In our lecture notes we had an example of an Pseudo-metric which is not an metric:
Let $R([a,b])$ the space of the Riemann-integrable functions on the interval $[a,b]$, where $a<b$, then $$ \rho(f,g)=\int_{a}^{b} |f(t)-g(t)|dt $$ is an Pseudo-metric, however no metric. Obviously it has to fail at the condition where: $$ \rho(f,g)=0 \implies f=g $$ Unfortunately, I can not find an counter-example to show this.
Can anybody help please?
$(\scr{R}[a, b],\rho)$ is a pseudometric space. Where, $\rho(f,g)=\int_{a}^{b} |f(t)-g(t)|dt$
It satisfies all properties of a metric except the definiteness property.i.e $\rho(f, g) =0 \iff f=g$
It satisfy only "if" part. i.e $\rho(f, g) =0 $ if $f=g$
But, $\rho(f, g) =0$ may not imply $f=g$.
For an example, $f:[a, b]\to \Bbb{R} $ where
$f(x)=\begin{cases} 0 &\text{ if } x\in [a, b)\\1 &\text{ if } x=b\end{cases}$
Clearly, $f\in \scr{R}[a, b]$ as $D_f=\{b\}$
And $\rho(f, 0) =0$ but $f\not\equiv 0$.
But , $C[a, b]$ as subspace of $(\scr{R}[a, b],\rho)$ is a metric space (Check!)
Here continuity plays a crucial role to make the pseodometric to a metric.