IGNOU MMT-004 Solved Assignment 2024 for M.Sc. MACS

IGNOU MMT-004 Solved Assignment 2024 | M.Sc. MACS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU MMT-004 Assignment Question Paper 2024

mmt-003-solved-assignment-2024-qp-2afdf4f2-c576-4553-8e87-8dfd8225b98e

mmt-003-solved-assignment-2024-qp-2afdf4f2-c576-4553-8e87-8dfd8225b98e

MMT-003 Solved Assignment 2024 QP
  1. a) Let X = C [ 0 , 1 ] X = C [ 0 , 1 ] X=C[0,1]\mathrm{X}=\mathrm{C}[0,1]X=C[0,1]. Define d : X × X R d : X × X R d:X xx X rarrRd: X \times X \rightarrow \mathbf{R}d:X×XR by d ( f , g ) = 0 1 | f ( t ) g ( t ) | d t , f , g X d ( f , g ) = 0 1 | f ( t ) g ( t ) | d t , f , g X d(f,g)=int_(0)^(1)|f(t)-g(t)|dt,f,ginX\mathrm{d}(\mathrm{f}, \mathrm{g})=\int_0^1|\mathrm{f}(\mathrm{t})-\mathrm{g}(\mathrm{t})| \mathrm{dt}, \mathrm{f}, \mathrm{g} \in \mathrm{X}d(f,g)=01|f(t)g(t)|dt,f,gX where the integral is the Riemann integral. Show that d d ddd is a metric on X X XXX. Find d ( f , g ) d ( f , g ) d(f,g)d(f, g)d(f,g) where f ( x ) = 4 x f ( x ) = 4 x f(x)=4xf(x)=4 xf(x)=4x and g ( x ) = x 3 , x [ 0 , 1 ] g ( x ) = x 3 , x [ 0 , 1 ] g(x)=x^(3),x in[0,1]g(x)=x^3, x \in[0,1]g(x)=x3,x[0,1].
b) Let ( X , d ) X , d ) X,d)X, d)X,d) be a metric space and a X a X a in Xa \in XaX be a fixed point of X X XXX. Show that the function f a : X R f a : X R f_(a):X rarrRf_a: X \rightarrow \mathbf{R}fa:XR given by f a ( x ) = d ( x , a ) f a ( x ) = d ( x , a ) f_(a)(x)=d(x,a)\mathrm{f}_{\mathrm{a}}(\mathrm{x})=\mathrm{d}(\mathrm{x}, \mathrm{a})fa(x)=d(x,a) is continuous. Is it uniformly continuous? Justify you answer.
2. a) Let A A AAA and B B BBB be any two subsets of a metric space (X, d), then show that
i) int A = { E A = { E A=uu{E\mathrm{A}=\cup\{\mathrm{E}A={E : is open and E A } E A } EsubeA}\mathrm{E} \subseteq \mathrm{A}\}EA}
ii) int ( A B ) = int A int B int ( A B ) = int A int B int(AnnB)=int Ann int B\operatorname{int}(\mathrm{A} \cap \mathrm{B})=\operatorname{int} \mathrm{A} \cap \operatorname{int} \mathrm{B}int(AB)=intAintB
iii) int ( A B ) int ( A B ) quad int(A uu B)supe\quad \operatorname{int}(A \cup B) \supseteqint(AB) int A A A nnA \capA int B B BBB
iv) A B ¯ A ¯ B ¯ A B ¯ A ¯ B ¯ bar(AnnB)sube bar(A)nn bar(B)\overline{\mathrm{A} \cap \mathrm{B}} \subseteq \overline{\mathrm{A}} \cap \overline{\mathrm{B}}AB¯A¯B¯.
b) Find the interior, boundary and closure of the following sets A A A\mathbf{A}A in R R R\mathbf{R}R with the usual metric and discrete metric.
