Consider the piecewise linear function \(f\) defined on \([0,+\infty)\) taking following values for all \(n \in \mathbb{N^*}\):
\[
f(x)=
\left\{
\begin{array}{ll}
0 & \mbox{if } x=0\\
0 & \mbox{if } x=n-\frac{1}{2n^3}\\
n & \mbox{if } x=n\\
0 & \mbox{if } x=n+\frac{1}{2n^3}\\
\end{array}
\right.
\]
The graph of \(f\) can be visualized in the featured image of the post. Continue reading An unbounded positive continuous function with finite integral
Look at pi-Base for a gold mine of topological examples.
Most of Cauchy existence theorems for a differential equation
\begin{equation}
\textbf{x}^\prime = \textbf{f}(t,\textbf{x})
\end{equation} where \(t\) is a real variable and \(\textbf{x}\) a vector on a real vectorial space \(E\) are valid when \(E\) is of finite dimension or a Banach space. This is however not true for the Peano existence theorem. Continue reading A continuous differential equation with no solution
Want to know more about counterexample definition? Look at Wikipedia.
We consider a decreasing sequence \((C_n)_{n \in \mathbb{N}}\) of non empty closed convex subsets of a Banach space \(E\).
If the convex subsets are closed balls, their intersection is not empty. To see this let \(x_n\) be the center and \(r_n > 0\) the radius of the ball \(C_n\). For \(0 \leq n < m\) we have \(\Vert x_m-x_n\Vert \leq r_n – r_m\) which proves that \((x_n)_{n \in \mathbb{N}}\) is a Cauchy sequence. As the space \(E\) is Banach, \((x_n)_{n \in \mathbb{N}}\) converges to a limit \(x\) and \(x \in \bigcap_{n=0}^{+\infty} C_n\). Continue reading An empty intersection of nested closed convex subsets in a Banach space
This is a strange function!
One example is the Dirichlet function \(1_{\mathbb{Q}}\).
\(1_{\mathbb{Q}}(x)=1\) if \(x \in \mathbb{Q}\) and
\(1_{\mathbb{Q}}(x)=0\) if \(x \in \mathbb{R} \setminus \mathbb{Q}\).
\(1_{\mathbb{Q}}\) is everywhere discontinuous because \(\mathbb{Q}\) is everywhere dense in \(\mathbb{R}\).
The function \(x \mapsto x \cdot 1_{\mathbb{Q}}(x)\) is continuous in \(0\) and discontinuous elsewhere.
