This is from my classnote:

Define $\displaystyle C[0,1]$ to be all continuous functions on [0,1], and define its metric be $\displaystyle p(f,g)= \int ^1 _0 \mid f-g \mid dx $

Show that if $\displaystyle f \neq g $, then $\displaystyle p(f,g) \neq 0 $

Proof.

if $\displaystyle f \neq g $, then $\displaystyle f(x_0) \neq g(x_0) $ for some $\displaystyle x_0 \in [0,1] $

Consider the function $\displaystyle h(x) = f(x)-g(x) $, we see that [tex]h(x) [tex] is continuous on $\displaystyle x_0$

Now, let $\displaystyle \epsilon = \mid f(x_0)-g(x_0) \mid $, then $\displaystyle \exists \delta > 0 \ s.t. \ \mid x - x_0 \mid < \delta $

with $\displaystyle \mid [ f(x) - g(x)]-[f(x_0)-g(x_0)] \mid < \frac { \epsilon }{2} $

Then we have $\displaystyle \mid f(x) - g(x) \mid \geq \frac {\epsilon }{2} $ on $\displaystyle [x_0 - \delta , x_0 + \delta ] $ (Why is that?)

Then we have $\displaystyle p(f,g)= \int ^1_0 \mid f-g \mid dx \geq \int ^{x_0+ \delta} _{x_0 - \delta } \mid f-g \mid dx $$\displaystyle \geq 2 \delta \frac { \mid f(x_0) - g(x_0) \mid }{2} > 0 $