PSEUDO METRIC & NORM OF A FUNCTION #pseudo #metric #norm #of #function
https://youtu.be/CoYUxNtFOt4?si=1NLs0hEpNkRjJvBo
Contents :
1. Norm of an element
2. Norm of a function
3. Pseudo Metric
NORM OF AN ELEMENT :
In each space there is defined a notion of the distance from an arbitrary element to origin, that is, a notion of the " size " of an arbitrary element. The size of an element x is a real number denoted by ||x|| and called its " norm ". For any arbitrary element x , the norm of x i.e. ||x|| has the following properties :
1. ||x|| ≥0 and
||x|| =0 iff x=0.
2. ||-x|| =||x||
3. ||x+y|| ≤ ||x|| +||y||.
Each metric arises as the norm of the difference between two elements d(x,y)= ||x-y||.
NORM OF A FUNCTION :
Let X be the set of all bounded continuous real valued functions defined on the closed interval [0,1].
Now we define the norm of a function f Ꜫ X as ||f|| = sup { |f(x)|/ x Ꜫ [0,1] }.
Also the metric on this x is defined as
for any f,g ꜪX, d(f,g) = ||f-g|| = sup { |f(x)-g(x) |/x Ꜫ [0,1] }
PSEUDO METRIC :
Let X be a non-empty set , and let d be a real function of ordered pair of elements of X which satisfies the following conditions.
i. d(x,y) ≥ 0 and
x=y implies d( x,y ) =0.
ii. d(x,y) =d(y,x)
iii. d(x,z) ≤ d(x,y)+d(y,z)
A function d with these properties is called a pseudo - metric on X and hence X is called a pseudo-metric space.
Note:
A metric is obviously a pseudo-metric . But every Pseudo- metric is need not be a metric.
why?
Let here we see the example :
Consider the metric space R2 and a function d :RxR→[0, ∞) by
d( (x1,y1),(x2,y2)
)= |x1-x2|
Is this function d a pseudo-metric
or metric?
Verification :
i.
Since the absolute value of any variable is non-negative , this
mapping d is also non-negative.
ii.
For any (x1,y1),
(x2,y2) in R2 , suppose (x1,y1) = (x2,y2)
→ x1= x2 and y1=y2
→ x1-x2
=0
→ | x1-x2 | = 0
→ d( (x1,y1),(x2,y2)
) = 0
iii.
d( (x1,y1),(x2,y2) ) =
| x1-x2 |
= | x2- x1|
= d( (x2,y2),(x1,y1)
)
∴ d holds symmetry.
iv.
d( (x1,y1),(x3,y3) ) =
|x1-x3|
= | x1-x2+x2-x3|
≤ | x1-x2|+|x2-x3|
= d( (x1,y1),(x2,y2)
) + d( (x2,y2),(x3,y3) )
∴ d( (x1,y1),(x3,y3) ) ≤ d( (x1,y1),(x2,y2)
) + d( (x2,y2),(x3,y3) )
∴ d satisfies transitivity.
So by all these properties d is a
Pseudo-metric on R2.
Now
what about d( (x1,y1),(x2,y2) ) = 0
implies (x1,y1) = (x2,y2)?
Verification :
For (3,5),(3,7) in R2, d(
(3,5),(3,7) ) =|2-2| =0
But (3,5) ≠
(3,7).
Hence this pseudo-metric is not a metric.
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#pseudo #metric #norm #of #function #symmetry #transtivity
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