Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Just for the sake of completeness, I begin defining the Sobolev space $H^m(\mathbb{R}^n), \; m \in \mathbb{N}$, as the following set: $H^m(\mathbb{R}^n) = \{u \in L^2 : P^{\alpha} F u \in L^2,\; \forall |\alpha| \leq m\}$, where $P^{\alpha}(x) = x^{\alpha}$, $\alpha$ is an $n$-multiindex and $Fu$ is the Fourier transform of $u$. We defined the weak derivative of an element $u \in H^m$ as follows: $\partial^{\alpha} u = F^{-1}(P^{\alpha}F u)$ (this was motivated by the validity of this formula in the Schwartz space).

Well, the problem arises when I try to prove the consistency of this definition in the case where both the weak and the strong (classic) derivative exist.

More precisely, let $u \in C^m$. If further I have the strong derivatives $D^{\alpha} u \in L^2$ for all $|\alpha| \leq m$, then a theorem states that $u \in H^m$ and $D^{\alpha} u$ is almost everywhere equal to $F^{-1}(P^{\alpha}F u)$. Very good, so far.

But what if the second hypothesis fails? i.e. what happens when it exists $|\alpha| \leq m$ for which $D^{\alpha}u \not \in L^2$? My question is if, in this case, I can state that $u \not \in H^m$, or equivalently if $u \in C^m \cap H^m$ implies the pointwise almost-everywhere equality of strong and weak derivatives. This should be a "good behaviour" that I expect, but I'm not sure of its validity.

Any elucidation is appreciated!

share|cite|improve this question
up vote 2 down vote accepted

Your question is easily answered if one uses a different definition of weak derivatives which is more general but turns out to be equivalent to your definition for functions in $H^m(\mathbb{R}^n)$.

Let $\Omega \subset \mathbb{R}^n$ be an open subset. The usual way to define a weak derivative of a function $u \in L^1_{\mathrm{loc}}(\Omega)$ is by using test functions and is motivated by integration by parts. Let $\alpha$ be a multiindex. A function $v \in L^1_{\mathrm{loc}}(\Omega)$ is called the weak $\alpha$-derivative of $u$ if for all test functions $\phi \in C^{\infty}_{\mathrm{c}}(\Omega)$ we have $\int_{\Omega} u D^{\alpha}\phi dx = (-1)^{|\alpha|} \int_{\Omega} v \phi dx$. A weak derivative in this sense, if exists, is determined uniquely a.e and if $u \in C^m(\Omega)$ then it can be seen by integration by parts that all the weak derivatives of order $\leq m$ exist and must agree with the classical derivatives.

Then, for $\Omega = \mathbb{R}^n$, we have two equivalent definitions of Sobolev spaces:

1) $H^k(\Omega) = \{u \in L^2(\Omega) \,\,|\,\, D^\alpha u \,\,\mathrm{exists\, and\,belongs\,to}\,\,L^2(\Omega)\,\, \forall|\alpha| \leq k \}$

2) $H^k(\Omega) = \{u \in L^2(\mathbb{R}^n) \,\,|\,\, P^\alpha F(u) \in L^2(\mathbb{R^n})\,\, \forall |\alpha| \leq k$}

From the first definition, it is immediately clear that if you have a function in $C^m(\mathbb{R}^n)$ with some $\alpha$ derivative of order $\leq m$ that doesn't belong to $L^2(\mathbb{R}^n)$ then the function does not belong to $H^m(\mathbb{R}^n)$. You can prove the equivalence of the definitions by yourself or check any reference on Sobolev spaces, for example the chapter on Sobolev spaces in Evan's PDE.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.