i) A = Q A = Q A=QA=\mathbf{Q}A=Q, the set of rationals in R R R\mathbf{R}R
ii) A = ] 1 , 2 ] ] 2 , 4 [ A = ] 1 , 2 ] ] 2 , 4 [ A=]1,2]uu]2,4[\mathrm{A}=] 1,2] \cup] 2,4[A=]1,2]]2,4[
3. a) Let ( X , d 1 ) X , d 1 (X,d_(1))\left(X, d_1\right)(X,d1) and ( Y , d 2 ) Y , d 2 (Y,d_(2))\left(Y, d_2\right)(Y,d2) be metric spaces. Show that f : X Y f : X Y f:X rarr Yf: X \rightarrow Yf:XY is continuous if and only if f ( A ¯ ) f ( A ) ¯ f ( A ¯ ) f ( A ) ¯ f( bar(A))sube bar(f(A))f(\bar{A}) \subseteq \overline{f(A)}f(A¯)f(A)¯ where A A AAA is any subset of X X XXX
b) Let ( X 1 , d 1 ) X 1 , d 1 (X_(1),d_(1))\left(\mathrm{X}_1, \mathrm{~d}_1\right)(X1, d1) and ( X 2 , d 2 ) X 2 , d 2 (X_(2),d_(2))\left(\mathrm{X}_2, \mathrm{~d}_2\right)(X2, d2) be two discrete metric spaces. Verify that the product metric on X 1 × X 2 X 1 × X 2 X_(1)xxX_(2)\mathrm{X}_1 \times \mathrm{X}_2X1×X2 is discrete.
c) Show that an infinite discrete metric space X X X\mathrm{X}X is bounded but not totally bounded.
4. a) Find the first derivative f ( a ) f ( a ) f^(‘)(a)\mathrm{f}^{\prime}(\mathbf{a})f(a) of the function f f f\mathrm{f}f defined by f : R 3 R 2 f : R 3 R 2 f:R^(3)rarrR^(2)f: \mathbf{R}^3 \rightarrow \mathbf{R}^2f:R3R2 given by f ( x , y , z ) = ( x y z , x + y + z 2 ) f ( x , y , z ) = x y z , x + y + z 2 f(x,y,z)=(xyz,x+y+z^(2))f(x, y, z)=\left(x y z, x+y+z^2\right)f(x,y,z)=(xyz,x+y+z2) where a = ( 1 . 1 , 2 ) a = ( 1 . 1 , 2 ) a=(1.-1,2)\mathbf{a}=(1 .-1,2)a=(1.1,2).
b) Let E E E\mathrm{E}E be an open subset of R n R n R^(n)\mathbf{R}^nRn and f : E R m f : E R m f:E rarrR^(m)f: E \rightarrow \mathbf{R}^mf:ERm be a function such that each of its components function f i f i f_(i)f_ifi are differentiable, then show that f f fff is differentiable. Is the converse of this result true? Justify your answer.
c) Near what points may the surface z 2 + x z + y = 0 z 2 + x z + y = 0 z^(2)+xz+y=0z^2+x z+y=0z2+xz+y=0 be represented uniquely as a graph of a differentiable function z = k ( x , y ) z = k ( x , y ) z=k(x,y)\mathrm{z}=\mathrm{k}(\mathrm{x}, \mathrm{y})z=k(x,y) ? Locate such a point.
5. a) Use the method of Lagrange’s multiplier method to find the shortest possible distance from the ellipse x 2 + 2 y 2 = 2 x 2 + 2 y 2 = 2 x^(2)+2y^(2)=2x^2+2 y^2=2x2+2y2=2 to the line x + y = 2 x + y = 2 x+y=2x+y=2x+y=2.
b) Find the directional derivative of the function f : R 4 R 3 f : R 4 R 3 f:R^(4)rarrR^(3)f: \mathbf{R}^4 \rightarrow \mathbf{R}^3f:R4R3 defined by
f ( x , y , z , w ) = ( x 2 y , x y z , x 2 + y 2 + z 2 ) f ( x , y , z , w ) = x 2 y , x y z , x 2 + y 2 + z 2 f(x,y,z,w)=(x^(2)y,xyz,x^(2)+y^(2)+z^(2))f(x, y, z, w)=\left(x^2 y, x y z, x^2+y^2+z^2\right)f(x,y,z,w)=(x2y,xyz,x2+y2+z2)
at a = ( 1 , 2 , 1 , 2 ) a = ( 1 , 2 , 1 , 2 ) a=(1,2,-1,-2)\mathrm{a}=(1,2,-1,-2)a=(1,2,1,2) in the direction v = ( 0 , 1 , 2 , 2 ) v = ( 0 , 1 , 2 , 2 ) v=(0,1,2,-2)\mathrm{v}=(0,1,2,-2)v=(0,1,2,2).
