Consider the set of continuous function on
\[[0,1]\]
.The set of functions is a vector space and we can define a norm on this space as
\[||f(x)|| = \sqrt{\int^1_0 (f(x))^2 dx}\]
We show that this satisfies all the properties of a norm.
1.
\[||f(x)|| \geq 0\]
and \[||f(x)|| =0 \leftrightarrow f(x)=0\]
everywhere on \[[0,1]\]
.\[||f(x)|| = \sqrt{\int^1_0 (f(x))^2 dx} \geq \sqrt{\int^1_0 0 dx} =0\]
Suppose
\[f(x) \neq 0 \]
for some interval \[[a,b] \subseteq [0,1]\]
so that \[|f(x)| > \epsilon > 0\]
for \[x \in [a,b]\]
Then
\[||f(x)|| > \sqrt{(b-a) \epsilon^2}= \epsilon \sqrt{b-a} \]
so that if \[||f(x))| =0, \: f(x)=0\]
2. \[||kf(x)|| =k ||f(x))|| \]
for any \[k, \: x\]
\[||k f(x))|| = \sqrt{\int^1_0 (kf(x))^2}= k \sqrt{\int^1_0 (f(x))^2 dx } = k ||f(x))||\]
3. \[||f(x)+g(x)|| \leq ||f(x)|| + ||g(x)||\]
\[\begin{equation} \begin{aligned} ||f(x)+g(x)||^2 &= \int^1_0 (f(x)+g(x))^2 dx \\ &= \int^1_0 (f(x))^2 + 2 f(x) g(x) + (g(x))^2 dx \\ & \leq \int^1_0 (f(x))^2 dx + 2 \int^1_0 |f(x)| |g(x)| dx + \int^1_0 (g(x))^2 dx \\ & \leq \int^1_0 (f(x))^2 dx + 2 \sqrt{\int^1_0 (f(x))^2 dx} \sqrt{\int^1_0 (g(x))^2 dx} + \int^1_0 (g(x))^2 dx \\ &= (||f(x)||+||g(x||)^2 \end{aligned} \end{equation}\]
Square rooting both sides gives
\[||f(x)+g(x)|| \leq ||f(x)|| + ||g(x)||\]