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Okay, here goes my attempt at this: part 1: Sn-1 is both complete and bounded (can we assume these two things, or do we have to prove them). Thus, Sn-1 must be a compact. Since f is a real function on compact Sn-1 into Rk, f is bounded. Thus, it follows that there exist b,c E Sn-1 such that...

Honestly, I have no clue about this one.

Homework Statement f is a polynomial with n variables (x1, x2, ... , xn) with real coefficients. Let Sn-1 = {x E Rn | x12 + x22 + ... + xn2 = 1} (n-1 unit sphere). Show that \exists b,c E Sn-1 such that m = f(b) \leq f(x) \leq f(c) \leq = M for all x E Sn-1. If f(x1, ... , xn) = a1x1 + a2x2...

hmmm. so f(x) = f(x-1) + f(1) = 2f(1) + f(x-2) = ... = f(x-x) + xf(1) = xf(x) but that only works when x is an integer. How would you do it for when x is rational?

Homework Statement Let f:R\rightarrowR satisfy f(x+y) = f(x) + f(y) for real numbers x and y. If we let f be continuous, show that \exists a real number b such that f(x) = bx. Homework Equations n/a The Attempt at a Solution Nooooo clue!

So how would you use boundedness to prove convergence? PS. MIT>CalTech

I'm sorry. It should've been 2an instead of an. I've edited it above

Homework Statement For n\geq1 let 2an \leq an-1 + an+1 Prove that an convergesHomework Equations n/aThe Attempt at a Solution 2an+1 \leq an + an+2 an+2 \geq 2an+1 - an How do I proceed? Ratio test?

Thanks for the answer! It makes a lot of sense. For the second part, I have a general idea of how to solve the problem. If X is dense in R, then inf|t-x| will always be zero. However, how can this be expressed with a formal proof?

Homework Statement Let X\subsetR be nonempty. Let f:R\rightarrowR be defined by f(x)=inft E X|t-x|. Show: 1. f is uniformly continuous 2. f\equiv0 if and only if X is dense in R Homework Equations none The Attempt at a Solution I am clueless as to how to go about this. Help!