Question

I need help. Question is from METRIC SPACE.

Answer 1

@eliassaab @ganeshie8 @SithsAndGiggles anyone please help

Answer 2

What are the four things a metric must satisfy?

Answer 3

1) d(x,y)>=0
2) d(x,y)=0 iff x=y
3) d(x,y)=d(y,x)
4) d(x,y)<=d(x,z)+d(z,y)

Answer 4

Right, so I would think you need only establish these conditions for the given metric, and that should be enough to prove that \((C[0,1],~d)\) is a metric space. I might be wrong though, I haven't taken a formal course in topology.

Answer 5

Yes we need to prove these 4 axioms only. And topology i too am not acquainted with.. :) I know only Real Analysis and Metric space just started

Answer 6

d(x, y) >= 0 is true as the absolute value is always non negative

Answer 7

yeah it is .:)

Answer 8

Actually i need help with establishing triangle's inequality, other two are pretty much trivial .:)

Answer 9

Okie

\[\rm \sup | f(x)-g(x)| = \sup | (f(x)-h(x) )+ (h(x) - g(x))|\]

Answer 10

Then can we write <= sup |f(x)-h(x)|+sup |h(x)-g(x)|?? I need justification too..

Answer 11

try using below :
\[\forall a \in [0, 1], ~\rm |f(a) - h(a)|\le \sup |f(x) - h(x)|\]

Answer 12

@ganeshie8 i cant get the solution... :(

Answer 13

\[\rm{ \begin{align}\forall a, |f(a)-g(a)| &\le |f(a)-h(a)|+ |h(a)-g(a)| \\~\\
&\le \sup |f(x)-h(x)|+ |h(x)-g(x)|
\end{align}}\]

Answer 14

you can replace left side of inequality with sup because you're iterating over the entire domain

Answer 15

****

\[\rm{ \begin{align}\forall a, |f(a)-g(a)| &\le |f(a)-h(a)|+ |h(a)-g(a)| \\~\\
&\le \sup |f(x)-h(x)|+ \sup |h(x)-g(x)|
\end{align}}\]

Answer 16

first line is just triangle inequality. the second line uses the fact that any distance has to be less than the sup

Answer 17

but how to show sup|fx)-g(x)|<=sup|f(x)-h(x)|+sup|h(x)-g(x)|.

Answer 18

thats exactly whats shown above right ?

Answer 19

rhs is thre, but the lhs is not there..

Answer 20

I see, let me think a bit..

Answer 21

rhs is also there :
Since you're iterating over the entire domain, sup is also included.

Answer 22

you can replace \(\rm \forall a, | f(a)-g(a)|\) by \(\rm \sup |f(x) - g(x)|\)

Answer 23

the value of distance between functions (|f(a)-g(a)|) will attain its max/sup for some x = a in the domain

Answer 24

oh yes, we will define it... Thanks.. I got it..:)

Answer 25

Yes :) Is this from analysis/topology ?

Answer 26

metric space is the onefold generalization of analysis and topology we will learn in masters.:).. it is from mathematical analysis

Answer 27

Ohk.. i shouldn't be attempting this question if it is from topology lol

Answer 28

@ikram002p is taking topology this sem

Answer 29

no it is not from topology... :)

Answer 30

haha no difference :P