# Extrema points on a Dirichlet variation

I was presented with the following function:

$$f(x) = \begin{cases} x^2, & \text{while } x\in \mathbb Q \\ 0, & \text{while } x\notin \mathbb Q \end{cases}$$ Which seems to me a variation of some sort of the Dirichlet function.

The question is, find all of the local extrema point of this function. Now, a friend offered the following solution: he claims that all point in which x is an irrational are local minimum, and that there's no local max. I agree that the function has no local maximum points,

But i'm not sure i follow the second part. What comes to my mind is, that between every couple of rational numbers "sits" a zero, which contradicts the defenition of extremum points (if i'm not mistaken), which leads me to think of 0 itself as the only minimum point. Any help would be useful, and thanks in advance!

• Irrational points are global minima. – Giuseppe Negro Feb 17 '18 at 19:26

Every irrational number and also the point $0$ are local and also global minimimum.
Notice that $f(x) \ge 0$, hence if you can attain $0$, you attains the smallest number in the neighborhood.
Now to prove that a non-zero rational point, $x$, cannot be a local minimum. $f(x)>0$. Consider any neighborhood of $x$, it must contain an irrational point, $y$, and hence $f(x) > f(y)$.