Why is point-wise evaluation in H^1 not possible?

404 Views Asked by At

In Section 1 of the paper "Finite element interpolation of nonsmooth functions satisfying boundary conditions" by Scott and Zhang it is stated:

However, the nodal value of the function may not be well-defined if the function under consideration is too "rough". For example, functions in the Sobolev space $H^1$ have no point-wise value in two or more dimensions. In [4], Clément defined an optimal-order interpolation operator using local averaging (regularizing) to define nodal values for functions even in $L^1$.

My Questions:

Why is point-wise evaluation in $H^1$ not possible?

Can you give me a concrete example how this will not work if I want to define an interpolation Operator point-wise.

How can the Clément operator compensate for this lack?

Any help is appreciated!

1

There are 1 best solutions below

1
On BEST ANSWER

First remember that elements of $L^2(X)$ are equivalence classes of functions that are equal outside a null set. This is the reason that in general you can not give function values at specific points, because a set containing a finite number of points is a null set, so changing the values there does not change the element of $L^2(X)$.

One needs more regularity one wants to single out the one representative of the class that allows to give definitive function values. If there is a continuous function in the equivalence class, this would be a good candidate.

The next fact to remember is that $H^s(X)$ is contained in $C^k(X)$, where $X\subset \Bbb R^n$ an open set, if $s>k+\frac n2$. So if $n=2$ then you need $s>1$ to get that all elements of $H^s(X)$ can be represented by continuous functions, or in other words, $H^1(X)\subset C(X)$ only for $n=1$.