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It is true in one dimension that $H^1$ is continuously embedded in $L^{\infty}.$

Now, consider a compact interval $[0,1]$ with a partition $I_n:=([m/n,(m+1)/n])_{m \in \left\{0,...,n-1 \right\}}$ and a function $f$ on $I_n$ that is constant on every interval $[m/n,(m+1)/n].$

Then we can define the discrete derivative $f'(x)=n\left(\tfrac{f\left(\tfrac{x+1}{n}\right)-f\left(\tfrac{x-1}{n}\right)}{2}\right)$ and modify it accordingly on the boundary of the partition.

If we then define the $H^1_n$ norm on the space of such functions $$\Vert f \Vert _{H^1_n} := \sqrt{ \frac{1}{n} \sum_{i=1}^n \left(\left\Vert f\left( \frac{i}{n} \right) \right\Vert^2 +\left\Vert f'\left( \frac{i}{n} \right) \right\Vert^2\right) },$$

it is obvious that $H^1_n$ embeds of course continuously into $L^{\infty}.$

However, I am wondering whether $\Vert f \Vert_{L^{\infty}} \le C \Vert f \Vert_{H^1_n}$ holds for a constant independent of $n$ and $f$? This is to say, is this a discrete approximation of the Sobolev embedding theorem?

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  • $\begingroup$ You probably meant $2/n$ instead of $2n$ in the denominator in the definition of the discrete derivative. Is your constant allowed to depend on $f$? If not there is no such constant (consider the function which is 1 on one interval and zero on all others). $\endgroup$ – user35593 May 7 at 15:05
  • $\begingroup$ @user35593 no, the constant must not depend on $f$ but in this case, the $H^1_n$ norm would blow up as well, no? Because of the derivative. $\endgroup$ – AlgebraicGeometer May 7 at 17:12
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    $\begingroup$ I think this works, and it should in fact become obvious when you switch to the more natural convention of interpolating the $n$ values of an element $f\in H^1_n$ linearly (rather than make the function piecewise constant). Then your discrete derivative is the actual derivative, and you're just computing the conventional $H^1$ norm. $\endgroup$ – Christian Remling May 7 at 17:19
  • $\begingroup$ @ChristianRemling but what is your argument that linear interpolation is the same as the piecewise constant one I am using? $\endgroup$ – AlgebraicGeometer May 7 at 18:21
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    $\begingroup$ What I'm trying to say is that the constant for the usual $L^{\infty}/H^1$ embedding will work for your $H^1_n$ norm also because that norm is the $H^1$ norm of the linearly interpolated function, and for the $L^{\infty}$ norm, it doesn't matter whether we interpolate linearly or piecewise constant. $\endgroup$ – Christian Remling May 7 at 18:32

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