6. a) Let A be a compact non-empty subset of a metric space (X, d) and let F be a closed subset of X X XXX such that A F = ϕ A F = ϕ A nn F=phiA \cap F=\phiAF=ϕ, then show that d ( A , F ) > 0 d ( A , F ) > 0 d(A,F) > 0d(A, F)>0d(A,F)>0 where d ( A , F ) = inf { d ( a , b ) : a A , b F } d ( A , F ) = inf { d ( a , b ) : a A , b F } d(A,F)=i n f{d(a,b):a in A,b in F}d(A, F)=\inf \{d(a, b): a \in A, b \in F\}d(A,F)=inf{d(a,b):aA,bF}.
b) Give an example of the following with justification
i) A vector-valued function f : R 3 R 3 f : R 3 R 3 f:R^(3)rarrR^(3)f: \mathbf{R}^3 \rightarrow \mathbf{R}^3f:R3R3 which is not differentiable at ( 0 , 0 , 0 ) ( 0 , 0 , 0 ) (0,0,0)(0,0,0)(0,0,0).
ii) A function which is Legesgue measurable on R R R\mathbf{R}R.
c) Show that the components of a metric space is either identical or pairwise disjoint.
7. a) Let Q Q Q\mathbf{Q}Q be the set of rationals with the metric defined on Q Q Q\mathbf{Q}Q by d : Q × Q R d : Q × Q R d:QxxQrarrRd: \mathbf{Q} \times \mathbf{Q} \rightarrow \mathbf{R}d:Q×QR, defined by d ( x , y ) = | x y | , x , y R d ( x , y ) = | x y | , x , y R d(x,y)=|x-y|,AA x,y inRd(x, y)=|x-y|, \forall x, y \in \mathbf{R}d(x,y)=|xy|,x,yR.
Show that { ( 1 + 1 n ) n } 1 + 1 n n {(1+(1)/(n))^(n)}\left\{\left(1+\frac{1}{\mathrm{n}}\right)^{\mathrm{n}}\right\}{(1+1n)n} is Cauchy sequence in Q Q Q\mathbf{Q}Q, but does not converge in Q Q Q\mathbf{Q}Q and { 1 3 n } 1 3 n {(1)/(3^(n))}\left\{\frac{1}{3^n}\right\}{13n} is a Cauchy sequence Q Q Q\mathbf{Q}Q which converges in Q Q Q\mathbf{Q}Q to the limit 0 .
b) Which of the following sets are totally bounded? Give reasons for your answer. Are they compact?
i) 2 N 2 N quad2N\quad 2 \mathbf{N}2N in ( N , d ) ( N , d ) (N,d)(\mathbf{N}, d)(N,d) where d d ddd is the discrete metric.
ii) [ 0 , 2 ] [ 5 , 10 ] [ 0 , 2 ] [ 5 , 10 ] quad[0,2]uu[5,10]\quad[0,2] \cup[5,10][0,2][5,10] in ( R , d ) ( R , d ) (R,d)(\mathbf{R}, d)(R,d) where d d ddd is the Euclidean metric.
c) Which of the following sets are connected sets in R 2 R 2 R^(2)\mathbf{R}^2R2 with the metric given against it? Justify your answer.
i) A = { ( x , y ) : 0 x 1 , 0 y 2 } A = { ( x , y ) : 0 x 1 , 0 y 2 } quadA={(x,y):0 <= x <= 1,0 <= y <= 2}\quad \mathrm{A}=\{(\mathrm{x}, \mathrm{y}): 0 \leq \mathrm{x} \leq 1,0 \leq \mathrm{y} \leq 2\}A={(x,y):0x1,0y2} under the standard metric.
ii) A = { ( x , y ) : x 2 + y 2 = 1 } A = ( x , y ) : x 2 + y 2 = 1 A={(x,y):x^(2)+y^(2)=1}\mathrm{A}=\left\{(\mathrm{x}, \mathrm{y}): \mathrm{x}^2+\mathrm{y}^2=1\right\}A={(x,y):x2+y2=1} under the discrete metric.
