# Confusions in Evans book regarding weak derivatives in Banach spaces

I am studying PDE using Evans' book and I have two main confusions (probably stupid questions to experts) regarding weak derivatives in Banach spaces.

First confusion: $$\def\u{\mathbf u}$$ $$\def\v{\mathbf v}$$

DEFINITION. Let $$\u \in L^1(0, T; X)$$. We say $$\v \in L^1(0, T; X)$$ is the weak derivative of $$\u$$, written $$\u'=\v,$$ provided $$\int_0^T \phi'(t) \u(t) \, dt = -\int_0^T \phi(t) \v(t) \, dt$$ for all scalar test functions $$\phi \in C_c^\infty (0, T)$$.

However, a subsequent theorem begins with an assumption of

THEOREM $$\mathbf 3$$ (More calculus). Suppose $$\u \in L^2(0, T; H_0^1(U))$$, with $$\u' \in L^2(0, T; H^{-1}(U))$$.

Now, the main confusion here is that $$H^{-1}$$ is only the dual space of $$H^1_0$$, not a subset nor a superset of $$H^1_0$$, whereas in the definition, both the function and its weak derivative take values in the same Banach space $$X$$. Is there some kind of hidden identification of spaces?

Second confusion:

THEOREM $$\mathbf 2$$ (Calculus in an abstract space). Let $$\u \in W^{1,p} (0, T; X)$$ for some $$1 \leq p \leq \infty$$. Then

(i) $$\u \in C([0,T]; X)$$ (after possibly being redefined on a set of measure zero), and
(ii) $$\u(t) = \u(s) +\displaystyle\int_s^t \u'(\tau) \, d\tau$$ for all $$0 \leq s \leq t \leq T$$.
(iii) Furthermore, we have the estimate $$\max_{0 \leq t \leq T} \|\u(t)\| \leq C\|\u\|_{W^{1,p} (0,T;X)}, \tag{7}$$ the constant $$C$$ depending only on $$T$$.

Proof. $$1$$. Extend $$\u$$ to be $$\mathbf 0$$ on $$(-\infty, 0)$$ and $$(T, \infty)$$, and then set $$\u^\varepsilon = \eta_\varepsilon * \u$$, $$\eta_\varepsilon$$ denoting the usual mollifier on $$\mathbb R^1$$. We check as in the proof of Theorem $$1$$ in $$\S5.3.1$$ that $$\u^{\varepsilon'} = \eta_\varepsilon * \u'$$ on $$(\varepsilon, T-\varepsilon)$$.

Then as $$\varepsilon \to 0$$, $$\begin{cases} \u^\varepsilon \to \u & \text{in } L^p(0, T; X), \\ (\u^\varepsilon)' \to \u' & \text{in } L^p (0, T; X). \end{cases} \tag{8}$$ Fixing $$0, we compute $$\boxed{\u^\varepsilon (t) = \u^\varepsilon (s) +\int_s^t \u^\varepsilon {}' (\tau) \, d\tau.}$$

Some kind of "fundamental theorem of calculus for Bochner integrals" seems to be used here. Am I correct that the functions are $$C^1$$ after mollification, so some version of the "fundamental theorem of calculus for Bochner integrals" can be applied? Here, at least the weak derivative and the function belong to the same space, as in the definition. But this is not the case in the proof of the subsequent theorem. Dual functions suddenly appear as derivatives:

Proof. $$1$$. Extend $$\u$$ to the larger interval $$[-\sigma, T+\sigma]$$ for $$\sigma>0$$, and define the regularizations $$\u^\varepsilon = \eta_\varepsilon * \u$$, as in the earlier proof. Then for $$\varepsilon$$, $$\delta>0$$, $$\frac{d}{dt} \|\u^\varepsilon (t) -\u^\delta (t)\|_{L^2(U)}^2 = 2 \bigl(\u^{\varepsilon'} (t) -\u^{\delta'} (t), \u^\varepsilon (t) -\u^\delta (t)\bigr)_{L^2(U)}.$$ Thus \begin{align} \|\u^\varepsilon (t) -\u^\delta (t)\|_{L^2(U)}^2 &= \|\u^\varepsilon (s) -\u^\delta (s)\|_{L^2(U)}^2 \\ &{}+2\int_s^t \langle \u^{\varepsilon'} (\tau) -\u^{\delta'} (\tau), \u^\varepsilon (\tau) -\u^\delta (\tau)\rangle \, d\tau \tag{11} \end{align}

None of these two "fundamental theorems of calculus" are derived explicitly and I am really confused!

• In your second displayed equation (under 'First confusion'), it is indeed weird if the LHS and RHS don't belong to the same space. How is then the weak derivative defined? I think this part is poorly written in Evans'. Did you find a source that explains this well? – Weltschmerz Dec 28 '20 at 20:34

For the first question, we have $$H_0^1(\Omega) \hookrightarrow L^2(\Omega) \hookrightarrow H^{-1}(\Omega),$$ so $$H_0^1(\Omega)$$ is indeed isomorphic to a subspace of $$H^{-1}(\Omega)$$. Evans notes this at the beginning of section 5.9.

Yes, there is a FTOC for Bochner integrals and that would imply the given (boxed) equation (for $$\mathbf{u}^\varepsilon$$). For more info on that, you can Google "Bochner integral fundamental theorem of calculus" and you should be able to readily find something of interest. For example, you may find the following helpful.

https://tinyurl.com/y2sxnvdg

I think looking up the FTOC for Bochner integrals (along with the above embeddings) should clear up the last question, but if you need more clarification/info I'm happy to help.

• I still don’t understand how the final FTOC (eq. 11) can be proved? What kind of identification has been used to derive this? – Richard Mar 24 '19 at 14:15
• He's just applying the FTOC from above with $\|\mathbf{u}^\varepsilon - \mathbf{u}^\delta\|_{L^2}$ replacing $\mathbf{u}^\varepsilon$. – Gary Moon Mar 24 '19 at 15:50
• Sure. That’s just the standard FTOC. It was the line before this that I was lost... he wrote down the formula of the derivative of the square of the $L^2$ norm without explanation! – Richard Mar 24 '19 at 16:03
• Yep, that's what I'm talking about for differentiating the inner product. Due to the smoothness of the mollified $\mathbf{u}$, all of the above computations work in the strong sense. I'm assuming the goal will then be to pass to the limit as $\varepsilon \downarrow 0$ and obtain a result about the weak derivative. – Gary Moon Mar 24 '19 at 17:25
• The first is viewing $(\mathbf{u}^\varepsilon)^\prime$ as just an $L^2$ function and taking an inner product. The next line is changing to thinking of $(\mathbf{u}^\varepsilon)^\prime$ as an element of $H^{-1} = (H_0^1)^*$ and considering the dual pairing (which is basically the same because the dual pairing is just integrating $(\mathbf{u}^\varepsilon)^\prime$ against an element of $H_0^1$). This change is fine since we have $L^2 \hookrightarrow H^{-1}$. – Gary Moon Mar 24 '19 at 17:32