# Identify the distrionbutional derivative with classical derivative?

I am reading Rudin's Functional Analysis and got quite confused by his proof for theorem 7.25, which he calls Sobolev's Lemma.

In proving the theorem, he defines the function $F$, and calculates its distributional derivative $D_i^k F$ (eq.3). Then he says this implies $D_i^k F$, originally defined as a distribution, is actually a $\mathcal{L}^2$.

This is what confuses me because $D_i^k$ is only the distributional derivative, how can one identify this with the function derivatives? Maybe we can redefine $F$ so that the distributional derivative agrees with the classical derivative but I still feel unconfortabla unless someone points out how this can be done without contradicting our original definition.

Thanks!

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## 1 Answer

He doesn't use the classical derivative of $F$, but takes only distributional derivatives. What he implicitly does is identifying a function $f \in L^1_{\mathrm{loc}}(\Omega)$ with the distribution $\Lambda_f : C^{\infty}_c(\Omega) \rightarrow \mathbb{R}$ given by $$\Lambda_f(\phi) = \int_\Omega f \phi.$$

So for example, one can think of $F$ in his proof as $\psi \cdot \Lambda_f$ - that is, a function multiplied by a distribution. Then, he uses the Leibniz formula for this case, to obtain

$$D^r_i F = D^r_i (\psi \Lambda_f) = \sum_{s=0}^r {r \choose s} (D_i^{r-s} \psi) D^s_i (\Lambda_f).$$

Here, $D^{r-s}_i \psi$ is a classical derivative of $\psi$ while $D^s_i(\Lambda_f)$ is the distributional derivative of the distribution $\Lambda_f$. However, because the distributional derivatives of $f$ (that is, of $\Lambda_f$) are (identified) with $g_{is}$, then you have

$$D^r_i F = D^r_i (\psi \Lambda_f) = \sum_{s=0}^r {r \choose s} (D_i^{r-s} \psi) \Lambda_{g_{is}} = \Lambda_{{\sum {r \choose s} (D^{r-s}_i \psi) g_{is}}}.$$

But this means that the distributional derivative $D^r_i F$ is represented by a function in $L^2(\Omega)$ and this is what he claims.

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