Posted by: scott | April 3, 2010

continuity and boundedness of linear functionals

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

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

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

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

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

Given \epsilon > 0, y \in 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 \in R, \frac{||y-z||}{N} = 1/N, which gives us |f((y-z)/N)| = |(1/N)f(y-z)| = |(1/N)(f(y)-f(z))| \leq M/N \to 0 as N \to \infty .

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

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

Suppose f: X \to R is continuous. Then we know that given \epsilon > 0, y \in X, \exists \delta > 0 s.t. ||y-z|| < \delta \Rightarrow |f(y)-f(z)| = |f(y-z)| < \epsilon .

For this fixed \delta , choose \delta_1 s.t. \delta_1 < \delta .

Then |f((y-z)/\delta_1 )| = |(1/\delta_1)f(y-z)| < \epsilon /\delta_1 and ||(y-z)/\delta_1 || = (1/\delta_1 )||y-z|| < \delta /\delta_1 , any vector x = (y-z)/\delta_1 s.t. ||x|| = ||(y-z)/\delta_1 || = 1 still gives us our bound |f((y-z)/\delta_1 )| < \epsilon / \delta_1 .

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

Let M be s.t. \forall x with ||x||=1, |f(x)| \leq M. Given \epsilon > 0, y \in X, choose \delta = \epsilon /M.

If ||z-y|| \leq \delta then |f(z)-f(y)| = |f(z-y)| = ||z-y|| |f((z-y)/||z-y||)| \leq \delta M = \epsilon , where we know that |F((z-y)/||z-y||)| \leq 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 \forall \epsilon > 0, y \in X, \exists \delta > 0 s.t. ||y-z|| < \delta \Rightarrow |f(y)-f(z)| = |f(y-z)| < \epsilon .

Then |f(y-z)| = ||y-z|| \frac{|f(y-z)|}{||y-z||} = ||y-z|| |f((y-z)/||y-z||)| < \epsilon
\Rightarrow |f((y-z)/||y-z||)| < \epsilon /||y-z|| \leq \epsilon /\delta.

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

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