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.