Solving the functional equation using some more rigorous method

Question

Let $ f: \mathbb{R} \rightarrow \mathbb{R} $ be a differentiable function such that $$ f(x + y) = f(x) + f(y) + 2xy \quad \forall x, y \in \mathbb{R} $$ and $ f'(0) = 3 $. Find $f(x)$.

Having no immediate intuition for the functional form, I posited that $ f(x) $ might be a polynomial of degree $n$.

• The left-hand side (LHS) clearly has degree $n $

• The right-hand side (RHS) must have degree at least $2$

This observation naturally led me to consider the simplest non-trivial case: a quadratic polynomial. Implementing this assumption yielded the correct solution that matched the provided answer key.

While this approach proved successful, I find it unsatisfactory from a theoretical standpoint. There should exist a principled justification for initially assuming a polynomial form, rather than relying on what amounts to an educated guess.

The fact that the quadratic solution matches the given answer suggests it may be the unique solution. However, I currently lack the tools to rigorously prove that no other functional forms satisfy the equation.

I would be particularly interested to learn if there exists a more systematic methodology to attack this problem - one that doesn't depend on making initial assumptions about the functional form but rather derives it deductively.

Any insights or guidance on these matters would be most valuable.

Answer 1

If you differentiate your equation with respect to $x$, you get that for all $x,y$, $f’(x+y)=f’(x)+2y$. Then evaluating in zero you get that for all y $f’(y)=3+2y$.

Then just integrate and $f(x)=3x+x^2+c$ for some $c$, and then $c=0$ as $f$ has to satisfy your first equation.

Answer 2

There's a method for this type of equations when we are not given differentiability and $f'(0)$, but only continuity: First, $y=0$ gives us the value $f(0)=0$. Setting $y=kx$, we obtain $$f((k+1)xX)-f(kx)=f(x)+2kx^2.$$ Summing from $k=0$ to $n-1$, the LHS telescopes, and we have $$f(nx)=nf(x)+n(n-1)x^2.\tag1$$ Substituting $x\to\frac{x}n$ and solving for $f(\frac{x}n)$ gives $$f\left(\frac{x}n\right)=\frac1n f(x)-\frac{n-1}{n^2}x^2.$$ Using this and (1) with $m$ instead of $n$, we obtain $$f\left(m\frac{x}n\right)=m\left(\frac1n f(x)-\frac{n-1}{n^2}x^2\right)+m(m-1)\left(\frac{x}n\right)^2,$$ i.e. $$f\left(\frac{m}{n}x\right)=\frac{m}{n}f(x)+\frac{m(m-n)}{n^2}x^2.$$ This is valid for positive integers $m,n$, so we have $$f(rx)=rf(x)+r(r-1)x^2\tag3$$ for positive rational $r$, and that's valid for $r=0$ as well. From the original equation with $y=-x$, we see that $f(-x)=-f(x)+2x^2$, so $$f(-rx)=-rf(x)+(-r)(-r-1)x^2,$$ and (3) is valid for every rational $r$. With $x=1$, (3) becomes $$f(r)=(f(1)-1)r+r^2\tag4$$ for rational $r$, and by continuity, for every real $r$. To make the solution unique, we'd need another condition, e.g. $f(1)=4$.

Answer 3

Even with no assumption on $f$ at all, we have $f(0) = 0$ and

$$f(0) = 0 = f(x)+f(-x)-2x^2$$

Therefore the even part of $f$ is $x^2$; in other words $f(x) = x^2 + g(x)$ where $g$ is an odd function. $g$ satisfies the functional equation

$$(x+y)^2 + g(x+y)= x^2 + g(x) + y^2 + g(y) + 2xy$$

in other words, $g(x+y) = g(x)+g(y)$. This strongly suggests $g$ is linear (and in your case since $g$ is differentiable, it is linear). Then you just figure out the coefficients.