Question on application of the dominated convergence theorem in showing that $|f|$ is weakly differentiable

Question

The following is an outline of the proof of a theorem found in Analysis by Lieb and Loss.

(Derivative of the Absolute Value) Let $f \in W^{1,p}(\Omega)$ real valued, $\Omega \subseteq \mathbb{R}^n$ open and $1 \leq p \leq \infty$. Then $|f| \in W^{1,p}(\Omega)$.

In the proof we approximate $|f|$ by $$K_\varepsilon := \sqrt{\varepsilon^2 + u^2} - \varepsilon$$ for $\varepsilon > 0$. Clearly, $K_\varepsilon \to u$ pointwise as $\varepsilon \to 0$. Now a part of the claim follows by using (I just write the significant part) the Lebesgue dominated convergence theorem, i.e. $$\int_\Omega|f| d\lambda = \lim_{\varepsilon \to 0}\int_\Omega K_\varepsilon$$ In the book it is written that $$|K_\varepsilon| \leq |f|$$ which is clear for me. However, $f \in L^p(\Omega)$, so why should this imply that we can apply the D.C.T.? I mean for applying the D.C.T. we should have $$|K_\varepsilon| \leq g$$ where $g$ is some positive integrable function. What am I missing?

Answer 1

Looking at my copy of Lieb and Loss, they don't claim $$\int_\Omega|f| d\lambda = \lim_{\varepsilon \to 0}\int_\Omega K_\varepsilon$$

They are claiming that $K_\epsilon$ converges in $W^{1,p}(\Omega),$ and they argue that:

• $K_\epsilon$ converges to $|f|$ in $L^p$
• $\nabla K_\epsilon$ converges to $\nabla|f|$ in $L^p$

where $\nabla |f|$ is an explicitly defined function. The convergence in $W^{1,p}(\Omega)$ then implies that $\nabla|f|$ is indeed the weak derivative of $|f|.$

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The principle they are using is often called $\mathbf{L^p}$ dominated convergence.

Suppose $g\in L^p,$ and $|g_n|\leq |g|$ a.e., and $g_n\to g$ pointwise a.e. Note $|g_n-g|^p\leq |2g|^p$ a.e. By dominated convergence,

$$0=\int_\Omega\lim_{n\to \infty}|g_n-g|^p d\lambda = \lim_{n\to \infty}\int_\Omega |g-g_n|^p$$

Therefore $g_n\to g$ in $L^p.$

This is discussed in Lieb and Loss, not under the heading "Dominated Convergence" but under "Missing term in Fatou's lemma."

Answer 2

The trick here are the embeddings between Sobolev spaces: a function $u \in W^{1,p}(\Omega)$ is in $L^p$ by definition, but we can usually obtain much more: for instance if $\Omega$ is open and bounded then $W^{1,p}(\Omega) \subset L^q(\Omega) $ for every $q \in [1, p^{*}]$, where $p^{*}$ is the conjugate Sobolev exponent of $p$.