# Discrete Sobolev embedding

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?

• 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). – user35593 May 7 at 15:05
• @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. – AlgebraicGeometer May 7 at 17:12
• 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. – Christian Remling May 7 at 17:19
• @ChristianRemling but what is your argument that linear interpolation is the same as the piecewise constant one I am using? – AlgebraicGeometer May 7 at 18:21
• 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. – Christian Remling May 7 at 18:32