7. (a) Let X = R2, and for x, y E R1 define d(x,y) by d(x, y) = d( (xi, x2)) (yi, y2)) = 'xi — yi I { Ix' I + lx2 — y2I + IA if x2 = y2 , if otherwise. Show that (X, d) is a metric space. (b) Let X=...

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7. (a) Let X = R2, and for x, y E R1 define d(x,y) by
d(x, y) = d( (xi, x2)) (yi, y2)) = 'xi — yi I { Ix' I + lx2 — y2I + IA
if x2 = y2 , if otherwise.
Show that (X, d) is a metric space. (b) Let X= R, and for x, y E R define d(x,y) = Ix' + Ix — yl + ly1 when x 0 y, and d(x, y) = 0 when x = y. Show that d is a metric on R. Hint: (a) Let x = (xi, 3c2),y = (yl, y2) and z = (z1, z2) be in X. We note firstly that ixi — Yii C d(x, y). If x2 = y2, then d(x, y) = 'xi — yi I C Ix' — zi I+ Izi — Yi I d(x, z) + d(z, y). If x2 y2, then z2 cannot be equal to both x2 and y2; so assume z2 x2. Then
d(x,y) = 'xi' + lx2 — y2I + 1)/11 :5. Ix' I + lx2 — z2I + Iz2 — y21 + 1)/11
{ (14 + lx2 — z2I + Izi I) + Izi — yi I

(b) Straightforward.
if y2= if y2
Z2 )
Z2


Answered Same DayDec 21, 2021

Answer To: 7. (a) Let X = R2, and for x, y E R1 define d(x,y) by d(x, y) = d( (xi, x2)) (yi, y2)) = 'xi — yi I...

David answered on Dec 21 2021
122 Votes
1. ( )
Sol:
Consider the function ( )
( )

, for x≥1.
Differentiating the function with respect to x, we get,
( )
( ) ( )
( )
Since 0 ( ) .
From the above inference
( ) .
Hence g(x) is a monotonically increasing function for the given conditions and the maximum
value of the function is obtained as x→


( )

( )


Dividing the numerator and denominator by xp.


( )



(


( (

))

( (

))
)

Hence the upper limit of g(x) is 1 i.e. g(x) ≤1.

( )


We know that a,b>0.
 Either of a/b >1 or b/a >1 or a/b=1.
Also, (


)
(


)

( (


)

)

( )

(


) ---------------------------①
If a/b=1
 (


) ( )



(

)

( (


)

)


( )


 ( )
If a/b>1
For x>1, we know that g(x) <1
Hence g(a/b) <1.
 ( ) (from eqn. 1)
Similar is the case when (b/a) >1.
From the above proof, we can say that,
( )
2.
( ) (∑| |
)

( ) ( )
Sol:
A metric space is an ordered pair (M,d) where M is a set and d is a metric on M, i.e., a function
such that for any , the following holds:
 (non-negative),
Since there is a modulus in the summation, the p
th
root of the summation is non – negative.
 iff (identity of indiscernibles),
Since the function is the p
th
root of the sum of non – negative numbers, for the function to be
zero, each individual term should be zero.
 , for all k in the index range.
 ( )
 (symmetry)
Since there is a modulus in the summation, the order of z and w doesn’t matter. Hence
symmetric
 (triangle inequality) .
This doesn’t hold.
Proof by counter example:
Let x=(1,1,0,…,0) , y=(0,1,0,…,0) and z=(0,0,0,…,0)
( )

( ) ( )
( ) ( )

( )
Since p<1
 ( ) ( ) ( )
This is a contradiction.
Hence the given dp doesn’t define a metric space.
3. ...
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