# Functional equation determination

Let a function $$f$$ be continuous and differentiable for all x such that it satisfies $$f(x+y)f(x-y)= f^2(x)$$ Given that $$f(0)$$ is nonzero and $$f(1)$$ is 1. how to find f. I tried replacing $$x$$ by $$y$$ and then by $$-x$$, but not able to proceed further.

• What happens if you put $x=0$? Also, you could add what the result is if you replace $x$ by $y$. – supinf Feb 8 at 8:52
• The tag (functional-analysis) is intended for questions about infinite dimensional vector spaces, there is a separate tag for (functional-equations); see the tag-wiki and the tag-excerpt. (The tag-excerpt is also shown when you are adding a tag to a question.) – supinf Feb 8 at 8:54

## 3 Answers

Hints: the answer is $$f(x)=e^{a(x-1)}$$ for some conatant $$a$$. First prove that $$f(t)=0$$ for some $$t$$ leads to $$f \equiv 0$$, a contradiction. Then prove that $$f(x) >0$$ for all $$x$$. Let $$g(x)=\ln\, f(x)$$ and show that $$g(\frac {a+b} 2)= \frac {g(a)+g(b)} 2$$. From this we get $$g(at+(1-a)s)=ag(t)+(1-a)g(s)$$ for dyadic rational $$a$$, hence for all $$a$$ (by continuity). Conclude that $$g(x)$$ has the form $$ax+b$$

• i dont understand your solution. – maveric Feb 8 at 9:21
• The basic idea is the following: assuming that $f$ is a positive function take logarithms to get $2g(x)=g(x+y)+g(x-y)$ where $g(x)=\log \, f(x)$. This is a much simpler equation to solve and it can be written as $g(\frac {a+b} 2)=\frac {g(a)+g(b)} 2$. We can solve this by iterating this equation. – Kavi Rama Murthy Feb 8 at 9:28
• yes. the only problem i have is how to prove fucntion is protive before taking log – maveric Feb 8 at 9:39
• @maveric If $f(t)=0$ for some $t$ then $(f(x))^{2}=f(t) f(2x-t)=0$ (by taking $y=t-x)$. Hence $f(x)=0$ for all $x$ but $f(1)=1$. Hence $f$ never vanishes. Since $f$ is continuous it must be always positive or always negative. But $f(1)=1$, so it is always positive. – Kavi Rama Murthy Feb 8 at 9:45
• thankx.yeah got it. – maveric Feb 8 at 9:50

With $$x=y$$, $$f(2x)f(0)=f^2(x)$$ and $$f$$ has a constant sign and is positive ($$f(1)=1$$).

Let $$h(x):=\log\frac{f(x)}{f(0)}.$$

Setting $$x=y$$, we have

$$h(2x)=2h(x),$$

then with $$u:=x+y,v:=x-y$$,

$$h(u)+h(v)=2h\left(\frac{u+v}2\right)=h(u+v)$$

so that $$h$$ must be linear.

After a little computation,

$$f(x)=f_0^{1-x}.$$

Hint.

Assuming $$f(0) \ne 0$$ we have that $$f(y)f(-y) = f^2(0)$$ then we conclude that $$f(0) \ne 0 \to f(x) > 0\to f(0) > 0$$

so making

$$\log_u f(x+y)+\log_u f(x-y) = 2\log_u f(x)$$

or

$$g(x+y) + g(x-y) = 2g(x)$$

and now making $$g(x) = a x + b$$

$$a(x+y)+b +a(x-y) + b = 2a x + 2 b$$

we have a good estimation as

$$\log_u f(x) = a x + b\to f(x) = u^{ax+b}$$

Of course if $$f(0) = 0$$ then $$f(x) = 0$$ is also a solution.

• This proves a possible solution. But is it the only one ? – Yves Daoust Feb 8 at 10:57