8. a) Consider Z Z Z\mathbf{Z}Z and let F 1 F 1 F_(1)\mathcal{F}_1F1 denote the class of subsets of Z Z Z\mathbf{Z}Z, given by F 1 = { A Z F 1 = { A Z F_(1)={AsubZ\mathcal{F}_1=\{\mathrm{A} \subset \mathbf{Z}F1={AZ : either A A A\mathrm{A}A is finite or A c A c A^(c)\mathrm{A}^{\mathrm{c}}Ac is finite }. Check whether F 1 F 1 F_(1)\mathcal{F}_1F1 is a σ σ sigma\sigmaσ algebra or not.
b) Let A be any set in R R R\mathbf{R}R, show that m ( A ) = m ( A + x ) m ( A ) = m ( A + x ) m^(**)(A)=m^(**)(A+x)m^*(A)=m^*(A+x)m(A)=m(A+x) where m m m^(**)m^*m denotes the outer measure.
c) Find the measure of the following sets.
i) E = n = 1 ( a 1 n , b ) E = n = 1 a 1 n , b quad E=nnn_(n=1)^(oo)(a-(1)/(n),b)\quad E=\bigcap_{n=1}^{\infty}\left(a-\frac{1}{n}, b\right)E=n=1(a1n,b)
ii) E = Q { 1 , 2 , 3 , 4 } E = Q { 1 , 2 , 3 , 4 } E=Quu{1,2,3,4}\mathrm{E}=\mathbf{Q} \cup\{1,2,3,4\}E=Q{1,2,3,4}
iii) E = ] 5 , 7 [ [ 7 , 7.5 ] E = ] 5 , 7 [ [ 7 , 7.5 ] E=]5,7[uu[7,7.5]\mathrm{E}=] 5,7[\cup[7,7.5]E=]5,7[[7,7.5].
9. a) Show that if f f fff is measurable, then the function f a ( x ) f a ( x ) f^(a)(x)f^a(x)fa(x) given by
f a ( x ) = { a if f ( x ) > a f ( x ) if f ( x ) a f a ( x ) = a if f ( x ) > a f ( x ) if f ( x ) a f^(a)(x)={[a,” if “f(x) > a],[f(x),” if “f(x) <= a]:}f^a(x)=\left\{\begin{array}{cc} a & \text { if } f(x)>a \\ f(x) & \text { if } f(x) \leq a \end{array}\right.fa(x)={a if f(x)>af(x) if f(x)a
is also measurable.
b) Verify Bounded Convergence Theorem for the sequence of functions { f n } f n {f_(n)}\left\{f_n\right\}{fn} where
f n ( x ) = 1 ( 1 + x / n ) n , 0 x 1 , n N f n ( x ) = 1 ( 1 + x / n ) n , 0 x 1 , n N f_(n)(x)=(1)/((1+x//n)^(n)),0 <= x <= 1,ninN\mathrm{f}_{\mathrm{n}}(\mathrm{x})=\frac{1}{(1+\mathrm{x} / \mathrm{n})^{\mathrm{n}}}, 0 \leq \mathrm{x} \leq 1, \mathrm{n} \in \mathbf{N}fn(x)=1(1+x/n)n,0x1,nN
c) Find the fourier series of the function f f fff defined by
f ( x ) = { x 2 , π < x 0 x 2 , 0 < x < π f ( x ) = x 2 , π < x 0 x 2 , 0 < x < π f(x)={[-x^(2)”,”-pi < x <= 0],[x^(2)”,”0 < x < pi]:}f(x)=\left\{\begin{array}{c} -x^2,-\pi<x \leq 0 \\ x^2, 0<x<\pi \end{array}\right.f(x)={x2,π<x0x2,0<x<π
  1. State whether the following statements are True or False. Justify your answers.
    a) The sequence { ( 1 n , 1 n ) : n N } 1 n , 1 n : n N {((1)/(n),(1)/(n)):ninN}\left\{\left(\frac{1}{\mathrm{n}}, \frac{1}{\mathrm{n}}\right): \mathrm{n} \in \mathbf{N}\right\}{(1n,1n):nN} is convergent in R 2 R 2 R^(2)\mathbf{R}^2R2 under the discrete metric on R 2 R 2 R^(2)\mathbf{R}^2R2.
