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 ?
I've never seen this stated, but I think when $\displaystyle || f ||_p = \left[ \int |f|^p d\mu \right]^{1/p},$
$\displaystyle \lim_{p \to \infty} \frac{|| f ||_p}{\mu(X)^{1/p}} = || f ||_{\infty}.$
Is this proof correct or is there a simpler one?
Let $\displaystyle M = || f ||_{\infty} = \text{ess }\sup |f(t)| = \inf\ \{ N:\ \mu\{ t:|f(t)| > N \} = 0\ \}.$
Define for $\displaystyle 0 < \epsilon < M,$ $\displaystyle g_{\epsilon}(t) = \min \{ |f(t)|,M-\epsilon \}.$
Since
$\displaystyle
\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
$\displaystyle \mu\{ t : g_{\epsilon}(t) = M -\epsilon \} > 0$ by definition of $\displaystyle M,$
$\displaystyle 0 < \lim_{p \to \infty} \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon}}{M-\epsilon}\right|^p d\mu \le 1.$
(Note: getting a positive expression here so the next limit equals 1 is the whole reason for introducing the function $\displaystyle g_{\epsilon}.$ This would not necessarily hold for $\displaystyle f.$)
Therefore
$\displaystyle \lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon}}{M-\epsilon}\right|^p d\mu \right]^{1/p} = 1,$
$\displaystyle \lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int |g_{\epsilon}|^p \ d\mu \right]^{1/p} = {M-\epsilon},$
$\displaystyle \lim_{p \to \infty} \frac{|| g_{\epsilon} ||_p}{\mu(X)^{1/p}} = {M-\epsilon},$
and letting $\displaystyle \epsilon \to 0,$
$\displaystyle \lim_{p \to \infty} \frac{|| f ||_p}{\mu(X)^{1/p}} = M = || f ||_{\infty}.$
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
$\displaystyle \lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int \left| \frac{f}{M}\right|^p d\mu \right]^{1/p} = 1.$
If $\displaystyle \mu\{ t : |f(t)| = M \} = 0$, then $\displaystyle |f(t)/M| < 1$ almost everywhere and
$\displaystyle \lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int \left| \frac{f}{M}\right|^p d\mu \right] = 0$.
Then the desired limit has the indeterminate form
$\displaystyle \lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int \left| \frac{f}{M}\right|^p d\mu \right]^{1/p}
= 0^0.$
My argument using $\displaystyle g_{\epsilon}$ 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.
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 $\displaystyle g_{\epsilon}$ is monotonic for $\displaystyle \epsilon$.
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
$\displaystyle \lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int \left| \frac{f}{M}\right|^p d\mu \right]^{1/p} = 1.$
This means given $\displaystyle \alpha > 0$, there exists $\displaystyle P$ such that $\displaystyle p \ge P$ implies
$\displaystyle \left| \left[ \frac{1}{\mu(X)} \int \left| \frac{f}{M}\right|^p d\mu \right]^{1/p} - 1 \right| < \alpha.$
We can try using as my "proof" does that for fixed $\displaystyle \epsilon > 0$
$\displaystyle \lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon}}{M-\epsilon}\right|^p d\mu \right]^{1/p} = 1$
and
$\displaystyle \lim_{\epsilon \to 0} g_{\epsilon} = f \text{ a.e.}$ but it does not help.
To see this, write using the triangle inequality that
$\displaystyle
\left|\left[\frac{1}{\mu(X)}\int \left| \frac{f}{M}\right|^p d\mu \right]^{1/p}-1\right| \le$
$\displaystyle \left|\left[ \frac{1}{\mu(X)} \int \left| \frac{f}{M}\right|^p d\mu \right]^{1/p} - \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon}}{M-\epsilon}\right|^p d\mu \right]^{1/p} \right|$ $\displaystyle +
\left| \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon}}{M-\epsilon}\right|^p d\mu \right]^{1/p} - 1 \right| .$
Now $\displaystyle \lim_{\epsilon \to 0} g_{\epsilon} = f \text{ a.e.}$ guarantees that for each $\displaystyle p$ we can find $\displaystyle \epsilon(p)$ such that
$\displaystyle \left|\left[ \frac{1}{\mu(X)} \int \left| \frac{f}{M}\right|^p d\mu \right]^{1/p} - \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon(p)}}{M-\epsilon(p)}\right|^p d\mu \right]^{1/p} \right| < \alpha/2.$
But there is no corresponding guarantee that
$\displaystyle \left| \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon(p)}}{M-\epsilon(p)}\right|^p d\mu \right]^{1/p} - 1 \right| < \alpha/2.$
For that we would need for given $\displaystyle p$ that
$\displaystyle \lim_{\epsilon \to 0} \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon}}{M-\epsilon}\right|^p d\mu \right]^{1/p} = 1.$
Instead we have for given $\displaystyle \epsilon$ that
$\displaystyle \lim_{p \to \infty} \left[ \frac{1}{\mu(X)} \int \left| \frac{g_{\epsilon}}{M-\epsilon}\right|^p d\mu \right]^{1/p} = 1.$
Oops.