ca.analysis and odes - The characteristic (indicator) function of a set is not in the Sobolev space H1

The answer posted by Tom, as written is actually not true. A function in $H^1$ will not in general be differential almost everywhere; it depends on the dimension. In one dimension however it is indeed true that $H^1$ functions are differentiable almost everywhere (they are in fact absolutely continuous). There are two ways of seeing it is not in $H^1$. The simple answer is that if you differentiate the characteristic function of say $[0,infty)$ then you will get the dirac measure. However let me just answer your question first:

Answer 1:
Take any smooth compactly supported $phi:mathbb{R} to mathbb{R}$. By definition of weak derivative we have
$int phi g^{prime} dx = - int phi^{prime} g dx$ where I've set $g=1_{[0,infty)}$. This would have to be true for all such $phi$ if the weak derivative existed. Now take $phi^{epsilon}$ to be supported in a neighborhood $(-epsilon,epsilon)$ of $0$. We are making the crucial assumption that $g^{prime}$ is an integrable and hence it follows that $int phi^{epsilon} g^{prime} to 0$ as $epsilon to 0$. However, $phi^{epsilon}$ is smooth and so
$int partial_xphi^{epsilon}(x)g(x)dx = phi^{epsilon}(0)$ since $phi$ was assumed to have compact support in $(-epsilon,epsilon)$. Now just fix $phi^{epsilon}(0)=1$ and we have that $phi^{epsilon}(0) to 0$ by the first integral equality. This is a clear contradiction.

Notice that in fact that this really shows that $g' dx = delta(x)$.

Answer 2:
Take $1_{[0,1]}$ instead so that it is an $L^2([0,1])$ function. This is in fact the fourier transform of a "sinc" function, $sin(k)/k$ up to some normalization constants. If we consider the $H^1$ norm in frequency space we would need $int_0^{infty} |k|^2frac{sin(k)^2}{|k|^2} < infty$ which is clearly false. This requires being at ease with the fourier transform so if you're not, answer 1 is probably best.

It is true in $mathbb{R}^n$ that if $u in W^{1,p}$ for $p > n$ then $u$ is a.e differentiable and equals a.e its weak gradient (see Evans chapter 5). This is to correct what Tom had said although perhaps we was thinking about the $n=1$ case in which case $2 > 1$.

Hope this helps!
Dorian