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

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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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### user101182

Updated on December 19, 2020

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 $L^2([-L,L])$?
Do you take into account that functions in $L^2$ are not necessarily continuous, not to say differentiable?
No. Consider $\Phi(x) = x^2 \sin (x^{-2})$.