4

Find all continuous functions $f: \mathbb{R} \to \mathbb{R}$ satisfying $f(x+y) = f(x)+f(y)+2xy$.

We have that $f(0) = 0$ and $f(x+1) = f(x)+f(1)+2x$ and thus $f(x+1) - f(x) = f(1)+2x$. Then we see that $f(0) = f(x)+f(-x)-2x^2 \implies f(x)+f(-x) = 2x^2$. Then I get stuck since I don't see what I can do next.

user19405892
  • 15,592
  • You have probably guessed that $x^2$ is one solution. Now define $g(x) := f(x) - x^2$ and see what you can deduce about $g(x)$. – Erick Wong Apr 12 '16 at 03:19

1 Answers1

2

Define $g(x)=f(x)-x^2$. Then by the above equation, $$g(x+y)+(x+y)^2=g(x)+x^2+g(y)+y^2+2xy \implies g(x+y)=g(x)+g(y)$$

To solve the latter, note that $g(0)=0$, and $g(cx)=cg(x)$ for all integers $c$. Let $\frac{p}{q}$ be any rational number, then replace $x$ by $\frac{x}{c}$ in the equation $g(cx)=cg(x)$ to get: $$g(x)=cg\biggl(\frac{x}{c}\biggr).$$

Now, $$ g\biggl(\frac{px}{q}\biggr) = pg\biggl(\frac{x}{q}\biggr) = \frac{p}{q}g(x) $$ Given that $g$ is continuous, it follows that $g(cx)=cg(x)$ for all real numbers (use the sequential definition of continuity). Thus, suppose $g(1)=k$ for some constant $k$, then for every $x$, $g(x)=kx$. It follows that $g(x)$ is always of this form, and hence $f(x)$ is of the form $kx+x^2$ for some constant $k$. Let's confirm $kx+x^2$ for fun: $$ f(x+y) = k(x+y)+(x+y)^2 = kx + x^2 + ky + y^2 + 2xy = f(x)+f(y)+2xy $$ Hence the answer.

Christopher Carl Heckman
  • 4,458
Sarvesh Ravichandran Iyer
  • 76,105
  • Isn't it circular to define $f(x)$ in terms of itself, i.e. $g(1)$? – user19405892 Apr 12 '16 at 17:19
  • I understand what you are saying. The point is , any value of $g(1)$ will work : of course if $g$ exists, $g(1)$ is some value, say $2$ or $3$ or something. Then I am saying take $k=$ that number, and the rest of the function values are decided. In short, $g(x)=xg(1)$, and $g(1)$ is some fixed value (I am not fixing it, but I know it is some value and am calling that k), so I know how $g$ behaves. The argument is thus not circular : My $k$ isn't fixed, it could be anything, and $f(x)$ isn't defined in terms of $g(1)$, because we don't know what exactly $g(1)$ is. – Sarvesh Ravichandran Iyer Apr 13 '16 at 00:24
  • Once again, we don't know what $f(1)$ is. I am just assuming that it is some random value, say k (it has to be some value, right? By giving it a variable name like k, I am not saying I know the value of the quantity. If I say shoes cost k dollars per pair, I don't know the cost of the shoes, but I do know that they cost something.) The point is, I am trying to explain the nature of the function by looking at it's behavior at one point (which I don't know, so I call it $k$). Please ask again if it's not clear, because I'd like it if you leave happy. – Sarvesh Ravichandran Iyer Apr 13 '16 at 01:09
  • But $g(1) = f(1)-1$. So what you are saying is that $f(x) = (f(1)-1)x+x^2$, which isn't a valid definition for $f(x)$. – user19405892 Apr 13 '16 at 01:12
  • Ok. I will rephrase, so that I am clear. Let $k$ be any real number. Define $g(x)=kx$ and $f(x)=kx+x^2$. Simply see that $f(x)$ satisfies our properties, and also see that $f(1)=k+1$, and $g(1)=k$. I rephrased my answer somewhat loosely: What I said was that any value of $k$ would work, but in truth I should have said : $g(x)$ and $f(x)$ defined as above work for any $k$. This time, my argument is not circular : I fixed $k$ independently of $g$ and $f$, defined $g$ and $f$ in terms of $k$ and then got the result. It just so turns out that $g(1)=k$ and $f(1)=k+1$. – Sarvesh Ravichandran Iyer Apr 13 '16 at 01:19