I learned Sobolev spaces & bounded variation spaces. I read this sentence:

Sobolev spaces include functions such that $\int |f^{(1)}(x)|^2 dx < \infty$, but do not include functions that are piecewise smooth. The space of bounded variation $\{f: \int |f^{(1)}(x)|dx < \infty\}$ include such functions.

So the only difference in two equations is square, I am not sure how this makes a big difference. Can anyone elaborate on this difference with example of step function $f(x)=0$ if $x<0$, $f(x)=1$ if $x\ge0$ for $x \in [-1,1]$?


1 Answer 1


An illuminating way to think about Sobolev spaces is that functions from such spaces are integrable, don't have jumps, are differentiable almost everywhere and their derivative is integrable. This is given a precise meaning using the weak derivative: $$ f = Dg \qquad \stackrel{def}{\Longleftrightarrow} \qquad \int_\Omega f \varphi = -\int_\Omega g \varphi' \quad \forall \varphi \in C^\infty(\Omega) \quad \forall \Omega \text{ compact} $$ If $f, g \in C^1(\mathbb{R})$, this is just integrating by parts, but you can also introduce removable signularities and the weak derivative will still work. Notice however, that if $g$ had a jump, such a function $f$ wouldn't exist. Then the Sobolev spaces are defined as follows: $$ W^{k,p}(\Omega) = \Big\{ f \in L^p(\Omega) \; \Big| \; D^n f \in L^p(\Omega) \;\; \forall n \in \{ 1, ... k \} \Big\} $$ The space you describe in your question by “$\int |f^{(1)}(x)|^2 \mathrm{d}x < \infty$” is the Sobolev space $W^{1,2}$. However, what you're missing is that this is the weak derivative – therefore it “automatically forbids” any jumps in $f$. The square in the integral isn't important at all for your question, the space $W^{1,1}$ is characterized by $\int |D^1 f(x)| \, \mathrm{d}x < \infty$, without the square. But it's still the weak derivative.

On the other hand, the space $BV$ of (not necessarily continuous) bounded variation functions is characterized by the fact that they allow for removable discontinuities and jumps [src], and by the fact that they have a derivative almost everywhere. The difference here is that this derivative is the classical derivative of a function. If $f \in BV$ has a jump in a point, it just doesn't have a derivative there. In other words, the integral $g(x) = \int_0^x f^{(1)}(x') \mathrm{d}x'$ would be equal to the function $f$ piecewise shifted by a constant, so that it doesn't have jumps anymore.

plots of f and g

This also explains, why one would want to use the weak derivative instead of the almost-everywhere derivative: it doesn't lose track of the jumps, which can be quite important eg. in physical models.

On a final note, this wouldn't happen if we were talking about continuous functions with bounded variation. The Sobolev space $W^{1,p}$ could be alternatively described as (a.e.) absolutely continuous functions with suitable growth conditions. Since the space of continuous bounded-variation functions is a proper subset of absolutely continuous functions.

  • $\begingroup$ wow thank you so much!!! really helpful and clear! $\endgroup$
    – Jayboy
    Feb 11, 2021 at 16:44
  • $\begingroup$ I'm really glad it helped you! I've recently been learning about this topic too, and I found it difficult to find comparisons like this. If my answer solved you issue, please accept it using the tick mark next to the upvote count :) $\endgroup$
    – m93a
    Feb 11, 2021 at 18:55

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