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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.

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  • $\begingroup$ What happens if you put $x=0$? Also, you could add what the result is if you replace $x$ by $y$. $\endgroup$ – supinf Feb 8 at 8:52
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    $\begingroup$ 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.) $\endgroup$ – supinf Feb 8 at 8:54
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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 $ax+b$

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  • $\begingroup$ i dont understand your solution. $\endgroup$ – maveric Feb 8 at 9:21
  • $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy Feb 8 at 9:28
  • $\begingroup$ yes. the only problem i have is how to prove fucntion is protive before taking log $\endgroup$ – maveric Feb 8 at 9:39
  • $\begingroup$ @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. $\endgroup$ – Kavi Rama Murthy Feb 8 at 9:45
  • $\begingroup$ thankx.yeah got it. $\endgroup$ – maveric Feb 8 at 9:50
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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,

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

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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.

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

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