ defined in the domain $\Omega$ $$\int_{\Omega}\phi^2dxdy\quad \text{and} \quad \int_{\Omega}(\phi_x^2+\phi_y^2)dxdy$$ are uniformly bounded it is possible to select a subsequence $\phi_n$ such that $$\lim_{m,n\to\infty}\int_{\Omega}(\phi_m-\phi_n)^2dxdy=0$$ What I tried. As opposed to Rellich's procedure I attempted to extend naively the above one-dimensional argument to the case of many variables to prove equicontinuity, hoping to invoke an generalised Arcela-Ascoli theorem to complete the task Consider the difference of the values of $\phi$ at two distinct points:

$$\begin{aligned}\phi\left(x_{2},y_{2}\right)-\phi\left(x_{1},y_{1}\right) & =\phi\left(x_{2},y_{2}\right)-\phi\left(x_{2},y_{1}\right)+\phi\left(x_{2},y_{1}\right)-\phi\left(x_{1},y_{1}\right)\\ & =\int_{y_{1}}^{y_{2}}\left.\frac{\partial\phi}{\partial y}\right|_{x=x_{2}}dy+\int_{x_{1}}^{x_{2}}\left.\frac{\partial\phi}{\partial x}\right|_{y=y_{1}}dx \end{aligned}$$

Writing $\phi_{x}\left(x,y_{1}\right)=\left.\frac{\partial\phi}{\partial x}\right|_{y=y_{1}}$ and $\phi_{y}\left(x_{2},y\right)=\left.\frac{\partial\phi}{\partial y}\right|_{x=x_{2}}$ and Applying Cauchy-Schwarz inequality:$$\begin{aligned}\phi\left(x_{2},y_{2}\right)-\phi\left(x_{1},y_{1}\right) & \le\sqrt{\int_{y_{1}}^{y_{2}}\phi_{y}^{2}\left(x_{2},y\right)dy\cdot\int_{y_{1}}^{y_{2}}1^{2}dy}\\ & +\sqrt{\int_{x_{1}}^{x_{2}}\phi_{x}^{2}\left(x,y_{1}\right)dx\cdot\int_{x_{1}}^{x_{2}}1^{2}dx}\\ & =\sqrt{\int_{y_{1}}^{y_{2}}\phi_{y}^{2}\left(x_{2},y\right)dy}\cdot\sqrt{y_{2}-y_{1}}\\ & +\sqrt{\int_{x_{1}}^{x_{2}}\phi_{x}^{2}\left(x,y_{1}\right)dx}\cdot\sqrt{x_{2}-x_{1}} \end{aligned}$$

Now transform the last expression as follows:$$\begin{array}{cc} \phi\left(x_{2},y_{2}\right)-\phi\left(x_{1},y_{1}\right) & \le\sqrt{\int_{x_{1}}^{x_{2}}\int_{y_{1}}^{y_{2}}\phi_{y}^{2}\left(x_{2},y\right)dxdy}\cdot\sqrt{\frac{y_{2}-y_{1}}{x_{2}-x_{1}}}\\ & +\sqrt{\int_{x_{1}}^{x_{2}}\int_{y_{1}}^{y_{2}}\phi_{x}^{2}\left(x,y_{1}\right)dxdy}\sqrt{\frac{x_{2}-x_{1}}{y_{2}-y_{1}}} \end{array}$$

Applying Hölder's inequality:$$\phi\left(x_{2},y_{2}\right)-\phi\left(x_{1},y_{1}\right)\le\sqrt{\iint_{g}\left[\phi_{x}^{2}\left(x,y_{1}\right)+\phi_{y}^{2}\left(x_{2},y\right)\right]dxdy}\frac{\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}}{\sqrt{S\left(g\right)}}$$

Since the integral of the squared gradient is uniformly bounded, we deduce:

$$\sqrt{S\left(g\right)}\phi\left(x_{2},y_{2}\right)-\sqrt{S\left(g\right)}\phi\left(x_{1},y_{1}\right)\le C\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$$

Where $S(g)$ is the area of the rectangle.

Where I got stuck. Now that I end up with an expression somewhat similar to the one that turns out in Rellich's article, I don't know how to proceed and whether it makes sense to. Will the generalised Arzela-Ascoli be really applicable here? What assumptions am I making, or should I make with regards to $\Omega$? If this draft argument is valid what would be the next step? Iterate the procedure in some way? This article by Terence Tao says that the answer to my attempt is "barely no", but what does it actually mean?

Many thanks in advance for those who reads this till the end.

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1 Answer 1

up vote 1 down vote accepted

In more than one dimension having an $L^2$ derivative is not enough to control even the $\infty$-norm of your functions. What is true is that you can control the $L^{\frac{2n}{n-2}}$ norm (for $n=2$ you can control all $p<\infty$) but in general this is a sharp result.

Moreover, even if you only want equicontinuity you would need something like Morrey's inequality, which in the 1-dimensional case is precisely the estimate you give, but in general this is only true for functions having derivatives (along with the function) in $L^p$ for $p>n$.

On the other hand Rellich's approach works because $H^1(\Omega)=\{ f\in L^2: \nabla f \in L^2\}$ embbeds compactly in $L^2(\Omega)$ (this is particular case of Rellich's compactness theorem) which is what you wrote.

To really understand what's going on here I recomend reading up on Sobolev spaces (Evans' "Partial Differential Equations" is a good introductory reference).

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Yes, I'd second reading about Sobolev spaces. The $L^2$ case is easiest, partly because it allows use of Fourier series and/or Fourier transforms in a straightforward fashion, and the corresponding Plancherel-Parseval results. "Sobolev imbedding" asserts qualitatively that sufficient $L^2$ differentiability implies classical differentiability, with a discrepancy that grows with dimension. It's cool and useful stuff. –  paul garrett Aug 1 '12 at 0:20
    
Thank you for a thorough answer. I did realize incidentally that some experience in Sobolev spaces was needed here as they keep turning up in all references. Will check out that book. –  Valentin Aug 1 '12 at 7:34

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