Analysis in $\mathbb{Q}$

Calculus Level 3

It is a well-known theorem that if $f:\mathbb{R}\to\mathbb{R}$ is a differentiable function such that $f'(t)=0$ for all $t\in\mathbb{R}$ , then $f$ is constant.

Is it true that if $f:\mathbb{Q}\to\mathbb{Q}$ is a differentiable function such that $f'(t)=0$ for all $t\in\mathbb{Q}$ , then $f$ is constant?

Notations :

$\mathbb R$ denotes the set of real numbers .
$\mathbb Q$ denotes the set of rational numbers .

Yes No

1 solution

Victor Moreno Diaz
Jul 24, 2016

Notice that the function $\displaystyle g:\mathbb{Q}\to\mathbb{Q}$ defined as $\displaystyle g(t):=\left\{\begin{matrix} -1 & t^2<2 \\ 1 & \text{otherwise} \end{matrix}\right.$ verify all the hypothesis of the theorem and however it's not constant.

Is it differientiable everywhere?

Stéphane Gasser - 4 years, 10 months ago

Exactly. Such a function is not differentiable by the definition used in the link, because the derivative is not defined for all real numbers.

A Former Brilliant Member - 4 years, 10 months ago

After some discussion with the problem author, I realized finally that the example function is continuous and differentiable at every rational point. And it is only defined on the rational numbers. Interesting.

Steven Perkins - 4 years, 10 months ago

Nice counter example.

A Former Brilliant Member - 4 years, 10 months ago

Nice use of the Dirichlet function,

Sharky Kesa - 4 years, 10 months ago

Prove that your function is differentiable

Ravi Dwivedi - 4 years, 10 months ago

I don't quite get it. If the discontinuity points are rationals, (e.g x=|2|) the function is not continuous at x=2 (we can take two sequences taking values only on rational numbers, one coming from the left, one from the right), then How can it be differentiable in x=2? Or are we somehow talking about some "weak"differentiability?

Gerónimo Rojas Barragán - 4 years, 10 months ago

I believe the discontinuity points are at +/- sqrt(2).

Steven Perkins - 4 years, 10 months ago

Oh, you are right! Thank you

Gerónimo Rojas Barragán - 4 years, 10 months ago

@Gerónimo Rojas Barragán – As @Steven Perkins pointed out the points in which the function should be discontinuous are $\pm\sqrt{2}$ . But those points are not rationals numbers, so it doesn't make sense to ask if $f$ is not continuous there.

Victor Moreno Diaz - 4 years, 10 months ago

Analysis in Q \mathbb{Q} Q

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Analysis in $\mathbb{Q}$