So a couple of days ago, a Jedi Master gave me a pretty straightforward problem to look at:

**[({|f(x)| : ||x||=1} is bdd) (f: X R is continuous)]**

And after thinking about it for a while, I came up with the following rationale:

**[({|f(x)| : ||x||=1} is bdd) (f: X R is continuous)]**

Suppose {|f(x)| : ||x||=1} is bdd. In other words, M s.t. {|f(x)| : ||x||=1} M < .

Given > 0, y X, choose z s.t. y-z=x. Then ||y-z||=1 and we know that an M exists.

Also, we know that for N R, = 1/N, which gives us |f((y-z)/N)| = |(1/N)f(y-z)| = |(1/N)(f(y)-f(z))| M/N 0 as N .

Thus, choosing N s.t. M/N < , we have that ||(y-z)/N|| = 1/N ||f(y-z)/N|| M/N. Notice that any N_1 > N also suffices for our inequality, since N_1 > N M/N_1 < M/N. Thus ||y-z|| 1/N ||f(y)-f(z)|| M/N < .

**[(f: X R is continuous) ({|f(x)| : ||x||=1} is bdd)]**

Suppose f: X R is continuous. Then we know that given > 0, y X, > 0 s.t. ||y-z|| < |f(y)-f(z)| = |f(y-z)| < .

For this fixed , choose s.t. < .

Then |f((y-z)/)| = |(1/)f(y-z)| < / and ||(y-z)/|| = (1/)||y-z|| < / , any vector x = (y-z)/ s.t. ||x|| = ||(y-z)/|| = 1 still gives us our bound |f((y-z)/)| < / .

Said Jedi Master then responded by telling me that one can rewrite the first part as follows:

Let M be s.t. x with ||x||=1, |f(x)| M. Given > 0, y X, choose = /M.

If ||z-y|| then |f(z)-f(y)| = |f(z-y)| = ||z-y|| |f((z-y)/||z-y||)| M = , where we know that |F((z-y)/||z-y||)| M because |F((z-y)/||z-y||)| = |F((z-y)/||z-y||)|, and ||((z-y)/||z-y||)|| = 1.

By the method applied, this actually also proves that f is uniformly continuous.

Using the exact same idea, the second part of the proof can be simplified to:

Suppose > 0, y X, > 0 s.t. ||y-z|| < |f(y)-f(z)| = |f(y-z)| < .

Then |f(y-z)| = ||y-z|| = ||y-z|| |f((y-z)/||y-z||)| <

|f((y-z)/||y-z||)| < /||y-z|| /.

I guess that’s why he’s a Jedi Master, and I’m just Scott.

## Leave a Reply