# about weak derivative of Bochner integrable function

Hi I am studying the following theorem ( the theorem can be found in the classical Evans PDE book in the apendix 5.9 )

Theorem : Suppose $u \in L^{2}(0 , T ; H^{1}_{0}(U))$ with $u^{‘} \in L^{2}(0 , T ; H^{-1}(U))) ….$

I am trying to understad what this mean : $u^{‘} \in L^{2}(0 , T ; H^{-1}(U)))$

In the same apendix we have the definition :

Definition: Let $u \in L^1(0,T; H^{1}_{0}(U))$. we say that $v \in L^1(0,T; H^{1}_{0}(U))$ is the weak derivative of $u$, written

$$u^{‘} = v$$

provided

$$\int_{0}^{T} {\phi}^{‘} (t) u(t) \ dt = – \int_{0}^{T} \phi (t) v(t) \ dt , \forall \varphi \in C_{c}^{\infty}(0,T)$$

this definiton dont help me to understand the meaning of $u^{‘} \in L^{2}(0 , T ; H^{-1}(U)))$
Someone can help me ? I tried all the day to understand the meaning of the expression $u^{‘} \in L^{2}(0 , T ; H^{-1}(U)))$ …

#### Solutions Collecting From Web of "about weak derivative of Bochner integrable function"

Let $X,Y$ be Banach spaces such that $X\subset Y$ and $u\in L^{p}((0,T);X)$. A more geral definition of weak derivative is (see Definition II.5.7 from this book): $v\in L^q((0,T);Y)$ is a weak derivative of $u$ if $$\int_0^T\varphi'(t)u(t)dt=-\int_0^T\varphi(t)v(t)dt,\ \forall\ \varphi\in C_0^\infty((0,T))$$

Note that this definition make sense, because $X\subset Y$ and you can see the left integral as a element of $Y$. We denote $\frac{du}{dt}=u’=v$. With this more general definition, your statement seems more clear.

Remark. Also take a look in section 5.2.3 of the same book I have cited, in particular, Theorem II.5.13.

See also Why is $\langle f, u \rangle_{H^{-1}, H^1} = (f,u)_{L^2}$ when $f\in L^2 \cap H^1$ and not $\langle f, u \rangle_{H^{-1}, H^1}=(f,u)_{H^1}$?.

The definition would be $v = u’$ if
$$\int_0^T \langle v(t),w\rangle \, dx \, \phi \, dt = \int_0^T \int_\Omega u \, w \, d x\, \phi’ \, dt$$
for all $\varphi \in C_0^\infty(0,T)$ and $w \in H_0^1(\Omega)$.