Exercise 4 chapter 7 Evans book

Question

I want to solve the following problem:

Suppose $H$ is a Hilbert space, if $u_{n}\rightharpoonup u$ in $L^{2}(0,T;H)$ and $u'_{n}\rightharpoonup v$ in $L^{2}(0,T;H^{'})$.

I want to prove that $v=u'$.

Thanks for your help.

Answer 1

Let's write down what we know: \begin{align} \int_0^T\langle u_n(t),h(t) \rangle\,dt \to &\ \int_0^T\langle u(t),h(t) \rangle\,dt \quad \forall h \in L^2(0,T;H') \tag 1\\ \int_0^T\langle u_n'(t),h(t) \rangle\,dt \to &\ \int_0^T\langle v(t),h(t) \rangle\,dt \quad \forall h \in L^2(0,T;H)\tag 2\\ \int_0^T\varphi'(t)u_n(t)\,dt = &\ -\int_0^T\varphi(t)u_n'(t)\,dt \quad \forall \varphi \in C^{\infty}_c((0,T)). \tag 3 \end{align}

Notice that for $\varphi \in C^{\infty}_c((0,T))$ and $w \in H$ we have that $t \mapsto \varphi(t)w \in L^2(0,T;H)$. In the following I am going to identify $H$ with its dual $H'$. If you prefer to treat them as different spaces you can modify the following argument introducing Riesz's linear isometry $J \colon H \to H'$. For $\varphi$ and $w$ be as above, we have:

\begin{align} \Big\langle\int_0^T\varphi'(t)u(t)\,dt,w\Big\rangle = &\ \int_0^T\langle\varphi'(t)u(t),w\rangle\,dt\\ = &\ \int_0^T\langle u(t),\varphi'(t) w\rangle\,dt\\ \overset{(1)}{=} &\ \lim_{n \to \infty} \int_0^T\langle u_n(t), \varphi'(t)w\rangle\,dt \\ = &\ \lim_{n \to \infty}\Big\langle\int_0^T\varphi'(t)u_n(t)\,dt,w\Big\rangle \\ \overset{(3)}{=} &\ \lim_{n \to \infty}\Big\langle-\int_0^T\varphi(t)u_n'(t)\,dt,w\Big\rangle \\ = &\ \lim_{n \to \infty} \int_0^T-\langle u_n'(t),\varphi(t)w\rangle\,dt \\ \overset{(2)}{=} &\ \int_0^T-\langle v(t),\varphi(t)w\rangle\,dt \\ = &\ \Big\langle-\int_0^T\varphi(t)v(t)\,dt,w\Big\rangle. \end{align}

Comparing the first and last step in the previous chain of equalities we get

$$\Big\langle\int_0^T\varphi'(t)u(t)\,dt + \int_0^T\varphi(t)v(t)\,dt ,w\Big\rangle = 0.$$ Since this holds for every $w \in H$ we can conclude that $$\int_0^T\varphi'(t)u(t)\,dt = -\int_0^T\varphi(t)v(t)\,dt.$$ Since this holds for every test function $\varphi$, we get the desired result.

Answer 2

Your problem is a corollary of the following

Theorem: Let $Y$ be a Banach space such that $Y'$ has the Radon Nikodym property, $X$ a Banach space continuously embeded in $Y$ and $1\leq p,q\leq\infty$. If $$\left\{\begin{align}u_k\rightharpoonup u\quad&\mbox{in}\quad L^p(0,T;X),\\ u_k'\rightharpoonup v\quad&\mbox{in}\quad L^q(0,T;Y),\end{align}\right.$$ then $u'=v$.

Sketch of the proof:

Since $X\hookrightarrow Y$, we get $L^p(0,T;X)\hookrightarrow L^1(0,T;Y)$ and $L^q(0,T;Y)\hookrightarrow L^1(0,T;Y)$. Thus,

$$\left\{\begin{align}u_k\rightharpoonup u\quad&\mbox{in}\quad L^1(0,T;Y),\\ u_k'\rightharpoonup v\quad&\mbox{in}\quad L^1(0,T;Y).\end{align}\right.$$

As $Y'$ has the Radon Nikodym property, it follows that

$$\left\{\begin{align}\int_0^T u_k\varphi'\,dt&\rightharpoonup \int_0^T u\varphi'\,dt &\mbox{in}\quad Y,\\ \int_0^T u_k'\varphi\,dt&\rightharpoonup \int_0^Tv\varphi\,dt &\mbox{in}\quad Y,\end{align}\right.$$ for all $\varphi\in C_c^\infty(0,T)$. But, by definition of weak derivative, we have $$\int_0^T u_k\varphi'\,dt=-\int_0^T u_k'\varphi\,dt,\quad \forall\ \varphi\in C_c^\infty(0,T)$$ so that (by the uniqueness of the weak limit) $$\int_0^T u\varphi'\,dt=-\int_0^T v\varphi\,dt,\quad \forall\ \varphi\in C_c^\infty(0,T).$$

Hence, $u'=v$.

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Addendum. To prove that the weak convergence in $L^1(0,T;Y)$ implies that weak convergence of integrals in $Y$, we need the surjectivity of a certain mapping from $L^\infty(0,T;Y')$ to $L^1(0,T;Y)'$ (as sketched here). This mapping is surjective if $Y'$ has the Radon Nikodym property (what happens, for example, if $Y$ is separable reflexive or if $Y$ is Hilbert).