    b) A subset in a metric space is compact if it is closed.
    c) Continuous image of a path connected space is path connected.
    d) The second derivative of a linear map from R n R n R^(n)\mathbf{R}^nRn to R m R m R^(m)\mathbf{R}^mRm never vanishes.
    e) If A f d m = A g d m A f d m = A g d m int_(A)fdm=int_(A)gdm\int_{\mathrm{A}} \mathrm{fdm}=\int_{\mathrm{A}} \mathrm{gdm}Afdm=Agdm for all A M A M Ain M\mathrm{A} \in \boldsymbol{M}AM, then f = g f = g f=g\mathrm{f}=\mathrm{g}f=g.
\(cos\:2\theta =1-2\:sin^2\theta \)

MMT-004 Sample Solution 2024

mmt-004-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

mmt-004-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

MMT-004 Solved Assignment 2024 SS
  1. a) Let X = C [ 0 , 1 ] X = C [ 0 , 1 ] X=C[0,1]\mathrm{X}=\mathrm{C}[0,1]X=C[0,1]. Define d : X × X R d : X × X R d:X xx X rarrRd: X \times X \rightarrow \mathbf{R}d:X×XR by d ( f , g ) = 0 1 | f ( t ) g ( t ) | d t , f , g X d ( f , g ) = 0 1 | f ( t ) g ( t ) | d t , f , g X d(f,g)=int_(0)^(1)|f(t)-g(t)|dt,f,ginX\mathrm{d}(\mathrm{f}, \mathrm{g})=\int_0^1|\mathrm{f}(\mathrm{t})-\mathrm{g}(\mathrm{t})| \mathrm{dt}, \mathrm{f}, \mathrm{g} \in \mathrm{X}d(f,g)=01|f(t)g(t)|dt,f,gX where the integral is the Riemann integral. Show that d d ddd is a metric on X X XXX. Find d ( f , g ) d ( f , g ) d(f,g)d(f, g)d(f,g) where f ( x ) = 4 x f ( x ) = 4 x f(x)=4xf(x)=4 xf(x)=4x and g ( x ) = x 3 , x [ 0 , 1 ] g ( x ) = x 3 , x [ 0 , 1 ] g(x)=x^(3),x in[0,1]g(x)=x^3, x \in[0,1]g(x)=x3,x[0,1].
Answer:
To show that d d ddd is a metric on X X XXX, we need to verify the following properties for all f , g , h X f , g , h X f,g,h in Xf, g, h \in Xf,g,hX:
  1. Non-negativity: d ( f , g ) 0 d ( f , g ) 0 d(f,g) >= 0d(f, g) \geq 0d(f,g)0
  2. Identity of indiscernibles: d ( f , g ) = 0 d ( f , g ) = 0 d(f,g)=0d(f, g) = 0d(f,g)=0 if and only if f = g f = g f=gf = gf=g
  3. Symmetry: d ( f , g ) = d ( g , f ) d ( f , g ) = d ( g , f ) d(f,g)=d(g,f)d(f, g) = d(g, f)d(f,g)=d(g,f)
  4. Triangle inequality: d ( f , h ) d ( f , g ) + d ( g , h ) d ( f , h ) d ( f , g ) + d ( g , h ) d(f,h) <= d(f,g)+d(g,h)d(f, h) \leq d(f, g) + d(g, h)d(f,h)d(f,g)+d(g,h)
Let’s verify each property:
  1. Non-negativity:
    For any f , g X f , g X f,g in Xf, g \in Xf,gX, the absolute value function | | | | |*||\cdot||| ensures that | f ( t ) g ( t ) | 0 | f ( t ) g ( t ) | 0 |f(t)-g(t)| >= 0|f(t) – g(t)| \geq 0|f(t)g(t)|0 for all t [ 0 , 1 ] t [ 0 , 1 ] t in[0,1]t \in [0, 1]t[0,1]. Therefore, the integral 0 1 | f ( t ) g ( t ) | d t 0 1 | f ( t ) g ( t ) | d t int_(0)^(1)|f(t)-g(t)|dt\int_0^1 |f(t) – g(t)| \, dt01|f(t)g(t)|dt is also non-negative. Hence, d ( f , g ) 0 d ( f , g ) 0 d(f,g) >= 0d(f, g) \geq 0d(f,g)0.
