How to check whether the following functions are norms on the corresponding spaces?

Question

$\|f\|$:=sup$_{x\in[0,1]}$$\quad \frac{\vert f(x)-f(0)\vert}{x}$ on the space C[0,1]

and the same function but on the subspace V of $C^1$[0,1] of functions which equal zero at x=$\frac{1}{2}$.

How is the first function not a norm on that space and the later one is a norm on V?

$\| f+g\|$=sup$_{x\in[0,1]}$$\quad \frac{\vert f(x)+g(x)- f(0)-g(0)\vert}{x}$= sup$_{x\in[0,1]}$$\quad \frac{\vert f(x)-f(0)+g(x)-g(0))\vert}{x}$ $\leq$ sup$_{x\in[0,1]}$ $\frac{\vert f(x)-f(0)\vert}{x}$ + sup $_{x\in[0,1]}$ $\frac{\vert g(x)-g(0)\vert}{x}$ = $\| f\|+\|g\|$

Answer 1

In order to prove that $||.||$ is a norm, you have to prove the three properties of norms :

1. $||f||=0 \Leftrightarrow f=0$.
2. $||\alpha f||=|\alpha | ||f||$, for all $\alpha \in \mathbb{R}$.
3. $ ||f+g|| \leq ||f||+||g||$.

So in your question, the first norm you gave defined on $C[0,1]$ is not a norm, since if you take $f(x)=2 $, then $f \in C[0,1]$ , and $||f||=\sup_{x \in [0,1]} \frac{|f(x)-f(0)|}{x}= \sup_{x \in [0,1]} \frac{|2-2|}{x}=0$, so $||f||=0$ but $f \neq 0$.

For the second norm, defined on $V= \{ f \in C^1[0,1], \text{ such that }f(\frac{1}{2})=0 \}$.

1. If $f=0$, then clearly $||f||=0$. Now suppose $||f=0||$, and required to show that $f=0$. Indeed, if $||f=0||$, then $\sup_{x \in [0,1]} \frac{|f(x)-f(0)|}{x}=0$, this implies $\frac{|f(x)-f(0)|}{x}=0$ for all $x\in [0,1]$, hence $|f(x)-f(0)|=0$ for all $x\in [0,1]$, so $ f(x)=f(0) $ for all $x\in [0,1]$. As $f(\frac{1}{2})=0$, then $f=0$.

The two other properties are left for you as an exercise, and I am ready for help.