2
$\begingroup$

Let $f:[0,1] \rightarrow R$ with $1 \leq f \leq 2$ set $$N(p)=\left( \int_0^1 f^p dx \right)^{\frac{1}{p}} \qquad p \neq 0$$ To find the three limits $\lim_{p\rightarrow \pm \infty} N(p)$ and $\lim_{p\rightarrow 0} N(p)$.

$\\$

One thing is clear: $\liminf_{p\rightarrow \infty} N(p) \geq 1$.

$\endgroup$
  • $\begingroup$ $\liminf_{p\rightarrow 0^{+}} N(p) \neq + \infty$ $\endgroup$ – Paul Sep 16 '15 at 18:32
  • $\begingroup$ right, sorry let me edit that part... $\endgroup$ – user16015 Sep 16 '15 at 18:33
  • $\begingroup$ Aren't the hints in Rudin enough? Look at the exercises in chapter 3 of Rudin's Real and Complex Analysis. $\endgroup$ – guest Sep 16 '15 at 18:37
  • $\begingroup$ If $f$ is any continuous real function then $\lim_{p \to + \infty } (\int_0^1| f(x)|^p dx )^{1/p} = \max \{ |f(x)| : 0 \leq x \leq 1\}.$ $\endgroup$ – DanielWainfleet Sep 16 '15 at 18:40
1
$\begingroup$

For $p>0$ $$ 1\leqslant N(p)=\left( \int_0^1 f^p dx \right)^{\frac{1}{p}} \leqslant (2^p)^{1/p}=2 $$ For $p<0$, let $q=-p$. $$ 1\leqslant N(p)=N(-q)=\frac1{\left( \int_0^1 \frac1{f^q} dx \right)^{\frac{1}{q}}} \leqslant 2 $$ Since $$ \liminf_{p\to0^+}N(p)\geqslant 1\quad\text{and }\quad\limsup_{p\to0^-}N(p)\leqslant 1 $$ We have $$ \liminf_{p\to0}N(p)=\limsup_{p\to0}N(p)=1\quad\text{or }\quad\lim_{p\to0}N(p)=1 $$ Moreover we have $$ \liminf_{p\to{\pm\infty}}N(p)=1 \quad\text{and }\quad\limsup_{p\to{\pm\infty}}N(p)=2 $$ It is easy to find particular $f$ to attain above limits.

If $f$ is continuous, then $$ \lim_{p\to{\pm\infty}}N(p)=\sup{\{f(x):x\in[0,1]\}}=2 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.