Cauchy's functional equation with polynomial: $ f ( u + a v ) = f ( u ) + f ( a v ) + P _ n ( u , v ) $

Question

Let $ f : \mathbb R \to \mathbb R $ be a continuous function that satisfies $$ f ( u + a v ) = f ( u ) + f ( a v ) + P _ n ( u , v ) $$ where $ a $ is a known constant and $ P _ n ( u , v ) $ is a polynomial in $ u $ and $ v $ of degree $ n $. Is $ f $ a polynomial of degree $ n $?

Edit: I'm looking for a solution that doesn't use derivatives but rather "clever" substitutions for $ u $ and $ v $.

score 2 · Accepted Answer · answered Sep 10 '14 at 20:48

First of all, let asume wlog $a=1$, otherwhise take $\tilde u = u$, $\tilde v = av$ $$f(\tilde u+\tilde v) = f(\tilde u) + f(\tilde v) + P\left(\tilde u, \frac{\tilde v}{a}\right)$$ and $P\left(\tilde u, \frac{\tilde v}{a}\right)$ is also a polynomial of degree $n$ in $\tilde u,\tilde v$. From the functional equation we have $$f(x-y) = f(x) + f(-y) + P(x,-y)$$ Lets take $g\in C_0^\infty(\mathbb R)$ such that $$\int_{\mathbb R} g(y) dy = 1$$,then $$f(x-y)g(y) = f(x)g(y) + f(-y)g(y) + g(y)P(x,-y)$$ Also $f$ is continous so $f\in L_{loc}^1(\mathbb R)$ hence $f*g\in C^\infty(\mathbb R)$ and we can integrate the last equation $$f*g(x) = f(x)+ \int_{\mathbb R}f(-y)g(y)dy + \int_{\mathbb R} g(y)P(x,-y) dy$$ $$f(x)=f*g(x) - \int_{\mathbb R}f(-y)g(y)dy - \int_{\mathbb R} g(y)P(x,-y) dy$$ the three function in the right hand side are $C^{\infty}$ so $f\in C^{\infty}$ and the proof continues as Aditya sugest for example.

score 0 · Answer 2 · edited Sep 08 '14 at 14:29

0

Rough Outline: $$f'(u+av)=f'(u)+P'_{n-1}(u,v)$$ Put $v=-u/a$: $$f'(u)=f'(0)-P'_{n}(u,-u/a)$$ $$f(u)=P^{\circ}_{n}(u)+c_1u+c_2\quad \text{where }c_1\text{ and }c_2\text{ are constants}$$ ^{even though $\bf P'_n$ has subscript $\bf n$ it is of degree $\bf n-1$, $\bf P^{\circ}_{n}(u,-u/a)=\int P'n(u)du$ which is of degree $\bf n$}

edited Sep 08 '14 at 14:29

user103828

2,368

answered Sep 07 '14 at 14:13

RE60K

17,716

A similar approach can be applied to the Cauchy functional equation but is it possible to do this without taking derivatives (like some solutions to the Cauchy functional equation where we substitute ``clever'' choices of $u$ and $v$)?.... I think some of your $v$'s should be replaced by $u$'s. – user103828 Sep 07 '14 at 14:37
Do you mean $P'_n(u,v)$? – almagest Sep 07 '14 at 14:39
When you put $v=-u/2$ the $P$ term becomes $P'_n(u,-u/2)$. So how do you integrate it back to $P_n(u,v)$? – almagest Sep 07 '14 at 14:41
Of course, all you are trying to do is to establish that $f$ is a polynomial of degree $n$, so all that is fixable. – almagest Sep 07 '14 at 14:43
@almagest tried to fix it$\require{cancel}\cancel{?}$ – RE60K Sep 07 '14 at 14:47
It is your answer, I think you should fix it! :) – almagest Sep 07 '14 at 14:47
@almagest I meant "I tried to fix it, does that look good?" – RE60K Sep 07 '14 at 14:48
Looks great, except that it is so tiny! :) – almagest Sep 07 '14 at 14:49
who said that $f(x)$ is differentiable too? – Shivang jindal Sep 07 '14 at 15:28
@Shivangjindal if we assume f is a polynomial then as p is also a polynomial hence .... – RE60K Sep 07 '14 at 16:07
hence nothing :) – Shivang jindal Sep 07 '14 at 16:35
@Shivangjindal question is "is f a polynomial of degree n" and our our assumption of taking f diffrentiable is just an assumption, which seems to satisfy the conditions for f, note that f may also be non-diffrentiable, but that does not answer "is there a ..." which talks about possibility, if you insist prove it is not! – RE60K Sep 07 '14 at 16:50
Then a better proof would be. Assume it to be polynmoial of degree and check the conditions right? – Shivang jindal Sep 07 '14 at 17:24
i)of degree?.....ii)yes checking would be easy – RE60K Sep 07 '14 at 17:26
i don't think this is what question means :) – Shivang jindal Sep 07 '14 at 17:28
@Shivangjindal 31 – RE60K Sep 07 '14 at 17:30
Let us continue this discussion in chat. – Shivang jindal Sep 08 '14 at 16:50

Cauchy's functional equation with polynomial: $ f ( u + a v ) = f ( u ) + f ( a v ) + P _ n ( u , v ) $

2 Answers2