Solution - MID SEM 1 Question 1
(i) To show:∀ >0,∃δ >0s.t.|x−y|< δ gives|φ(x)−φ(y)|<
Given: |φ(x)−φ(y)| ≤(x−y)2 (1)
Chooseδ =√ .
Using (1),|φ(x)−φ(y)| ≤(x−y)2 < δ2 = Hence proved.
(ii)|φ(x)−φ(y)| ≤(x−y)2
⇐⇒ −(x−y)2 ≤φ(x)−φ(y)≤(x−y)2 dividing both sides by(x−y)
−(x−y)≤ φ(x)−φ(y)x−y ≤(x−y) on taking limits
limx→y[−(x−y)]≤limx→y φ(x)−φ(y)
x−y ≤limx→y(x−y) Now,limx→y[(x−y)] = 0
Therefore by squeeze theorem φ0(x) = 0
(iii) Choose anyx < y. By Intermediate Value Theorem, φ(x)−φ(y)x−y =φ0(z)for somez ∈(x, y) Sinceφ0(z) = 0∀z, we haveφ(x) = φ(y).
Question 2
Since the objective function is increasing in both x1 and x2 in the neighbourhood of 0 (check partials of the objective function with respect tox1 andx2), the maximum can not be obtained at the boundary points (that is eitherx1 = 0or x2 = 0). If it exists, maximum will be an interior solution.
π(p, w1, w2) = maxx1,x2>0[p(√
x1+√
x2)−(w1x1+w2x2)].
FOC: ∂x∂π
1 = 2√px
1 −w1
∂π
∂x2 = 2√px
2 −w2
1
Critical Point: (x∗1, x∗2) = ((2wp
1)2,(2wp
2)2) SOC: ∂∂x2π2
1 =−p4(x1)32.
∂2π
∂x22 =−p4(x2)32
∂2π
∂x2∂x1 = 0
H(x1, x2) =
−p4(x1)−32 0 0 −p4(x2)−32
EvaluatingH(x1, x2)at the critical point:
H(x∗1, x∗2) =
−p4(2wp1)3 0 0 −p4(2wp2)3
H(x1, x2)is a negative definite (Show either by the signs of principle leading minors orzTHz <0 for z 6= 0). Hence, it is a strict local maximum. This is also Global Maximum as the objective function decreases inx1andx2(check partials) asx1, x2 → ∞
Question 3
(i) Counterexample -A= (0,1)andα = 1. Here,a < αforalla∈Abut SupA= 1 =α.
(ii) Let > 0be arbitrary. Then since{xn} → x, there existN ∈ N such that for alln ≥ N, we have|xn−x|< . Then ifn≥N, the reverse triangle inequality shows:
||xn| − |x|| ≤ |xn−x|<
We have shown that for all > 0, there existsN ∈ N such that ifn = N, then||xn| − |x|| < . Therefore{|xn|} → |x|.
Question 4
(i) There are many example which will work. For instance,In = (0,n1). Note thatI1 ⊃I2 ⊃I3 ⊃
· · · ⊃In. Also,∩∞n=1In =φ. The intersection is an empty set as required.
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(ii)f :< → <
f(x1, x2) =
log(x) ifx >0 0 ifx≤0
This function is NOT one-one. f(−1) = f(−2) = 0and −1 6= −2. Also, this function is onto, take any point in the Range, we can always find its pre-image in the domain. Lety =log(x), then 10y =xand we can find always findxfor anyyin<.
Question 5
(i)∀ >0, ∃x∈B(p)s.t.x∈L. (Sincepis a limit point of L)
Sincex∈L,xis a limit point of E (Definition ofL)
Take0 =−d(x, p)
⇒B0(x)⊂B(p) (1)
∃y∈B0(x)s.t. y∈E. (Sincexis a limit point ofE)
⇒y∈B(p)s.t. y∈E (from (1))
Hence proved.
(ii) We first prove thatE∪Lis a closed set.E∪Lis a closed set iff it contains all its limit points.
Letsbe a limit point ofE∪L⇒Eithersis a limit point ofE or a limit point ofL. Ifsis a limit point ofLthen by (1),sis also a limit point ofE. Thereforesmust be a limit point ofE.
Hences ∈Lands ∈E ∪L.
Now we show thatE∪Lthe smallest closed set containingE. Take anyL0 ⊂L. L0does not have all limit points ofE. HenceE∪L0 is not a closed set.
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