Section 9.2 Banach-Space Valued Functions
Banach-space valued function become important when we solve partial differential equations (PDEs). As an example let us look at a reaction diffusion equation for a physical or biological quantity \(u(x,t)\text{:}\)
\begin{equation}
\frac{\partial u}{\partial t} = D\Delta u + f(u).\label{RDE}\tag{9.2.1}
\end{equation}
If \(u(x,t)\) denotes the solution, then for each time \(t\geq 0\) we have a function of space \(x\text{:}\) \(u(t): x\mapsto u(x,t)\text{,}\) hence this map is in some functions space and we write \(u(t)\in X\) for all \(t\geq 0\text{.}\) Typical function spaces in the context of PDEs are given below. For example \(X=C^0([0,T], L^p(\Omega))\) has the norm
\begin{equation*}
\|u\|_X = \sup_{0\leq t\leq T} \|u(t)\|_{L^p(\Omega)}.
\end{equation*}
The space \(X=L^p([0,T], L^q(\Omega)) \) has the norm
\begin{equation*}
\|u\|_X = \left(\int_0^T \|u(t)\|_{L^q(\Omega)}^p dt\right)^{\frac{1}{p}}.
\end{equation*}
The norm in \(L^p(0,T; L^p(\Omega))\) can be written in two ways:
\begin{equation*}
\|u\|_X = \left(\int_0^T \|u(t)\|_p^p dt\right)^{\frac{1}{p}} = \left(\int_0^T\int_\Omega |u(x,t)|^p dx dt \right)^{\frac{1}{p}} = \|u\|_{L^p([0,T]\times\Omega)} .
\end{equation*}
The space \(X=L^2([0,T], L^2(\Omega))\) is a Hilbert space with inner product
\begin{equation*}
\langle u,v\rangle_X = \int_0^T\int_\Omega u(x,t) v(x,t) dx dt.
\end{equation*}
There is a Fundamental Theorem of Calculus for Banach-space valued functions. Theorem 9.2.1.
Let \(B\) be a Banach space, \(X= W^{1,p} ([0,T], B)\text{,}\) \(u\in X\) and \(1\leq p\leq \infty\text{.}\) Then
\begin{equation}
u(t) = u(s) +\int_s^t \frac{du}{dt}(\tau) d\tau, \qquad 0\leq s\leq t\leq T.\label{FundTheorem1}\tag{9.2.2}
\end{equation}
Furthermore \(u\in C^0([0,T], B) \) and
\begin{equation*}
\|u\|_{C^0([0,T], B)} \leq C\|u\|_X.
\end{equation*}
Proof.
\begin{equation*}
\frac{d u_h(t)}{dt} = \left(\frac{du(t)}{dt}\right)_h
\end{equation*}
and
\begin{equation*}
u_h\to u, \qquad\mbox{and}\qquad \frac{du_h}{dt} \to \frac{du}{dt} \qquad \mbox{in }\quad L^p \qquad \mbox{for } h\to 0.
\end{equation*}
For \(h>0\) the function \(u_h\) is continuously differentiable and we use the Fundamental Theorem of Calculus:
\begin{equation*}
u_h(t) = u_h(s) +\int_s^t \frac{d u_h}{dt}(\tau) d\tau .
\end{equation*}
Taking the limit as \(h\to 0\) gives equation (9.2.2). For the sup-norm estimate we chose \(t=0\) and \(s=t\) and find
\begin{equation*}
\|u(0)\|_B \leq \|u(t)\|_B +\int_0^t \|u'(\tau)\|_B d\tau,
\end{equation*}
which integrated from \(0\) to \(T\) becomes
\begin{align*}
T\|u(0)\|_B \amp\leq\amp \int_0^T \|u(t)\|_B dt + \int_0^T \int_0^t \|u'(\tau)\|_B d\tau dt\\
\amp\leq \amp \|u\|_{L^1([0,T],B)} + T\|u'\|_{L^1([0,T],B)}\\
\amp\leq \amp T^{\frac{1}{q}}\|u\|_{L^p([0,T],B)} + T^{\frac{1+q}{q}} \|u'\|_{L^p([0,T],B)},
\end{align*}
where we used H\"olders inequality and the fact that \(\frac{1+q}{q} = \frac{1}{q} +1 \) in the last step. We use (9.2.2) again and the previous estimates to get
\begin{align*}
\|u(t)\|_B \amp\leq \amp \|u(0)\|_B +\int_0^t \|u'(\tau)\|_B d\tau\\
\amp\leq \amp \|u(0)\|_B +T^{\frac{1}{q}}\|u'\|_{L^p([0,T],B)}\\
\amp\leq\amp T^{\frac{1-q}{q}} \|u\|_{L^p([0,T],B)} + 2 T^{\frac{1}{q}} \|u'\|_{L^p([0,T],B)}\\
\amp\leq\amp C \|u \|_{W^{1,p}([0,T],B), }
\end{align*}
which proves the last estimate of the Theorem.Example 9.2.2.
\begin{equation*}
u_t = D\Delta u + f(u)
\end{equation*}
we often have that
\begin{equation*}
u\in L^2([0,T], H_0^1(\Omega))\quad\mbox{ and } \quad u_t \in L^2([0,T], H^{-1}).
\end{equation*}
Since \(H^{1}\subset H^{-1}\) it follows that
\begin{equation*}
u\in H^1([0,T], H^{-1}) .
\end{equation*}
Then, by the above Theorem 9.2.1 we find
\begin{equation*}
u\in C^0([0,T]; H^{-1}).
\end{equation*}
In this context, the reaction diffusion equation presents itself as an equality in \(H^{-1}\text{:}\)
\begin{equation*}
\underbrace{u_t}_{H^{-1}} = \underbrace{D\Delta \underbrace{u}_{H^1}}_{H^{-1}} + \underbrace{f(u)}_{H^{-1}}.
\end{equation*}
Of course, we made no assumptions here on the growth term \(f(u)\text{.}\) But it is clear what the assumptions of \(f\) should be. One possibility is to assume that the map \(f: H^1\to H^{-1}\) is globally Lipschitz continuous.Theorem 9.2.3.
Let \(X\subset\subset H \subset Y\) be Banach spaces, and \(H\) reflexive. Suppose \(\{ u_n\}\subset L^2([0,T], X)\) is uniformly bounded and the time derivatives \(\{\partial_t u_{n}\} \subset L^p([0,T], Y) \) are uniformly bounded for some \(p>1\text{.}\) Then there exists a subsequence that converges strongly in \(L^2([0,T],H)\text{.}\)Example 9.2.4.
\begin{equation*}
u_n\subset L^2([0,T], H_0^1) \qquad u_{n,t} \subset L^2([0,T], H^{-1})
\end{equation*}
is uniformly bounded in those spaces. Then chosing \(X=H_0^1, H=L^2, Y= H^{-1}\) the previous Theorem provides us with a convergent subsequence
\begin{equation*}
u_{n_k} \to u \qquad \mbox{in} \qquad L^2((0,T)\times \Omega) .
\end{equation*}
This limit is then a candidate for a solution of the original reaction diffusion equation. This is the essential step of the Galerkin method. However, the set of basis functions need to be chosen carefully. Also, more work is needed to show that the time derivative converges, and that the limit function is continuous in the right spaces. Here we refer to other textbooks such as \cite{robinson}, where the entire proof of the Galerkin method occupies about 6 pages. We develop a simpler case in Exercise 6.4.3. 