# Limits of the p-norm

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 $\lim_{p\rightarrow 0} N(p)$.

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

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

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