4
votes
Accepted
Showing that $\mathbf{X}^{2} + \mathbf{X} = \mathbf{A}$ has a solution
Your approach is not correct, for the simple reason that you haven't showed $\Phi$ is a Banach contraction. You have provided upper bounds for both $\|X - Y\|_\infty$ and $\|\Phi(X) - \Phi(Y)\|_\infty$...
- 44.6k
3
votes
Accepted
Using Banach's Fixed Point Theorem on an Integral Equation
Let $A\colon \Bbb R\to \Bbb R$, $A(x)=\frac x{\sqrt{h^2+x^2}}$ and
$$F\colon C([0,1])\to C([0,1]),\quad F(f)(x)=\int_0^1 K(x,y)A(f(y))dy.$$
Observe that $A$ is Lipschitz function satisfying with a ...
- 4,430
3
votes
Accepted
Is there a constructive proof of Brouwer's fixed-point theorem that does not rely on triangulation?
There is no constructive proof of Brouwer’s fixed point theorem at all.
In particular, the following is not provable constructively:
Intermediate Value Theorem: Let $f : [-1, 1] \to \mathbb{R}$ be a ...
- 22.5k
2
votes
Fixed-Point iteration method fails on converging on equation.
$$\Phi(z) = \int_{-\infty}^z\frac{e^\frac{-x^2}{2}}{\sqrt{2\pi}}\,dx=\frac{1}{2} \left(1+\text{erf}\left(\frac{z}{\sqrt{2}}\right)\right)$$
$$\phi(z) = \frac{e^\frac{-z^2}{2}}{\sqrt{2\pi}}$$ Let $c=\...
- 220k
2
votes
Showing that $\mathbf{X}^{2} + \mathbf{X} = \mathbf{A}$ has a solution
I have a short nonconstructive proof that works in $n$ dimensions.
Let $F:\mathcal{B}\times\mathcal{B}\mapsto\mathcal{B}$ be defined as $F(A,X) = X^2+X-A$ where $\mathcal{B}$ is the Banach space of ...
- 3,815
1
vote
Accepted
Prove that in a given coloring of a square, there exists a sub-square with a certain coloring.
The proof does use Sperner's lemma after all:
Treat colour $4$ as a variation of $3$, then flatten the two sides $33'$ and $3'1$ into a single side and add an edge in each square as shown. Then by ...
- 90.7k
1
vote
Accepted
Existence of a fixed point to a vector-valued mapping
Yes, the function $F$ is not a contraction, so we can't use Banach fixed point theorem. There are of course many generalisations of it but I would rather go for topological fixed point theorems.
First ...
- 4,430
1
vote
Accepted
If $f\in \mathcal{C}(\mathbb{S}^n,\mathbb{S}^n)$ and $\mathrm{deg}_2(f)=0$, then $f(x_0)=x_0$ and $f(y_0)=-y_0$ for some $x_0, y_0$.
This is a community-wiki answer illustrating @Rishi's comments under the question I asked. This is the first time that I write a community-wiki answer, please feel free to edit it if anything is ...
Only top scored, non community-wiki answers of a minimum length are eligible