Condition about functional equation has a solution

Question

Problem :

Let $P(x, y)$ be a polynomial respect to $x, y$.

Find a condition about $P$ where functional equation of $f$

$$f(x+y)=f(x)+f(y)+P(x,y)$$

has a solution. (Where $f\colon \mathbb{R}\to\mathbb{R}$ be a differentiable function.)

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This problem came from some functional equation problems below :

P1 :

Find a differentiable $f$ which satisfies $$f(x+y)=f(x)+f(y)+2xy$$

S1 :

Set $x=y=0$ then we get $f(0)=0$.

From definition of derivative,

$$f'(x)=\lim_{y\to 0}\frac{f(x+y)-f(x)}{y} = \lim_{y\to 0} \frac{f(y)+2xy}{y} = 2x+f'(0)$$

This implies $$f(x)=x^2 + cx$$

for some constant $c$.

Plugging this to original functional equation,

$$\text{LHS} = x^2 + 2xy + y^2 + cx + cy$$ $$\text{RHS} = x^2 + y^2 + cx + cy + 2xy$$

which are same, so we find a solution of original functional equation.

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P2 :

Find a differentiable $f$ which satisfies $$f(x+y)=f(x)+f(y)+3x^2y$$

S2 :

Similar way, $f(0)=0$ and from definition of derivative,

$$f'(x)=f'(0)+3x^2$$

which implies $$f(x)=x^3 + cx$$

for some constant c. However If we plug this to original functional equation

$$\text{LHS} = x^3 + 3x^2y + 3xy^2 + y^3 + cx + cy$$ $$\text{RHS} = x^3 + 3x^2y + y^3 +cx + cy$$

which are not same.

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So, I want to know a condition of $P(x, y)$ which makes a functional equation of $f$ (which likes below form) has a solution :

$$f(x+y)=f(x)+f(y)+P(x,y)$$

Where $f$ is a differentiable function.

I find that WLOG $P(0, 0) = 0$

Because, if $P(0, 0) = a \neq 0$, add $a$ both side then we can write as

$$(f(x+y)+a) = (f(x)+a) + (f(y)+a) + Q(x, y)$$

Where $Q(x, y)$ is a polynomial respect to $x, y$ which satisfies $Q(0, 0) = 0$.

Substitution $g(x) = f(x)+a$ changes this equation to :

$$g(x+y)=g(x)+g(y)+Q(x, y)$$

So I can assume that WLOG $P(0,0)=0$.

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Added

Conjecture :

$P(x,y)$ must be constant or middle term(Cross -terms) of binomial expansion, i.e.

$$ (x+y)^2 = x^2 + y^2 + \mathbf{2xy}$$ or $$(x+y)^3 = x^3 + y^3 + \mathbf{3x^2y + 3xy^2}$$

I don't have a proof but I made some try and common thing of $P(x,y)$ is this.

Thanks for your help!

Answer 1

The complete description of the required polynomials $P(x,y)$ is provided by the following simple theorem.

Theorem. Let $P(x, y)\in \mathbb R[x,y]$ be a polynomial and $f\colon \mathbb{R}\to\mathbb{R}$ be a differentiable function such that $f(x+y)=f(x)+f(y)+P(x,y)$ for each $x,y\in\mathbb R$. Then $f$ is a polynomial.

Proof. For each $x,y\in \mathbb R$, $y\ne 0$ we have $\frac{f(x+y)-f(x)}{y}=\frac{f(y)+P(x,y)}{y}$. Since there exists $f'(x)=\lim_{y\to 0}\frac{f(x+y)-f(x)}{y}$, we have that $f(0)+P(x,0)=0$ and $f'(x)=f'(0)+P_y(x,0)$. Since the latter expression is a polynomial with respect to $x$, so are $f'(x)$ and $f(x)$. $\square$

Answer 2

We can use L'Hôpital's rule to solve for $f(x).$ \begin{eqnarray} f(x+y)&=&f(x)+f(y)+P(x,y)\\ f(0)&=&-P(0,0)\\ f'(x)&=&\lim_{y\to0}\frac{f(x+y)-f(x)}y\\ &=&\lim_{y\to0}\frac{f(y)+P(x,y)}{y}\\ f'(x)&=&f'(0)+P_y(x,0)\\ f(x)&=&\int_0^xP_y(t,0)dt+f'(0)x-P(0,0) \end{eqnarray} Substituting this into the functional equation, we get that $P(x,y)$ must satisfy \begin{eqnarray} P(x,y)-P(0,0)&=&\int_0^{x+y}P_y(t,0)dt-\int_0^xP_y(t,0)dt-\int_0^yP_y(t,0)dt\\ \left.P(x,t)\right|_0^y&=&\int_x^{x+y}P_y(t,0)dt-\int_0^yP_y(t,0)dt\\ \int_0^yP_y(x,t)dt&=&\int_0^y\left(P_y(t+x,0)-P_y(t,0)\right)dt\\ \int_0^y(\left.P_y(s,t)\right|_0^x+P_y(0,t))dt&=&\int_0^y\left.P_y(s,0)\right|_t^{t+x}dt\\ \int_0^y\int_0^xP_{xy}(s,t)dsdt+P(0,0)&=&\int_0^y\int_0^xP_{xy}(s+t,0)dsdt\\ P(0,0)&=&\int_0^y\int_0^x(P_{xy}(s+t,0)-P_{xy}(s,t))dsdt\\ &=&\int_0^y\int_0^x\left.-P_{xy}(s+t-r,r)\right|_0^tdsdt\\ &=&\int_0^y\int_0^x\int_0^t(P_x-P_y)_{xy}(s+t-r,r)drdsdt \end{eqnarray} For example, $P(x,y)=2xy$ is a solution. $$0=\int_0^y\int_0^x\int_0^t0drdsdt$$ But $P(x,y)=3x^2y$ isn't. $$0\ne3xy^2=\int_0^y\int_0^x\int_0^t6drdsdt$$