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Open mapping theorem in functional analysis


In this article, we give an application of the open mapping theorem in functional analysis. This fundamental theorem in functional analysis plays a key role in the study of evolution equations.

In the sequel, we propose a nice functional analysis worksheet on some deep applications in different domains of mathematical analysis.

An application of the open mapping theorem

Exercise: Let $f$ be an application from $\mathbb{R}^n$ to $\mathbb{R}^n$. Suppose that there exists $C>0$ such that \begin{align*}\forall (x,y)\in \mathbb{R}^n\times \mathbb{R}^n ,\quad \langle f(x)-f(y),x-y\rangle\ge C\|x-y\|^2.\end{align*}We propose to demonstrate that $ f $ is a homeomorphism from $ \mathbb{R}^n $ on $ \mathbb{R}^n $. To do so, consider the differential equation \begin{align*}\tag{Eq} \dot{x}(t)=-f(x(t)).\end{align*}

  • Let $x$ and $y$ two solutions of $(Eq)$ defined on $[0,T]$ with $T>0$. If we put $\varphi=\|x-y\|^2,$ prove that \begin{align*} \varphi(t)\le \varphi(0)e^{-2Ct},\quad \forall t\in [0,T].\end{align*} Deduce that \begin{align*} \|\dot{x}(t)\|\le \|\dot{x}(0)\| e^{-Ct}.\end{align*}
  • Let $x$ be a maximal solution of the Cauchy problem $\dot{x}=-f(x)$ and $x(0)=0$. prove that $x$ is defined on $[0,+\infty)$, global solution, and $x$ admits a finite limit $\ell$ as $t\to +\infty$ such that $f(\ell)=0$.
  • Give a conclusion

Solution: 1) As the application $x-y$ is differentiable, then $\varphi$ is differentiable and \begin{align*}\varphi'(t)&=2\langle x(t)-y(t),x'(t)-y'(t)\rangle\cr & \le -2C\varphi(t).\end{align*} This implies that \begin{align*} \varphi(t)\le \varphi(0)e^{-2Ct}.\end{align*}

On the other hand, let we fix $s\in [0,T[$. It is not difficult to see that the application $x_s:[0,T-s]\to \mathbb{R}^n$ such that $t\mapsto x_s(t)=x(t+s)$ is a solution of the equation $(Eq)$. Then by taking $y=x_s$ in the above estimate, we get \begin{align*}\forall t\in [0,T-s],\quad \|x(t+s)-x(t)\|\le \|x(t)-x(0)\| e^{-C t}.\end{align*} This inequality is true for any $(s,t)$ with $0\le s<T$ and $0\le t\le T-s$. Taking $s\in (0,T-t)$, so \begin{align*}\left\|\frac{x(t+s)-x(t)}{s}\right\|\le \left\|\frac{x(t)-x(0)}{s}\right\| e^{-C t}.\end{align*} Now by letting $s\to 0,$ we obten the result.

2) Let $x:[0,T[\to \mathbb{R}^n$ the maximal solution of the Cauchy problem with initial condition $x(0)=0$. If $T<+\infty$, then $\|x(t)\|$ explodes if $ t $ is close to $T$. This means that there exists $\delta>0$ such that $\|x(t)\|$ is not bounded on $[T-\delta,T[$. On the other hand, for $t\in [0,T]$, \begin{align*} \|x(t)\|&\le \int^t_0 \|\dot{x}(\sigma)\|d\sigma\cr & \le \int^t_0 \|\dot{x}(0)\| e^{-C \sigma}d\sigma\cr & \le \frac{\|\dot{x}(0)\|}{C}.\end{align*} This is a contradiction. Thus $T=+\infty$. Then \begin{align*} \|\dot{x}(t)\|\le \|\dot{x}(0)\| e^{-C t},\qquad\forall t\in [0,+\infty).\end{align*} This implies that the improper integral \begin{align*}\int^{+\infty}_0 \|\dot{x}(t)\|dt\end{align*} converges. Thus $x$ has a limit $\ell$ at $+\infty$. On the other hand, as $\dot{x}(t)\to 0$ as $t\to 0,$ then $f(\ell)=0$.

3) If we replace $ f $ by $ f-y $ which satisfies the same hypotheses, we see that $ f $ is a surjective from $\mathbb{R}^n$ to $\mathbb{R}^n$. According to the first hypothesis on $f,$ we deduce that $f$ is also injective. Then it realizes a bijection from $\mathbb{R}^n$ to $\mathbb{R}^n$ . By using Cauchy-Schwarz inequality we have \begin{align*} \|f(x)-f(y)\|\ge C \|x-y\|,\qquad \forall (x,y)\in \mathbb{R}^n\times \mathbb{R}^n.\end{align*} From this we deduce that \begin{align*} \|f^{-1}(x)-f^{-1}(y)\|\le \frac{1}{C} \|x-y\|,\qquad \forall (x,y)\in \mathbb{R}^n\times \mathbb{R}^n.\end{align*} This implies that $f^{-1}$ is continuous.

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