Let $X := [0, 1]$ , the closed unit interval. Then
$$\|f\|_2 := \left(\int_{0}^{1} |f(t)|^2 dt \right)^{\frac{1}{2}}$$
defines a norm on the set of all continuous real/complex valued functions on $[0, 1]$. Two properties of norm were easy to verify. But i am stuck in triangle inequality of the norm. For $1\le p<\infty$, is $\|f\|_p := \left(\int_{0}^{1} |f(t)|^p dt \right)^{\frac{1}{p}}$ also a norm ?
- non-negativty is trivial and $\|f\|_2 = 0 \iff $ $f=0$.
- $\|\alpha f\|_2 $ = $\left(\int_{0}^{1} |\alpha f(t)|^2 dt \right)^{\frac{1}{2}}$ $= |\alpha|\left(\int_{0}^{1} | f(t)|^2 dt \right)^{\frac{1}{2}}=|\alpha |\cdot\|f\|_2$ for all $\alpha \in \mathbb{R}$
- $\|f+g\|_2 $ $= \left(\displaystyle\int_{0}^{1} |f(t)+g(t)|^2 dt \right)^{\frac{1}{2}}$ $\le\left(\displaystyle\int_{0}^{1} \left(|f(t)|+|g(t)|\right)^2 dt \right)^{\frac{1}{2}} \le$ ?