# Quick question on definition of derivative in the sense of distribution

Consider a function f such that on $(-\infty,0)$ and $(0,\infty)$, f is differentiable. At 0, there is a point of discontinuity.

e.g. $f(x) = 0$ for $x\leq 0$ and $f(x)=x$ for $x>0$

Then if we are to calculate its weak derivative, what should be its weak derivative at 0? I asked this question in lecture and I was told that whether it is equal to 0 or 1, it does not matter, because they are equal almost everywhere.

My question is therefore, does that mean weak derivatives are defined uniquely, almost everywhere? For a classical real differentiable function, does that mean its weak derivative is equal to its classical derivative almost everywhere?

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In general, weak derivatives are distributions, i.e., linear functionals on a space of smooth, compactly supported test functions. A (locally integrable) function $f$ is identified with the distribution $\phi \mapsto \int \phi f$. So two almost everywhere equal functions define the same distribution, and are the same as weak derivatives. This also means that questions about the value of weak derivatives at a certain point don't have well-defined answers, unless you have continuous versions.

In your example, the weak derivative is the function which is $0$ on $(-\infty,0)$ and $1$ on $(0,\infty)$, and you can give it any value at $0$, or even modify it on any set of measure $0$. Now if you want to differentiate this derivative again, then its weak derivative is not representable by a function anymore, due to the jump discontinuity at $0$. In this case, you would get the "Dirac $\delta$-function" which really is not a function, but the distribution associated to the measure $\mu$ with mass $1$ on $0$ and mass $0$ everywhere else. (The distribution associated to a measure $\mu$ is the functional $\phi \mapsto \int \phi \, d\mu$, in this case $\phi \mapsto \phi(0)$.) Now you can even differentiate this Dirac $\delta$ one more time, in which case you get a distribution which is not even representable by a measure anymore. It would be the functional $\phi \mapsto \phi'(0)$, the third derivative of your original function in the sense of distributions.

If a function is differentiable everywhere with locally integrable derivative, then this derivative is also the weak derivative in the sense of distributions. However, note that there are differentiable functions whose derivative is not locally integrable, and that there are continuous almost everywhere differentiable functions with derivatives that are not their weak derivatives. (The classical example here is the "devil's staircase", whose derivative is almost everywhere $0$, and whose weak derivative is a measure on the Cantor set.)

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I have seen this stuff before, but was confused. This is very well explained here, thank you very much. – Lost1 Oct 18 '12 at 0:15
A quick additional question: Are all weak derivatives associated with a measure? – Lost1 Oct 18 '12 at 0:20