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It is well-known in $H^2(\mathbb R^3)$ embeds into $L^{\infty}(\mathbb{R}^3).$ Now consider a function $u \in \ell^{\infty}(h\mathbb Z^3)$ and a grid of points $x \in h\mathbb{Z}^3.$

We then define the finite-difference Laplacian

$$(\Delta_hu)(x):=\frac{\left(\sum_{i=1}^3 f(x+he_i)+f(x-he_i)\right)-6 f(x)}{h^2}$$

I wonder, is it true that for some universal $C>0$

$$\Vert u \Vert_{\ell^{\infty}(h\mathbb Z^3)} \le C (\Vert \Delta u \Vert_{\ell^2(h\mathbb Z^3)}+ \Vert u \Vert_{\ell^2(h\mathbb Z^3)})?$$

Here,

$$\Vert u \Vert^2_{\ell^2(h\mathbb Z^3)} = \sum_{x \in h\mathbb Z^3} h^3 \vert u(x) \vert^2 $$

The intuition behind this estimate is that we discretely approximate the continuous setting with $L^2$ and $L^{\infty}$ norms.

Please let me know if you have any questions.

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    $\begingroup$ $\|u\|_{\ell^\infty(I)}=\sup\{|u(i)|: i\in I\} \le \|u\|_{\ell^2(I)}$. $\endgroup$ – Jochen Wengenroth Jul 18 at 10:09
  • $\begingroup$ @JochenWengenroth sorry, the norm here is not the standard $l^2$ norm but a weighted one, so that we 'approximate' the continuous world... $\endgroup$ – Solid State Physicist Jul 18 at 11:37
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Yes, this is true, and there is a proof which closely tracks your intuition. As you know, this estimate can be proved in the continuum by applying the Sobolev embedding twice, first to get $\nabla u \in L^p$ for $p<\frac{2d}{d-2}=6$, and then once more to get $u\in L^\infty$. So for simplicity let me discuss how to get discrete versions of the Sobolev embedding for only one derivative. You can put these together in the same way to get your bound.

We will transfer the Sobolev inequalities in the continuum to the lattice by brute force. First, extend your function $u$ define on the lattice $h\mathbb{Z}^d$ by making it be constant on all cubes of side length $h$ which are centered on a point of $h\mathbb{Z}^d$. Next, mollify this piecewise constant function with the standard mollifier, on length scale $h/10$. Name the resulting smooth function $v$, which is now defined in $\mathbb{R}^d$. You have the following pointwise bounds: \begin{equation} \left\| v \right\|_{L^\infty(z+[-h/2,h/2]^d)} \leq \sup_{z' \sim z} | u(z')| \end{equation} and \begin{equation} |D_h(z)|:=\frac1h\sup_{z'\sim z}|u(z) - u(z')| \leq \frac{C}{h^d} \int_{z+[-h/2,h/2]^d)} |\nabla v|. \end{equation} Here $\sim$ means nearest-neighbor in the $h\mathbb{Z}^d$ lattice. The first bound is pretty easy, the second is true because near the boundary between nearest neighbor cubes, there will be a positive-measure set of points for which the mollifier picks up the difference between $u(z)$ and $u(z')$, and therefore on this set $|\nabla v|$ will be at least proportion to this difference. The proportion of this set in the cube is lower bounded by a constant (which does not depend on $h$).

Now, applying the (continuum) Sobolev inequality to $v$ and putting everything together gives a (discrete) Sobolev inequality for $u$.

There is one more minor point I will mention, which is that you have defined the $H^2$ norm with respect to the Laplacian only (and not the full set of mixed second derivatives). But you can perform a discrete integration by parts (twice) to bound the $\ell^2$ of $D^2_hu$, the full set of (possibly mixed) second-order differences, by the $\ell^2$ norm of $\Delta_h u$ (mimicking the usual proof in the continuum).

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