Does such function exist?

Let $0<a<1$. Does there exist a function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that:

$$\left|f(x)-f(y)\right |\geq \left|x-y\right|^a, \;\; \forall x, y \in \mathbb{R}$$

Solution

Discontinuous function

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be defined as:
$$f(x)= \left\{\begin{matrix}
 1&,  &x \in \mathbb{Q} \\
 0&,  & x \in \mathbb{R} \setminus \mathbb{Q}
\end{matrix}\right.$$

Prove that $f$ is discontinuous.

Sum of periodical functions

Let $f$ be a polynomial. Show that $f$ can be written as the sum of $\deg f +1$ periodical functions but not as the sum of less periodical functions.

Solution

Monotony of a function

Let $F:(0, +\infty) \rightarrow \mathbb{R}$ be defined as:

$$ F(x)= \int_{1/x^2}^{ \int_{0}^{x}\exp (s^2)\, {\rm d}s}e^{t^2}\, {\rm d}t $$

What is the monotony of the function?

Solution