Proof C[a,b] with the Norm ||⋅||∞ is a Normed Space

Clearly the set of continuous function $C[a,b]$ is a vector space. Now, we have to prove that the function ${\left||\cdot |\right|}_{\mathrm{\infty }}:C[a,b]\to \mathbb{R}$ such that 

${\left|\left|f\right|\right|}_{\mathrm{\infty }}={\displaystyle \underset{a\le x\le b}{max}|f(x)|}$ for every function $f\in C[a,b]$

is really a norm.

The first property we have to prove is:

${\left|\left|f\right|\right|}_{\mathrm{\infty }}=0$ if and only if $f=0$ is the zero function. 

Obviously, if $f$ is the zero function, 

${\left|\left|f\right|\right|}_{\mathrm{\infty }}={\displaystyle \underset{a\le x\le b}{max}|f(x)|={\displaystyle \underset{a\le x\le b}{max}\left|0\right|=0}}$.

Now, assume  $f$ is not a zero function, therefore there exists $c\in [a,b]$ such that $f(c)\ne 0$. Next, we obtain

${\left|\left|f\right|\right|}_{\mathrm{\infty }}={\displaystyle \underset{a\le x\le b}{max}|f(x)|\ge |f(c)|>0}$

$\Rightarrow {\left|\left|f\right|\right|}_{\mathrm{\infty }}\ne 0$.

Thus, we conclude that ${\left|\left|f\right|\right|}_{\mathrm{\infty }}=0$ if and only if $f$ is the zero function. 

Next, we have to prove for every $f\in C[a,b]$ and $k\in \mathbb{R}$,

${\left|\left|kf\right|\right|}_{\mathrm{\infty }}=\left|k\right|{\left|\left|f\right|\right|}_{\mathrm{\infty }}$.

Take any $f\in C[a,b]$ and $k\in \mathbb{R}$. Since $f$ is a continous function then $\left|f\right|$ is also continous on $[a,b]$. Now, since $\left|f\right|$ is continous on the compact set $[a,b]$, $\left|f\right|$ has maximum (this is also telling us that  ${\left||\cdot |\right|}_{\mathrm{\infty }}$ is well defined). Let $c\in [a,b]$ such that $|f(c)|$ is the maximum value.

Given $x\in [a,b]$, we obtain

$|kf(x)|=\left|k\right||f(x)|\le \left|k\right||f(c)|$,

and 

$|kf(c)|=\left|k\right||f(c)|$.

It shows that $\left|k\right||f(c)|$ is the maximum value of $\left|kf\right|$. In other words

${\left|\left|kf\right|\right|}_{\mathrm{\infty }}={\displaystyle \underset{a\le x\le b}{max}|kf(x)|=\left|k\right||f(c)|=\left|k\right|{\left|\left|f\right|\right|}_{\mathrm{\infty }}}$.

Hence, the second property has been proved.

Lastly, we have to prove the triangle inequality. It is easy to see that for any given $f,g\in C[a,b]$, we obtain for every $x\in [a,b]$,

$|f(x)+g(x)|\le |f(x)|+|g(x)|\le {\displaystyle \underset{a\le x\le b}{max}|f(x)|+{\displaystyle \underset{a\le x\le b}{max}|g(x)|={\left|\left|f\right|\right|}_{\mathrm{\infty }}+{\left|\left|g\right|\right|}_{\mathrm{\infty }}}}$.

Therefore,  ${\left|\left|f\right|\right|}_{\mathrm{\infty }}+{\left|\left|g\right|\right|}_{\mathrm{\infty }}$ is an upper bound of $|f+g|$. Thus, we have proved the triangle inequality

${\left||f+g|\right|}_{\mathrm{\infty }}={\displaystyle \underset{a\le x\le b}{max}|f(x)+g(x)|\le {\left|\left|f\right|\right|}_{\mathrm{\infty }}+{\left|\left|g\right|\right|}_{\mathrm{\infty }}}$.

We have proved all of the properties that necessary for ${\left||\cdot |\right|}_{\mathrm{\infty }}$ to be a norm on the vector space $C[a,b]$. So, $C[a,b]$ with the norm ${\left||\cdot |\right|}_{\mathrm{\infty }}$ is a normed space. $◼$