Is the derivative of a function square integrable if the function itself is square integrable in an interval?

1,513

The comments already give plenty of evidence to the contrary, to which I'll add the example of $f(x)=|x|^{-1/3}$ with derivative (which exists except at one point) $f'(x)=\frac13 |x|^{-4/3}\mathrm{sgn}\,x$.

Perhaps more important is to consider in what sense the derivative should be understood here. One definition, naturally adapted to the function space at hand, is the $L^2$ derivative: we say that $f'=g$ in the sense of $L^2$ derivative if the divided difference functions $$f_h(x)= \frac{f(x+h)-f(x)}h \tag{1}$$ converge to $g$ in $L^2$ as $h\to 0$. (One may have to step away from the endpoints of $[-L,L]$ for (1) to be defined, but this is no big deal.)

Not every $L^2$ function has an $L^2$ derivative; for example $f(x)=\mathrm{sgn}(x)$ does not (even though its classical derivative exists almost everywhere). Indeed, for this function we have $\|f_h\|_{L^2} = |h|^{-1/2}$ which obviously rules out $L^2$ convergence to anything.

As a matter of fact, an $L^2$ function with $L^2$ derivative has an absolutely continuous representative, and a lot of other nice properties. Such functions form the Sobolev space $W^{1,2}$, also denoted by $H^1$.

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Author by

user101182

Updated on December 19, 2020

Comments

user101182 almost 3 years

If a function $\Phi(x)$ is square integrable in an interval $[-L, L]$, is the function $\Phi'(x)$, the derivative of $\Phi$ with respect to $x$, necessarily square integrable in the same range? In other words if $\Phi(x)$ is in $L^2([-L,L])$ is $\Phi'(x)$ also in $L^2([-L,L])$?
- TZakrevskiy about 10 years
  
  Do you take into account that functions in $L^2$ are not necessarily continuous, not to say differentiable?
- Daniel Fischer about 10 years
  
  No. Consider $\Phi(x) = x^2 \sin (x^{-2})$.