# Continuous Functions Problems

I'm not sure how to solve continuous functions problems. So, could someone please explain these three problems with a solution?

Find all continuous functions $f$ for $x>0$ such that $f(xy)=xf(y)+yf(x)$.

Find all continuous functions $f$ for $x>0$ such that $f(x+y)=\frac{f(x)f(y)}{f(x)+f(y)}$.

Find all continuous real-valued functions $f$ satisfying $f(x+y)=f(x)+f(y)+f(xy)$

Are there any special function properties that maybe I don't know to solve these? Thanks!

• Is $y$ any real number, or is it positive, too, or is it something else? – Stefano Mar 12 '17 at 19:12
• o yeah, it is also >0 – user406996 Mar 12 '17 at 19:13
• One typical strategy: try to prove something about $f$ on the integers, then extend it to the rationals, then (by continuity) extend it to all reals. For instance, start by plugging in $x=1$ and $y=1$ in the first functional equation to find $f(1)$. Then plug in $x=y$, for instance, etc. – Clement C. Mar 12 '17 at 19:23
• For the first part, I am noticing that $f(y) = f(1y) = 1f(y) + yf(1) = f(y) + yf(1) \Longrightarrow yf(1) = 0 \Longrightarrow f(1) = 0$. – joeb Mar 12 '17 at 19:27
• I got that, too, but I don't think that helps much :( – user406996 Mar 12 '17 at 19:29

Here is a hint for the first one. First, I'll give a "trick," then explain where it came from.

For any solution $f\colon(0,\infty)\to\mathbb{R}$ to the first functional equation, define $g\colon \mathbb{R}\to \mathbb{R}$ by $$g(x) \stackrel{\rm def}{=} e^{-x} f(e^x).$$ Then $g$ is continuous on $\mathbb{R}$, given $g$ we can easily get back $f$ since $f(x) = x g(\ln x)$ for $x>0$, and, for any $x,y\in\mathbb{R}$, \begin{align} g(x+y) &= e^{-(x+y)}f(e^{x+y}) = e^{-(x+y)}f(e^{x}e^y) = e^{-(x+y)}\left(e^xf(e^y)+e^yf(e^x)\right) \\ &= e^{-y}f(e^{y})+e^{-x}f(e^{x}) = g(x)+g(y) \end{align}

Therefore, it is sufficient to solve the following:

Find all continuous functions $g\colon\mathbb{R}\to\mathbb{R}$ such that $$g(x+y)=g(x)+g(y), \qquad \forall x,y\in\mathbb{R}.$$

Hopefully, you should have a much easier time to do this — this is a standard exercise.

Now, where did that come from? First, $f$ was defined on the positive reals only. This is a bit fishy: between this and the multiplicative aspects of the functional equation, it strongly hints that there should be some logarithms/exponentials involved. And indeed, setting $x=e^u$, $y=e^v$ for any $u,v\in\mathbb{R}$, we get $$f(e^{u+v})=f(e^ue^v)=f(xy) = xf(y)+yf(x) = e^u f(e^v) + e^v f(e^u)$$ for any $u,v$. This looks a bit nicer, but we'd like to "decouple" $u$ and $v$ in the equation. The simplest way to do so is to multiply both sides by $e^{-u} e^{-v}$, getting $$e^{-(u+v)}f(e^{u+v}) = e^{-v} f(e^v) + e^{-u} f(e^u)$$ which now looks much more manageable, and suggests to "set" $g(u) = e^{-u}f(u)$.

• Oops, I'm sorry I missed a detail. g(x+y)=g(x)+g(y) when g(x)=cx...? (c is a constant) – user406996 Mar 12 '17 at 19:53
• @user406996 Yes, it's a standard exercise to show that any such $g$ must be linear. See Cauchy's functional equation. – Clement C. Mar 12 '17 at 19:55
• How will this help solve the question, though? I don't see a good application, yet. – user406996 Mar 12 '17 at 20:03
• If you have a continuous solution $g$ to the Cauchy's functional equation, you get a solution $f$ to the original problem by setting $f(x) = xg(\ln x)$ (cf. the answer). And vice-versa. So solving the original or the new problem is equivalent. – Clement C. Mar 12 '17 at 20:05
• I don't see it :( Is there another possible route to solving this? Or is this the only way? – user406996 Mar 12 '17 at 20:11