1. ## Metric of C[0,1]

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 $$h(x) [tex] is continuous on \displaystyle x_0 Now, let \displaystyle \epsilon = \mid f(x_0)-g(x_0) \mid , then \displaystyle \exists \delta > 0 \ s.t. \ \mid x - x_0 \mid < \delta  with \displaystyle \mid [ f(x) - g(x)]-[f(x_0)-g(x_0)] \mid < \frac { \epsilon }{2}  Then we have \displaystyle \mid f(x) - g(x) \mid \geq \frac {\epsilon }{2}  on \displaystyle [x_0 - \delta , x_0 + \delta ]  (Why is that?) Then we have \displaystyle p(f,g)= \int ^1_0 \mid f-g \mid dx \geq \int ^{x_0+ \delta} _{x_0 - \delta } \mid f-g \mid dx \displaystyle \geq 2 \delta \frac { \mid f(x_0) - g(x_0) \mid }{2} > 0  2. Originally Posted by tttcomrader Then we have \displaystyle \mid f(x) - g(x) \mid \geq \frac {\epsilon }{2}  on \displaystyle [x_0 - \delta , x_0 + \delta ]  (Why is that?)$$
By the triangle inequality, and properties of the absolute value we have:

$\displaystyle \vert f(x_0)-g(x_0) \vert - \vert f(x)-g(x) \vert \leq \vert f(x)-g(x) - [f(x_0)-g(x_0)] \vert < \frac{ \epsilon }{2}$ now, using the first and last parts of the inequality we have: $\displaystyle \vert f(x_0)-g(x_0) \vert -\frac{ \epsilon }{2} < \vert f(x)-g(x) \vert$ and using the fact that $\displaystyle \vert f(x_0)-g(x_0) \vert = \epsilon$ you're finished