If $X$ is a complete metric space, then a contraction mapping is defined by $T:X\rightarrow X$ where $d(T(x),T(y))\leq cd(x,y)$ where $0\leq c < 1$ and $d$ is the appropriate metric/norm.

I know if we are considering spaces such as $[0,1]$ and the function $f$, it means the function $f$ has a fixed point, namely $f(x)=x$ where $x\in [0,1]$. How should interpret $T:L^p([a,b])\rightarrow L^p([a,b])$ via an operator $T(g)$? Does this mean there is a unique function $g$ in $L^p([a,b])$ where $T(g)=g$? In other words, the operator inputs a function from the domain space and outputs a function from the same space? I am struggling to get a clear intuition for this.

Next, could we prove the Fourier transform $\mathcal{F}:L^2([0,1])\rightarrow L^2([0,1])$ is a contraction mapping (on a suitable subset) via the following argument:

$(1)$ WTS $||\mathcal{F}f-\mathcal{F}g||_{L^2}\leq c||f-g||_{L^2}$

$(2)$ consider the closed unit ball in $L^2$, i.e. $B_{1}=\{f\in L^2([0,1]):||f||_{L^2}\leq 1\}$

Take the Fourier transform $\mathcal{F}:B_1\rightarrow B_1$ as $\mathcal{F}=\int_{0}^{1}f(x)e^{-2\pi i x} dx$ and show this is a contraction on $B_1$ via the definition.

$(3)$ conclude the Fourier transform maps $L^2$ functions to other $L^2$ functions on the unit ball.

Finally, what other interesting operators and contraction mappings can we consider for $L^p$ and other function spaces?

  • $\begingroup$ It should be $0\le c<1$. $\endgroup$ – Chee Han May 7 '17 at 1:27
  • $\begingroup$ @CheeHan edited. $\endgroup$ – Kernel_Dirichlet May 7 '17 at 1:29
  • $\begingroup$ Interpreting an operator has nothing to do with fixed points. You should think of fixed points as an additional structure that an operator might have. $\endgroup$ – Chee Han May 7 '17 at 1:30
  • $\begingroup$ That makes sense, but so then should I just view an operator (that has fixed points) as taking a function from one function space $X$ and outputting a function in the same space $X$? $\endgroup$ – Kernel_Dirichlet May 7 '17 at 1:35
  • 1
    $\begingroup$ No, what you just said follows if you have an operator $T$ that maps from $X$ to $X$. If, in addition to this, $T$ has a fixed point $x\in X$, this means that $T$ maps this particular element $x$ to itself again. $\endgroup$ – Chee Han May 7 '17 at 1:39

Yes, an operator $T:L^p([a,b])\to L^p([a,b])$ takes a function $f$ and maps it into another function $T(f)$ in the same space. So having a fixed point, would mean that $T(f)=f$, that is, that the function $f$ is mapped into itself. For example, you could take $T(f)(x)=\int_a^x f(t)\,dt$ and so a fixed point for $T$ would be a function $f$ such that $f(x)=\int_a^x f(t)dt$ for every $x\in [a,b]$. The Fourier transform is defined for functions in $L^2(\mathbb{R})$ and not $L^2([0,1])$. Parseval identity gives $\Vert\mathcal{F(f)}\Vert_{L^2}=\Vert f\Vert_{L^2}$ and so the Fourier transform cannot be a contraction.

  • $\begingroup$ Very helpful! So basically the Fourier transform is an operator on $L^2(\mathbb{R})$ and not a contraction because we have $c=1$ by Parseval's identity? $\endgroup$ – Kernel_Dirichlet May 7 '17 at 2:00
  • $\begingroup$ correct. It is Lipschitz continuous with Lipschitz constant one and not a contraction, $\endgroup$ – Gio67 May 7 '17 at 2:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.