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Let $X,Y$ be Banach spaces with $X \subset Y$. Recall that $u \in L^1(0,T;X)$ has weak derivative $g \in L^1(0,T;Y)$ if $$\int_0^T u(t)\phi'(t) = -\int_0^T g(t)\phi(t) \qquad\forall \phi \in C_c^\infty(0,T).$$

Suppose that $u$ also has a weak derivative $h \in L^1(0,T;Z)$ where $Y \subset Z$.

In Boyer and Fabrie's book on Navier-Stokes, page 95, he states that if $Y \subset Z$ is dense and $Z'$ is separable, then $g=h$. My question, why is the density and separability needed? Isn't the argument this simple:

Since $g$ and $h$ are weak derivatives of $u$, we have $$\int_0^T (g(t)-h(t))\phi(t) = 0\qquad\forall \phi \in C_c^\infty(0,T)$$ and by the fundamental lemma of the calculus of variations, it follows that $g(t) = h(t)$ in $Z$ for almost every $t$.

Isn't this enough? What do I miss? Does anyone know another source for this uniqueness claim where the derivatives lie in different spaces?

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    $\begingroup$ The point for separability of $Z'$ might be the reduction to the scalar case by applying continuous linear functionals to the equation $\int_0^T(g(t)-h(t))\phi(t)dt$. Then the exceptional null sets would depend on the functional and you only can take the union of countably many. $\endgroup$
    – Jochen Wengenroth
    Commented Aug 11, 2020 at 17:53
  • $\begingroup$ @JochenWengenroth Do you mean that he assumes separability for simplicity, because the separable case of the fundamental lemma of calculus of variations is well known? Even though separability is not needed. $\endgroup$
    – StopUsingFacebook
    Commented Aug 12, 2020 at 5:27

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$\def\bbR{\mathbb R}\def\inc{\subseteq}$The requirements on density or separability are superfluous because of the following

Lemma. Let $J$ be a real open interval, and let $E$ be any real or complex Banach space. Let the function $f$ in $L^1(J,E)$ be such that $\int_J(\varphi\,f)=0_E$ holds for all compactly supported smooth $\varphi:J\to\bbR$. Then $f(t)=0_E$ holds for almost all $t\in J$.

Proof. Let $x\mapsto\|x\|$ be a norm for $E$. Having $f$ a.e. the limit of a sequence of simple functions, there are a Lebesgue null set $N_1\inc J$ and a separable closed linear subspace $S$ in $E$ such that $f(t)\in S$ holds for all $t\in J\setminus N_1$. By Lemma 8.15.1 (p. 573) in R. E. Edward's Functional Analysis there is a countable set $D$ in the unit ball of the dual of $E$ such that $\|x\|=\sup\{|u(x)|:u\in D\}$ holds for all $x\in S$. By classical results, it follows existence of a Lebesgue null set $N_0\inc J$ such that $u\circ f(t)=0$ holds for all $t\in J\setminus N_0$ and $u\in D$. It follows that $\|f(t)\|=0$ holds for all $t\in J\setminus N_0$.

I do not know whether there is a published reference where the above Lemma would be explicitly stated or proved.

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  • $\begingroup$ It is also proved in Corollary 6.33, page 202 of these notes by Hunter. $\endgroup$
    – StopUsingFacebook
    Commented Aug 13, 2020 at 8:17
  • $\begingroup$ Thanks for the reference. Hunter seems to first generalize Lebesgue's differentiation theorem to Banach space valued functions, and then get the result as a corollary to it. $\endgroup$
    – TaQ
    Commented Aug 26, 2020 at 0:52

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