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!


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.


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