  2. Identity of indiscernibles:
    If f = g f = g f=gf = gf=g, then f ( t ) g ( t ) = 0 f ( t ) g ( t ) = 0 f(t)-g(t)=0f(t) – g(t) = 0f(t)g(t)=0 for all t [ 0 , 1 ] t [ 0 , 1 ] t in[0,1]t \in [0, 1]t[0,1], so d ( f , g ) = 0 1 | f ( t ) g ( t ) | d t = 0 1 0 d t = 0 d ( f , g ) = 0 1 | f ( t ) g ( t ) | d t = 0 1 0 d t = 0 d(f,g)=int_(0)^(1)|f(t)-g(t)|dt=int_(0)^(1)0dt=0d(f, g) = \int_0^1 |f(t) – g(t)| \, dt = \int_0^1 0 \, dt = 0d(f,g)=01|f(t)g(t)|dt=010dt=0.
    Conversely, if d ( f , g ) = 0 d ( f , g ) = 0 d(f,g)=0d(f, g) = 0d(f,g)=0, then 0 1 | f ( t ) g ( t ) | d t = 0 0 1 | f ( t ) g ( t ) | d t = 0 int_(0)^(1)|f(t)-g(t)|dt=0\int_0^1 |f(t) – g(t)| \, dt = 001|f(t)g(t)|dt=0. Since the integrand is non-negative, it must be zero almost everywhere, implying that f ( t ) = g ( t ) f ( t ) = g ( t ) f(t)=g(t)f(t) = g(t)f(t)=g(t) for almost all t [ 0 , 1 ] t [ 0 , 1 ] t in[0,1]t \in [0, 1]t[0,1]. Since f f fff and g g ggg are continuous, they must be equal everywhere on [ 0 , 1 ] [ 0 , 1 ] [0,1][0, 1][0,1], so f = g f = g f=gf = gf=g.
  3. Symmetry:
    By the properties of the absolute value function, | f ( t ) g ( t ) | = | g ( t ) f ( t ) | | f ( t ) g ( t ) | = | g ( t ) f ( t ) | |f(t)-g(t)|=|g(t)-f(t)||f(t) – g(t)| = |g(t) – f(t)||f(t)g(t)|=|g(t)f(t)| for all t [ 0 , 1 ] t [ 0 , 1 ] t in[0,1]t \in [0, 1]t[0,1]. Therefore, d ( f , g ) = 0 1 | f ( t ) g ( t ) | d t = 0 1 | g ( t ) f ( t ) | d t = d ( g , f ) d ( f , g ) = 0 1 | f ( t ) g ( t ) | d t = 0 1 | g ( t ) f ( t ) | d t = d ( g , f ) d(f,g)=int_(0)^(1)|f(t)-g(t)|dt=int_(0)^(1)|g(t)-f(t)|dt=d(g,f)d(f, g) = \int_0^1 |f(t) – g(t)| \, dt = \int_0^1 |g(t) – f(t)| \, dt = d(g, f)d(f,g)=01|f(t)g(t)|dt=01|g(t)f(t)|dt=d(g,f).
