# Math Help - P norm and essential supremum

1. ## P norm and essential supremum

If ||f ||p= {òx |f|p dm}1/p is p norm of measurable complex function
' f ' , m denoting
the measure on set X ,what is the relationship between lim|| f ||p (p®¥) (if exists)
and || f ||¥ the essential supremum of f ?

2. I've never seen this stated, but I think when $|| f ||_p = \left[ \int |f|^p d\mu \right]^{1/p},$

$\lim_{p \to \infty} \frac{|| f ||_p}{\mu(X)^{1/p}} = || f ||_{\infty}.$

Is this proof correct or is there a simpler one?

Let $M = || f ||_{\infty} = \text{ess }\sup |f(t)| = \inf\ \{ N:\ \mu\{ t:|f(t)| > N \} = 0\ \}.$

Define for $0 < \epsilon < M,$ $g_{\epsilon}(t) = \min \{ |f(t)|,M-\epsilon \}.$

Since

$
\lim_{p \to \infty} \left| \frac{g_{\epsilon}(t)}{M-\epsilon}\right|^p = \begin{cases}
1 & \text{if } g_{\epsilon}(t) = M-\epsilon \\
0 & \text{otherwise} \\
\end{cases}
$

and

$\mu\{ t : g_{\epsilon}(t) = M -\epsilon \} > 0$ by definition of $M,$

$0 < \lim_{p \to \infty} \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon}}{M-\epsilon}\right|^p d\mu \le 1.$

(Note: getting a positive expression here so the next limit equals 1 is the whole reason for introducing the function $g_{\epsilon}.$ This would not necessarily hold for $f.$)

Therefore

$\lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon}}{M-\epsilon}\right|^p d\mu \right]^{1/p} = 1,$

$\lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int |g_{\epsilon}|^p \ d\mu \right]^{1/p} = {M-\epsilon},$

$\lim_{p \to \infty} \frac{|| g_{\epsilon} ||_p}{\mu(X)^{1/p}} = {M-\epsilon},$

and letting $\epsilon \to 0,$

$\lim_{p \to \infty} \frac{|| f ||_p}{\mu(X)^{1/p}} = M = || f ||_{\infty}.$

3. is there a simpler one?

Not that I know of.

I only have the same question every time: Why are the limits for epsilon and for p interchangeable?

4. Originally Posted by Rebesques
Not that I know of.

I only have the same question every time: Why are the limits for epsilon and for p interchangeable?
Good question. I chewing on it.

5. Originally Posted by Rebesques
I only have the same question every time: Why are the limits for epsilon and for p interchangeable?
Thanks for questioning the proof. I think the limits are not interchangeable and the proof is flawed.

The goal of the proof is to show

$\lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int \left| \frac{f}{M}\right|^p d\mu \right]^{1/p} = 1.$

If $\mu\{ t : |f(t)| = M \} = 0$, then $|f(t)/M| < 1$ almost everywhere and

$\lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int \left| \frac{f}{M}\right|^p d\mu \right] = 0$.

Then the desired limit has the indeterminate form

$\lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int \left| \frac{f}{M}\right|^p d\mu \right]^{1/p}
= 0^0.$

My argument using $g_{\epsilon}$ to try to get around this is invalid.

I don't see the solution and it seems to me to be a non-trivial complication. I find this odd because the proof (or a simpler version of it) is left as an exercise for the reader in a textbook I have.

6. I also think the solution is simple (maybe we have the same textbook? )

I e-mailed a friend (who happens to be a better mathematician ) and he says you are justified to interchange the limits, as $g_{\epsilon}$ is monotonic for $\epsilon$.

Other than this, the expert says the limits can switch, as long as the quantity is nonnegative; If one of the double limits is infinity, the other will be the same. This last one, I must admit, has got me thinking

7. Originally Posted by Rebesques
I also think the solution is simple (maybe we have the same textbook? )

I e-mailed a friend (who happens to be a better mathematician ) and he says you are justified to interchange the limits, as $g_{\epsilon}$ is monotonic for $\epsilon$.

Other than this, the expert says the limits can switch, as long as the quantity is nonnegative; If one of the double limits is infinity, the other will be the same. This last one, I must admit, has got me thinking
I think your friend is not looking at the problem in enough detail.

The basic goal is to prove

$\lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int \left| \frac{f}{M}\right|^p d\mu \right]^{1/p} = 1.$

This means given $\alpha > 0$, there exists $P$ such that $p \ge P$ implies

$\left| \left[ \frac{1}{\mu(X)} \int \left| \frac{f}{M}\right|^p d\mu \right]^{1/p} - 1 \right| < \alpha.$

We can try using as my "proof" does that for fixed $\epsilon > 0$

$\lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon}}{M-\epsilon}\right|^p d\mu \right]^{1/p} = 1$

and

$\lim_{\epsilon \to 0} g_{\epsilon} = f \text{ a.e.}$ but it does not help.

To see this, write using the triangle inequality that

$
\left|\left[\frac{1}{\mu(X)}\int \left| \frac{f}{M}\right|^p d\mu \right]^{1/p}-1\right| \le$

$\left|\left[ \frac{1}{\mu(X)} \int \left| \frac{f}{M}\right|^p d\mu \right]^{1/p} - \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon}}{M-\epsilon}\right|^p d\mu \right]^{1/p} \right|$ $+
\left| \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon}}{M-\epsilon}\right|^p d\mu \right]^{1/p} - 1 \right| .$

Now $\lim_{\epsilon \to 0} g_{\epsilon} = f \text{ a.e.}$ guarantees that for each $p$ we can find $\epsilon(p)$ such that

$\left|\left[ \frac{1}{\mu(X)} \int \left| \frac{f}{M}\right|^p d\mu \right]^{1/p} - \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon(p)}}{M-\epsilon(p)}\right|^p d\mu \right]^{1/p} \right| < \alpha/2.$

But there is no corresponding guarantee that

$\left| \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon(p)}}{M-\epsilon(p)}\right|^p d\mu \right]^{1/p} - 1 \right| < \alpha/2.$

For that we would need for given $p$ that

$\lim_{\epsilon \to 0} \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon}}{M-\epsilon}\right|^p d\mu \right]^{1/p} = 1.$

Instead we have for given $\epsilon$ that

$\lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon}}{M-\epsilon}\right|^p d\mu \right]^{1/p} = 1.$

Oops.