Show that $L^p$ "space" for $0<p<1$ does not define a norm

Question

Why do $L^p$ "space" for $0<p<1$ does not define a norm? Any property which gets violated?

Answer 1

$||(1,0)+(0,1)|| = ||(1,1)|| = 2^{1/p} \geq 2 = 1+1 = ||(1,0)||+||(0,1)||$ where in the critical step we use the fact that p is less than one. So we have a counterexample to the triangle inequality.

Basically, the main reason these are not norms is that the unit ball is not convex, which means you can pick two points in the unit ball, like (0,1) and (1,0), and draw a line between them and have points on that line with norm greater than one: for example $||(1/2,1/2)|| = {(2/2^{p})^{1/p}}=2^{(1-p)/p}$ which is always greater than one if the exponent is positive, (which it is if $0<p<1$).

Answer 2

If the underlying measure space is $\mathbb{R}^n$ then consider the element $(1,0,\cdots,0)$ and $(0,1,\cdots,0)$. Adding them we get $(1,1,0,\cdots,0)$. While checking the triangle inequality, one can notice that it doesn't holds and hence there is no norm defined on $l^p(\mathbb{R}^n)$. But this is an example of $l^p$ spaces, where we are taking the $p-$ norm which is defined as $$ \|x\|_p = \Bigg( \sum_{i=1}^{n}|x_i|^p \Bigg)^{1/p} $$ for $x = (x_1,x_2, \cdots, x_n)$.

But the actual way of giving an example where the triangle inequality doesn't hold is as follows, where we take elements from $L^p(\mathbb{R}^n)$:

Let $A_1$ and $A_2$ be two measurable disjoint sets which are elements of the underlying measure space (by this I mean the underlying $\sigma-$ algebra). Consider $f_1 = \chi_{A_1}$ and $f_2 = \chi_{A_2}$, where $\chi_A$ denotes the indicator function or the characteristic function. Then one can check that for $0 < p < 1$, $$ \|f_1 + f_2\|_p \geq \|f_1\|_p + \|f_2\|_p $$