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If $f$ be a real function satisfying $f(x+y+z)=f(x)f(y)f(z), \forall x,y,z \in \mathbb{R}.\space$ If $f(2)=4$ and $f'(0)=3$, how to find the values of $f(0)$ and $f'(2)$?

For the first part,I assumed $x=y=z$ then $f(3x)=f(x)^3$,which gives $3f'(3x)=3f(x)^2f'(x)$, substituting $x=0,$$$3f'(0)=3f'(0)f(0)^2 \Rightarrow f(0)= \pm 1 $$

Is this process yield the correct result? And how to find the $f'(2)$?

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Hint: Begin by looking at the value of $f(0 + 0 + 0)$. –  Dilip Sarwate Oct 23 '11 at 15:25
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One way to answer this question is to show that no function with the given properties actually exists. –  Chris Eagle Oct 23 '11 at 15:38
    
@Tretwick: where did you find such a problem? –  user17090 Oct 23 '11 at 16:13
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2 Answers

up vote 5 down vote accepted

There is no such function. If the functional equation is satisfied and is differentiable at zero, then either $f(x) = 0$ for all $x$ or $f(x) = \pm e^{ax}$ for some $a$.

To see this, letting $y = z = 0$ gives $f(x) = f(x)f(0)^2$ for all $x$, so unless $f(x)$ is the zero function you have to have $f(0) = \pm 1$. Since replacing $f(x)$ by $-f(x)$ doesn't change whether or not the functional equation is satisfied, replace $f(x)$ by $-f(x)$ if necessary and assume $f(0) = 1$.

Setting $z = 0$ gives $$f(x+y) = f(x)f(y)$$ For fixed $x$, taking the derivative of the right hand side with respect to $y$ at $y = 0$ gives $f(x)f'(0)$, while on the left hand side the same limit gives $f'(x)$. So for all $x$ you have $$f'(x) = f(x)f'(0)$$ This is a standard first-order differential equation. $f(x)$ has to be $C^1$ since the right-hand side is continuous (in fact we have already differentiated it), and the solution is $$f(x) = Ce^{f'(0) x}$$ Since we are assuming $f(0) = 1$ for now, $C = 0$ and therefore for $a = f'(0)$ you have $$f(x) = e^{ax}$$ Taking in account that we may have replaced $f(x)$ by $-f(x)$ above, you thus have $$f(x) = \pm e^{ax}$$ To see that $f(2) = 4$ and $f'(0) = 3$ can't happen, observe that $f'(0) = \pm a$, so that $a = \pm 3$, but $f(2) = \pm e^{ax} = \pm e^{\pm 6}$, which is never $4$.

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f(x+y+z) = f(x) f(y) f(z)

Suppose there exists an x such that f(x) is non-zero (It does exist according to the question. For instance f(2) = 4). Then substitute z=-y.

f(x+y-y) = f(x) = f(x) f(y) f(-y)

Since f(x) is non-zero, f(x) may be eliminated. Then:

f(y) f(-y) = 1

For every y. This also means that f(y) is nonzero for every y. Since f(y) is differentiable, we can deduce that f(y) is either positive or negative for every y. And, since we know that f(2)=4, hence f(x) is positive.

Since f(x) is positive it may be written as:

f(x) = exp[ g(x) ]

Now recall that f(x+y+z) = f(x) f(y) f(z):

exp[ g(x+y+z) ] = exp[ g(x) ] exp[ g(y) ] exp[ g(z) ] = exp[ g(x) + g(y) + g(z) ]

g(x+y+z) = g(x) + g(y) + g(z)

g(0) = g(0) *3. Hence g(0) = 0

g(x+y) = g(x) + g(y). Hence g(x) is linear.

The solution for g(x) is: g(x) = kx.

Substitute:

  • f(x) = exp[ kx ]
  • f'(x) = exp[ kx ] * k

Substitute:

  • f(2)=4
  • f′(0)=3

This gives:

  • exp[ 2k ] = 4
  • exp[ 0 ] * k = 3

Which does not comply

Hence - there is no such a function.

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