Answer to Question #150758 in Functional Analysis for Ashweta Padhan

Question

Show that (C[a, b], ||f||2) is a norm

Answer 1

First of all, for $f\in\mathcal{C}([a;b])$ we see that

$||f||_2 = (\int_a^b |f(x)|^2 dx)^{1/2}$

is well defined and is in $[0;+\infty)$ .

Now let's verify the norm properties:

1. Separation, $||f||_2 = 0 \leftrightarrow f=0$ . If $f=0, ||f||_2=0$ is obvious. Now suppose $f\neq 0, |f|^2 \neq 0$ is a continuous positive function and thus $\int_a^b |f(x)|^2 dx \neq 0$, therefore $||f||_2\neq 0$ .
2. Linearity, $||\alpha f||_2 = |\alpha| \cdot||f||_2$ . It follows from linearity of an integral : $(\int_a^b |\alpha f|^2(x) dx)^{1/2} = (|\alpha|^2 \int_a^b |f(x)|^2 dx)^{1/2}=|\alpha|\cdot||f||_2$
3. Triangle inequality, $||f+g||_2 \leq ||f||_2 + ||g||_2$ .

$\int_a^b|f+g|^2(x) dx \leq \int_a^b|f|^2(x) dx +\int_a^b|g|^2(x)dx + 2 \int_a^b |fg| (x) dx$

We have $(\int_a^b |fg|(x) dx)^2 \leq (\int_a^b |f(x)|^2 dx) \cdot (\int_a^b |g(x)|^2 dx)$, as

$\frac{|fg|}{(\int_a^b |f(x)|^2 dx)^{1/2}\cdot(\int_a^b |g(x)|^2 dx)^{1/2}} \leq \frac{\frac{|f|^2}{\int_a^b |f(x)|^2 dx}+\frac{|g|^2}{\int_a^b |g(x)|^2 dx}}{2}$ (by applying $2|ab|\leq a^2+b^2$ to $\frac{|f|}{(\int_a^b |f(x)|^2 dx)^{1/2}}$ and $\frac{|g|}{(\int_a^b |g(x)|^2 dx)^{1/2}}$ ) and thus by integrating on $[a;b]$ we find :

$\frac{\int_a^b|fg| (x)dx}{(\int_a^b |f(x)|^2dx)^{1/2}\cdot(\int_a^b |g(x)|^2 dx)^{1/2}} \leq 1$

Therefore $(\int_a^b |fg|(x) dx)^2 \leq (\int_a^b |f|^2 dx) \cdot (\int_a^b |g|^2 dx)$ .

And we conclude :

$||f+g||_2^2 = \int_a^b|f+g|^2 \leq \int_a^b|f|^2 +\int_a^b |g|^2 +2\int_a^b |fg|$

$||f+g||_2^2 \leq \int_a^b|f|^2 +\int_a^b|g|^2 +2(\int_a^b|f|^2)^{1/2}(\int_a^b|g|^2)^{1/2}$

$||f+g||_2^2 \leq (||f||_2+||g||_2)^2$

And thus $||f+g||_2 \leq ||f||_2 + ||g||_2$ .