Inspired By Arturo Presa

Calculus Level 5

What can we say about the function $f: \mathbb{R}^+ \rightarrow \mathbb{R}$ which satisfies

$f(xy) = f(x) + f(y) ?$

It is continuous but not necessarily differentiable It can only be the zero function It is differentiable but not necessarily the zero function None of the rest

1 solution

Otto Bretscher
Oct 23, 2015

Interesting, thought-provoking problem! Thanks, @Calvin Lin

Considering a basis of $\mathbb{R}$ as a vector space over $\mathbb{Q}$ , we can prove the existence of a discontinuous linear transformation $g :\mathbb{R}\rightarrow\mathbb{R}$ , using basic concepts of linear algebra.

According to the axiom of choice, the vector space $\mathbb{R}$ has a basis $B$ over $\mathbb{Q}$ . We can define a linear transformation $g :\mathbb{R}\rightarrow\mathbb{R}$ by setting $g(r_1b_1+....+r_nb_n)=r_1+...+r_n$ where the $r_j$ are rational and the $b_j$ are in $B$ . Note that $g(x+y)=g(x)+g(y)$ by definition.

Then $f= g \circ \ln$ is a function with the required property that $f(xy)=f(x)+f(y)$ . Also, $f$ is discontinuous since its range is $\mathbb{Q}$ (the intermediate value theorem does not hold). Any rational number $r$ is in the range since $f(e^{rb})=g(rb)=r$ for any basis element $b$ .

Moderator note:

Nice way of proving that it is not a continuous function, because the values of $g$ (but not $f$ ) are rational.

Please prove the details too Sir

Anirban Mandal - 5 years, 7 months ago

Ok here are some details... feel free to ask for more.

I will freely use basic concepts of linear algebra.

Otto Bretscher - 5 years, 7 months ago

Yes sir pls explain in detail

Deepak Kumar - 5 years, 7 months ago

Inspired By Arturo Presa

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