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