Let $F$ the set of all continuous real functions with domain $[0,a]$. Which of the following are metrics on $F$?

  1. $d(f_1,f_2)$ is the maximum value of $|f_1(x)-f_2(x)|$ for $x\in[0,a],$
  2. $d(f_1,f_2)=\int^a_0|f_1(x)|-|f_2(x)|$
  3. $d(f_1,f_2)=\int^a_0 |f_1(x)-f_2(x)|$
  4. $d(f_1,f_2)=\int ^a_0|f_1(x).f_2(x)|$

My attempt:

for (3)

(1) $d(f_1,f_2)\ge 0\; \forall x\in [0,a]$

(2) $d(f_1,f_2)=0 \iff \int ^a_0|f_1(x)-f_2(x)| \iff|f_1(x)-f_2(x)|=0 \iff f_1(x)=f_2(x)$

(3) $d(f_1,f_2)=\int^a_0 |f_1(x)-f_2(x)|=\int^a_0 |f_2(x)-f_1(x)|=d(f_2,f_1)$

(4) $d(f_1,f_2)=\int^a_0 |f_1(x)-f_2(x)|\le d(f_1,f_2)=\int^a_0 |f_1(x)-f_3(x)|+ \int^a_0 |f_3(x)-f_2(x)|\le d(f_1,f_3)+d(f_3,f_2)$

So (3) is metric

(4) is not metric since it not satisfies property $d(f_1,f_2)=0\iff f_1=f_2$

However, I am not sure about that.

Can any one help with (1) and (2)?


(1) is a metric. Just check that it has all the properties of a metric.

(2) isn't. For instance, if $f_1$=0 and $f_2=1$, then $d(f_1,f_2)<0$.

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  • $\begingroup$ ....thank your fast response........is my reason for (4) is true sir?.....@José Carlos Santos $\endgroup$ – Inverse Problem Apr 11 '18 at 10:15
  • 1
    $\begingroup$ @SureshPonnada Yes, you are right about (4). But you should have added an example ($f_1=0$ and $f_1=1$, for instance). $\endgroup$ – José Carlos Santos Apr 11 '18 at 10:16
  • $\begingroup$ thank you sir once again .......@José Carlos Santos $\endgroup$ – Inverse Problem Apr 11 '18 at 10:17

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