Is every norm in R^n a continuous function?

Yes, because in finite dimensional spaces all norms are topologically equivalent. 


Yes, and this is true more generally (same reason as John Cook) for norms over a finitedimensional vector space over a field complete with respect to an absolute value. It doesn't work for infinitedimensional spaces. 


I agree that this looks like a homework question, but since some people have already bitten, I'd just like to point out what might be said for infinitedimensional spaces. So suppose you have an infinitedimensional real or complex vector space, equipped with a norm  .  What does it mean for a function on V to be continuous? Well, you have to specify a topology on V, and it's natural to use the one defined by the norm. But then it's an immediate corollary of the triangle inequality that the norm function is continuous with respect to the topology it defines. (In some sense, if this weren't true, then we wouldn't bother studying normed vector spaces!) However, V might also carry some weaker topology (such as a w*topology induced by some predual) and then the norm will not in general be continuous with respect to that topology. (Silly remark: equip R^n with the indiscrete topology, i.e. the one with only two members. Then the usual norm is not continuous. Of course, that's a ridiculous topology to put on the space. I have a feeling that every Hausdorff topology on R^n for which translations and dilations are continuous, is equivalent to the usual one, but I'd need to check in something like Rudin's book to be sure.) 


The answer is "yes". The exaplanation depends on what is meant by "continuous". Let's agree that "the norm   is continuous" means "if (x_n) is a sequence of vectors such that the coordinates x_n converge to the coordinates of some vector x, then x_n converges to x". Let e_1, ..., e_n be the standard basis of R^n. First one shows that for every i, the function t > t*e_i is continuous (t real). Then, after writing every vector as a_1 * e_1 + ... + a_n * e_n, the result follows easily from the triangle inequality. 

