This function again attains its maximum modulus at the center of delta_2 and hen

Question

This function again attains its maximum modulus at the center of delta_2 and hence is constant on delta_2. Iterating this procedure we obtain that f(z) = f(a) for all z A. By the identity theorem we have that f(z) = f(a) for all z U. Let V be the volume measure on R^2n and hence on C^n. Suppose delta centered at a C^n, and f is a function holomorphic on a neighborhood of delta Prove f(a) = 1/V(delta) integral_delta f(eta) dV(eta). That is, f(a) is an average of the values on a polydise centered at a. Pane the maximum principle by using the Cauchy formula instead. (Him: use previous exercise) Prove a several variables of the Schwarz's lemma; Suppose f is in a neighborhood of D^n, f(0) = 0, and for some k N we have whenever |alpha|

Explanation / Answer

ans)Since Ran(f) R, we can choose B = Ran(f) and it is obvious that f is onto B. In order to prove that f is one-to-one, pick arbitrary x, y A with f(x) = f(y) and we will show that x = y. By our assumption, f(x) = 3x x2 = 3y y2 = f(y) and so 3x(y 2) = 3y(x 2). Expanding and collecting everything to one side of the equation we get that 3xy 6x 3xy + 6y = 0, i.e. 6x = 6y so x = y, proving our claim. In order to determine what the set B is, we choose arbitrary y R and find conditions under which there is an x A with f(x) = y. Specifically, we need to isolate x in the equation f(x) = y and eliminate the cases where the result is undefined in y. 3x x2=y 3x=y(x 2) 3x=xy 2y (y 3)x=2y x= 2y y3 This equation is undefined when y = 3 but defined everywhere else, hence B = R {3}. (b) Find a formula for f 1 (x)

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