Is $L_\infty$ norm the smallest or largest? I am a little bit confused. For a $L_p$ function norm, is it true that for any $ p<\infty $, 
$$ \|f\|_p>\|f\|_\infty$$
Is the statement true for any domain? I want to know more inequality about norm also.
 A: This is actually a very deep question. If you weren't confused, you'd be missing something big.
The structure of the space $L^p(X,\mu)$ is difficult. Here $\mu$ is a measure, but if you don't have experience with that just think of $\int_X |f|^p dx$ where $X$ is any nice space, like $\mathbb{R}$ or $[0,2\pi]$, or whatever motivated this question. Note that you have two things to be confused about: what is $X$ and what is $\mu$? 
I find to learn this stuff it's best to memorize a few basic examples to refer to whenever you encounter a new theorem. That way, you connect what it's saying to what it actually means (like H\"older's or Jensen's inequality -- to tell you the truth I have no deep understanding of these inequalities). 
The skeletal summary of working examples is as follows


*

*If the space you're working on is "finite" (i.e. $\mu(X) < \infty$), then $L^\infty \subset L^1$. This follows because $\int_X |f| \leq \sup |f|\mu(X).$ On the other hand you can take, e.g. $1/\sqrt{x}$ on $[0,1]$ which is in $L^1$ but not $L^\infty$. Interpolate to get $L^1 \supset L^2 \supset L^{3.15} \supset \dots \supset L^\infty$. (See Inclusion of $L^p$ spaces, bounded or unbounded)

*If the space you're working on does not have set of arbitrarily small size (like balls of radius $\epsilon > 0$), then $L^1 \subset L^\infty$. Consider $a:= \sum_{n=0}^\infty |a_n| < \infty$. Clearly $|a_n| \leq a < \infty$ for each $n$, so $a \in L^\infty$. This is known as counting measure. Once again, everything in between: $L^1 \subset L^2 \subset \dots \subset L^\infty$. 

A: It could be less or more.  Here are some examples:
On the less than side let
f(x) = x,  $0\le x \le 1/2$
$f(x) = 1,  1/2 < x \le 1$.  
Then $||f||_{L^ \infty} = 1 > ||f||_{L^p}$ for any p.
On the greater than side consider the interval [0,2] and let
f(x) = 1  for $0 \le x \le .001$ and
f(x)= .99 for $.001 < x \le 2$
The $\int_0^2 |f(x)|dx > 1.98$ but $||f||_{L^ \infty}$ = 1.
