Question #123529
Prove this
Let {xm} be a sequence in Kn, say xm = (x1m,...,xnm). Then
lim m infinity
xm = (x1,..., xn)
with respect to || ||2 if and only if
lim
m infinity
xim = xi
for i = 1,..., n.
1
Expert's answer
2020-06-23T20:18:56-0400

let us start with, limmxm=(x1,x2,x3,.........,xm)\lim_{m \to \infty} {x_m} = (x_1,x_2,x_3, . . . . . . . . . , x_m) = {x}


for ϵ>0,\epsilon > 0, n0Nn_0 \isin \N such that

xnx<ϵ|| {x_n} - {x} || < \epsilon for \forall nn0n \geq n_0


so nn0n \geq n_0 ​ , we can say that


(x1m,x2m,x3m,.........,xnm)(x1,x2,x3,......,xn)<ϵ||(x_{1m},x_{2m}, x_{3m}, . . . . . . . . . ,x_{nm}) - ( x_1,x_2,x_3,......,x_n )|| < \epsilon

(x1mx1),(x2mx2),.....(xnmxn)<ϵ|| (x_{1m}-x_1),(x_{2m}-x_2),.....(x_{nm}-x_n)|| < \epsilon

(x1mx1)2+(x2mx2)2+.....(xnmxn)2<ϵ\sqrt { (x_{1m}-x_1)^2+(x_{2m}-x_2)^2+.....(x_{nm}-x_n)^2} < \epsilon

(ximxi)2<ϵ\sqrt{{(x_{im} - x_i)}^2} < \epsilon

(ximxi)<ϵ||{(x_{im} - x_i)} ||< \epsilon

limxxim=xi\lim_{x \to \infty} {x_{im}} = x_i


Conversely, assume that limxxim=xi\lim_{x \to \infty} {x_{im}} = x_i

since, for ϵ>0,\epsilon > 0, \exists n0iNn_{0i} \isin \N such that

ximxi<ϵ|| x_{im} - x_i|| < \epsilon for every value of n>n0in > n_{0i} (1)

for each i=0,1,2,3,4,........,ni=0,1,2,3,4,. .......,n

Consider the maximum of (x01,x02,........,x0n)(x_{01},x_{02},........,x_{0n})

then (1) will hold true simultaneously for each i.

ximxi<ϵ|| x_{im} - x_i || < \epsilon \forall nn0imaxn\geq n_{0i}|_{max}

(ximxi)2<ϵ2(x_{im}-x_i)^2 < \epsilon^2 for each i, nn0imaxn\geq n_{0i}|_{max}

(x1mx1)2+(x2mx2)2+......+(xnmxn)2<nϵ2(x_{1m}-x_1)^2+(x_{2m}-x_2)^2+......+(x_{nm}-x_n)^2 < n\epsilon^2

xmx<nϵ|| x_m - x || < \sqrt{n}\epsilon for each values of i, nn0imaxn\geq n_{0i}|_{max}


limmxm=\lim_{m \to \infty} {x_m} = {x}

Hence proved


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