Spring 2017, problem 43
Comments
Let x=y. Then (f(x))2=f(0).
On the other hand,
f(x)f(0)=f(x)⟺f(x)(f(0)−1)=0.
As f(x)≢ 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}.
Letfbe a real-valued function which satisfies (a) for all realx,y,f(x+y) +f(x−y) = 2f(x)f(y). (b) there exists a real numberx0such thatf(x0) =−1. Prove thatfis periodic.
Solution:Swappingxandyyields that function is even. Yet plugging inx=y= 0 we getf(0) = 0or 1. If it is 0, then plugging iny= 0 yieldsf≡0, done.Otherwise,f(0) = 1, and plug inx=y=x0/2 to getf(x0)+1 = 2f(x0/2)2, implying thatf(x0/2) = 0.Now plugging iny=x0/2, we get thatf(x+x0/2) =−f(x−x0/2), so function inverts sign everyx0.Hence periodic with period 2x0 for more visit
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.
Nice & elegant version of my solution, Hubert......thanks!
Call P(x,y) the given property above, a with fa≠0 and α=f0,
P(a,o)⇒ αfa=fa⇒ α=1,
P(x,x)⇒ f2x=α≠0,
P(x,x2)⇒ fx.fx2=fx2≠0⇒ fx=1 which works
f≡1.