Spring 2017, problem 43

Determine all functions $f$ from the real numbers to the real numbers, different from the zero function, such that $f(x)f(y)=f(x-y)$ for all real numbers $x$ and $y$.

Comments

1 year ago

Call $P(x,y)$ the given property above, $a$ with $f_a\neq0$ and $\alpha=f_0$,

  • $P(a,o)\Rightarrow~\alpha f_a=f_a\Rightarrow~\alpha=1,$

  • $P(x,x)\Rightarrow~f^2_x=\alpha\neq 0,$

  • $P(x,\frac x2)\Rightarrow~f_x.f_{\frac x2}=f_{\frac x2}\neq 0\Rightarrow~f_x=1$ which works

$f\equiv1$.

1 year ago

Let $x=y$. Then $(f(x))^{2}=f(0)$.

On the other hand,

$f(x)f(0)=f(x)\iff f(x)(f(0)-1)=0.$

As $f(x)\not\equiv 0$ then $ f(0)=1$, and so $f(x) = \pm1$.

Claim: $ \not\exists x: f(x)= -1$.

Indeed, suppose $\exists x: f(x)=-1$. As $f(x)f(\frac{x}{2})=f(\frac{x}{2})$ we have $f(x)=1$, because $f(\frac{x}{2})\neq0$. A contradiction.

Hence, we conclude that $f(x)\equiv 1$.

Ex: $f(x)= (cos(x))^{2}+ (sin(x))^{2}$.

1 year ago

Assuming $f(x)$ is differentiable, let's differentiate $f(x)f(y) = f(x-y)$ with respect to $x$ and $y$:

$f'(x) f(y) = f'(x-y)$;

$f(x) f'(y) = -f'(x-y)$

which can be equated as: $\frac{f'(x)}{f(x)} = -\frac{f'(y)}{f(y)} = A$ (where $A $ is any real constant). Integrating this differential equation now produces:

$ln|f(x)| = Ax + B \Rightarrow f(x) = e^{Ax + B}$.

Taking $x = y = 0$, we can determine an initial condition for $f(x)$ above: $f(0)f(0) = f (0 - 0) \Rightarrow f(0)^2 = f(0) \Rightarrow f(0) = 0, 1$. Checking these into our solution for $f(x)$ yields:

$0 = e^{B}$ (not permissible), $1 = e^{B } \Rightarrow B = 0$ (permissible)

and thus the required set of functions is $f(x) = e^{Ax}$. Checking this solution against the original functional equation yields:

$e^{Ax} \cdot e^{Ay} = e^{A(x-y)} => e^{A(x+y)} = e^{A(x-y)} \Rightarrow Ax + Ay = Ax - Ay \Rightarrow Ay = -Ay$

which forces $A = 0$. Ultimately, $f(x) = e^{0x} = \boxed{1}$ is the only nonzero function that solves $f(x)f(y) = f(x-y)$ for $x,y \in \mathbb{R}$.

Assuming differentiability, $f$ is continuous,

by IVT, $f^2_x=1\Rightarrow~f\equiv \pm1$;

$f_1.f_0=f_1\Rightarrow~f\equiv1$.

Hubert 1 year ago

Nice & elegant version of my solution, Hubert......thanks!

TME102475 1 year ago

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