標題:
(Pure Mathematics) A question of a continuous function
發問:
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Let R denote the set of all real number and f : R → R a continuous function not identically zero such that f ( x + y ) = f ( x ) f ( y ) for all x , y ∈RShow that(a)(1) f ( x ) ≠ 0 ∈R(a)(2) f ( x ) > 0 ∈R(a)(3) f ( 0 ) = 1(a)(4) f ( - x ) = [ f ( x ) ] ^ - 1(b) Prove... 顯示更多 Let R denote the set of all real number and f : R → R a continuous function not identically zero such that f ( x + y ) = f ( x ) f ( y ) for all x , y ∈R Show that (a)(1) f ( x ) ≠ 0 ∈R (a)(2) f ( x ) > 0 ∈R (a)(3) f ( 0 ) = 1 (a)(4) f ( - x ) = [ f ( x ) ] ^ - 1 (b) Prove that for any rational number r , f ( r x ) = [ f ( x ) ] ^ r Hence prove that there exists a constant ' a ' , such that f ( x ) = a ^ x ?x∈R 更新: ★ What is the meaning of ‘ not identically zero ’?
最佳解答:
First of all, we have to know the meaning of “not identically zero”. Let g : R → R, g is an identically zero function if g(x) = 0 for all x belong to R. (a) (1) Suppose there exists x1 in R such that f(x1) =0, then for any x belong to R, (x-x1) also belongs to R, by the given properties, f(x) = f(x-x1+x1) = f[(x-x1) +x1] = f(x-x1) f(x1) = [f(x-x1)](0) =0. It means that for all x in R, f(x) =0, and that f is an identically zero function, which contradicts with the given condition that f is not an identically zero function. Hence the assumption is wrong in that there does not exist any x1 such that f(x1)=0. In other words, for all x belonging to R, f(x)≠0. (a)(2) For all x belonging to R, (x/2) also belongs to R, we therefore have f(x) = f(x/2 +x/2) = f(x/2) f(x/2) = [f(x/2)]^2 ≧ 0. This, coupled with (a)(1), means that f(x) ＞ 0 for all x belonging to R. (a)(3) For all x belonging to R, f(x) = f(x+0) = f(x)f(0). Since f(x) ≠0 by (a)(1), this implies that f(0) =1. (a)(4) For all x belonging to R, 1= f(0) = f(x + (-x)) = f(x) f(-x). Therefore, f(x) = f(-x) ^(-1) (b) for any rational number r, r may always be expressed in the form of q/p where both p and q are integers. For all x belonging to R, given that both x and r are finite (once they are determined), we have f(rx) = f((qx)/p) = f [(x/p) + (x/p) + …. (x/p)] (there are total of q terms in bracket) = [f(x/p)]^q. -------------- (1) On the other hand, f (x) = f(x/p + x/p + …x/p) (number of terms in bracket is p) = [f(x/p)]^p. It means that f(x/p) = f(x)^(1/p). Substitute it into (1), we have f(rx) = [f(x)]^(q/p) = [f(x)]^r. The next part is very easy. For all x belonging to R, there are two cases, x is rational or x is irrational. If x is rational, then f(x) = f[(x)(1)] = [f(1)]^x by what we have just proven. Then take f(1) as the constant a required. If x is irrational, there always exist an indefinite sequence of rational number {An} whose limit is x (the choice of the sequence does not matter). Then {f(An)} is also a convergening sequence whose limit is f(x). For each An in this sequence, given their rationality, f(An) = [f(1)]^An. It suffices to show that {f(1)^An} has a limit of f(1)^x. Now, when n --à ∞, An ---à x, hence given f(1) is a fixed constant (and therefore finite), f(1)^An ---à f(1)^x. Since there is only one limit for each sequence, we have f(x) = f(1)^x. So, the required a is f(1).
其他解答:
(a1) Suppose there is a x0 such that f(x0) = 0 Then f(y + x0) = f(x0) f(y) = 0 for any y contradict f is not identity zero so f(x) ≠ 0 for all x ∈R (a2) f(x) = f(x/2 + x/2) = [f(x/2)]^2 > 0 (For the last inequality sign, since f is real-valued we must have k^2 >= 0, in addition to (1), f(x) ≠ 0) (a3) f(0) = f(0)f(0) then f(0)[ f(0) - 1 ] = 0 so f(0) = 0 or 1 if f(0) = 0, then f(y + 0) = f(y) f(0) = 0 for any y, contradict f is not identically zero so the case f(0) = 0 must be rejected, and hence f(0) = 1 (a4) 1 = f(0) = f(x + (-x)) = f(x)f(-x) as f not zero at any x, we have f(-x) = 1 / f(x) (b) Let r = p/q for some p, q relative prime with q being non-zero first by MI, we proved that f(nx) = [f(x)]^n for all positive integers n, the case when n = 0 is trivial, (for negative n, just use 4 to extend the fact holds for negative n.) then f(p/q * x) = f( p * x/q) = [f(x/q]^p = [f(x)]^(p/q) Note: f(x/n) = f(x) ^(1/n) is obtained by considering f(x) = f(n * x/n) = f(x/n) ^(1/n) so last equality holds So we have for rational r, we proved f(rx) = f(x) ^ r let f(x0) = a, since for any real number x, there is a sequence of rational number {rn} converging to a, so taking limit on n to infinity, we have lim f(rn * x0) = lim f(x0) ^(rn) f[ lim (rn) * x0] = f(x0) ^ (lim (rn) f(x *x0) = a ^ x Let x*x0 = y we have f(y) = a ^ (y/xo) = b^y as required
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