# An unbounded positive continuous function with finite integral

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

# Link to $$\pi$$-Base

Look at pi-Base for a gold mine of topological examples.

# A continuous differential equation with no solution

Most of Cauchy existence theorems for a differential equation

\textbf{x}^\prime = \textbf{f}(t,\textbf{x})
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

# Wikipedia Counterexample definition

Want to know more about counterexample definition? Look at Wikipedia.

# An empty intersection of nested closed convex subsets in a Banach space

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

# A nowhere continuous function

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.