Dear Dinu,

Setting y=1/x, we have f(siny/y)=f(y)cosy, for every y in R\{0}. We know that the limit of siny/y at 0 exists and is equal to 1, and f is continuous at 1. As a result, limy->0f(siny/y)=f(1). Further, since both the cosine function and f are continuous at 0, we have limy->0(f(y)cosy)=f(0)cos0=f(0). Since both limits limy->0f(siny/y) and limy->0(f(y)cosy) exist, and since the equation f(siny/y)=f(y)cosy holds in a deleted neighborhood of 0, we have limy->0f(siny/y)=limy->0(f(y)cosy), and thus f(1)=f(0).

Next, setting x=1/(kpi) for every nonzero integer k, the equation

f(xsin(1/x))=f(1/x)cos(1/x) yields f((1/kpi)sin(kpi))=f(kpi)cos(kpi), whence

f(0)=f(kpi)(-1)k. Hence, we have f(0)=f(kpi)(-1)k and f(0)=f((k+1)pi)(-1)k+1, and combining the last two equations yields f(kpi)(-1)k=f((k+1)pi)(-1)k+1, whence f(kpi)=-f((k+1)pi), and thus f(kpi)f((k+1)pi)=-(f((k+1)pi))2, which is equal to zero if f((k+1)pi)=0, and negative otherwise. In the second case, applying Bolzano's theorem for f to the interval [kpi,(k+1)pi], we have that f vanishes at an interior point of the previous interval. Hence, in any case, f vanishes in the interval [kpi,(k+1)pi], for every nonzero integer k. Note that for k=-1, we have that f vanishes in [-pi,0], Now, by the equation f(0)=f(kpi)(-1)k, we have, for k=1, that f(0)=-f(pi), whence f(0)f(pi)=-(f(pi))2, and by the previous argument, we derive that f also vanishes in [0,pi]. Consequently, f vanishes in [kpi,(k+1)pi] for every integer k.

Actually, we can show that f(kpi)=0 for every integer k.

Proof

Setting x=2/((2n-1)pi) for every positive integer n, we have

sin(1/x)=sin((2n-1)pi/2)=(-1)n+1 and cos(1/x)=0, and the given equation yields f(2(-1)n+1/((2n-1)pi))=0, whence limn->infinityf(2(-1)n+1/((2n-1)pi))=0.

Besides, the sequence (2(-1)n+1/((2n-1)pi)) converges to 0, and f is continuous at 0. As a result, limn->infinityf(2(-1)n+1/((2n-1)pi))=f(0), and thus f(0)=0. Now, since f(0)=f(kpi)(-1)k for every nonzero integer k (see previous post), we have that f(kpi)=0 for every nonzero integer k. Therefore, f(kpi)=0 for every integer k.

Note that combining the previous statement and the statement f(1)=f(0), we derive that f(1)=0.

Spiros Konstantogiannis Congratulations dear Spiros! An wonderful approach!

Dear Dinu, thank you.

It seems f=0 is the unique continuous function from R to R satisfying above property, i.e. f(siny/y)=f(y)cosy, for every y in R\{0}. Difficult, maybe impossible, to prove that !

We can modify the problem in the following way:

Find all continuous functions f:R->R satisfying f(xsin(z/x))=f(1/x)cos(z/x) for all x in R\{0} and z in R.

For z=1 we obtain the initial condition, then f(0)=0, as Spiros Konstantogiannis proved.

Now we take a fixed arbitrary not null z in R, and, with x=n>0, we have

f(nsin(z/n))=f(1/n)cos(z/n) => f(z[sin(z/n)/(z/n)])=f(1/n)cos(z/n). When n tends to infinity, using continuity of f, we obtain f(z)=0