  4. Triangle inequality:
    For any f , g , h X f , g , h X f,g,h in Xf, g, h \in Xf,g,hX and for all t [ 0 , 1 ] t [ 0 , 1 ] t in[0,1]t \in [0, 1]t[0,1], we have | f ( t ) h ( t ) | = | f ( t ) g ( t ) + g ( t ) h ( t ) | | f ( t ) g ( t ) | + | g ( t ) h ( t ) | | f ( t ) h ( t ) | = | f ( t ) g ( t ) + g ( t ) h ( t ) | | f ( t ) g ( t ) | + | g ( t ) h ( t ) | |f(t)-h(t)|=|f(t)-g(t)+g(t)-h(t)| <= |f(t)-g(t)|+|g(t)-h(t)||f(t) – h(t)| = |f(t) – g(t) + g(t) – h(t)| \leq |f(t) – g(t)| + |g(t) – h(t)||f(t)h(t)|=|f(t)g(t)+g(t)h(t)||f(t)g(t)|+|g(t)h(t)| by the triangle inequality for real numbers. Integrating both sides over [ 0 , 1 ] [ 0 , 1 ] [0,1][0, 1][0,1] gives:
    0 1 | f ( t ) h ( t ) | d t 0 1 ( | f ( t ) g ( t ) | + | g ( t ) h ( t ) | ) d t = 0 1 | f ( t ) g ( t ) | d t + 0 1 | g ( t ) h ( t ) | d t = d ( f , g ) + d ( g , h ) 0 1 | f ( t ) h ( t ) | d t 0 1 ( | f ( t ) g ( t ) | + | g ( t ) h ( t ) | ) d t = 0 1 | f ( t ) g ( t ) | d t + 0 1 | g ( t ) h ( t ) | d t = d ( f , g ) + d ( g , h ) int_(0)^(1)|f(t)-h(t)|dt <= int_(0)^(1)(|f(t)-g(t)|+|g(t)-h(t)|)dt=int_(0)^(1)|f(t)-g(t)|dt+int_(0)^(1)|g(t)-h(t)|dt=d(f,g)+d(g,h)\int_0^1 |f(t) – h(t)| \, dt \leq \int_0^1 (|f(t) – g(t)| + |g(t) – h(t)|) \, dt = \int_0^1 |f(t) – g(t)| \, dt + \int_0^1 |g(t) – h(t)| \, dt = d(f, g) + d(g, h)01|f(t)h(t)|dt01(|f(t)g(t)|+|g(t)h(t)|)dt=01|f(t)g(t)|dt+01|g(t)h(t)|dt=d(f,g)+d(g,h)
    Thus, d ( f , h ) d ( f , g ) + d ( g , h ) d ( f , h ) d ( f , g ) + d ( g , h ) d(f,h) <= d(f,g)+d(g,h)d(f, h) \leq d(f, g) + d(g, h)d(f,h)d(f,g)+d(g,h).
Since all four properties are satisfied, d d ddd is a metric on X X XXX.
Next, let’s find d ( f , g ) d ( f , g ) d(f,g)d(f, g)d(f,g) where f ( x ) = 4 x f ( x ) = 4 x f(x)=4xf(x) = 4xf(x)=4x and g ( x ) = x 3 g ( x ) = x 3 g(x)=x^(3)g(x) = x^3g(x)=x3 for x [ 0 , 1 ] x [ 0 , 1 ] x in[0,1]x \in [0, 1]x[0,1]:
d ( f , g ) = 0 1 | 4 x x 3 | d x d ( f , g ) = 0 1 | 4 x x 3 | d x d(f,g)=int_(0)^(1)|4x-x^(3)|dxd(f, g) = \int_0^1 |4x – x^3| \, dxd(f,g)=01|4xx3|dx
To evaluate this integral, we can split it into regions where the integrand is positive or negative:
  1. For 0 x 2 0 x 2 0 <= x <= 20 \leq x \leq 20x2, 4 x x 3 4 x x 3 4x >= x^(3)4x \geq x^34xx3, so | 4 x x 3 | = 4 x x 3 | 4 x x 3 | = 4 x x 3 |4x-x^(3)|=4x-x^(3)|4x – x^3| = 4x – x^3|4xx3|=4xx3.
  2. For 2 < x 1 2 < x 1 2 < x <= 12 < x \leq 12<x1, 4 x < x 3 4 x < x 3 4x < x^(3)4x < x^34x<x3, so | 4 x x 3 | = x 3 4 x | 4 x x 3 | = x 3 4 x |4x-x^(3)|=x^(3)-4x|4x – x^3| = x^3 – 4x|4xx3|=x34x.
Thus, we can rewrite the integral as:
d ( f , g ) = 0 2 ( 4 x x 3 ) d x + 2 1 ( x 3 4 x ) d x d ( f , g ) = 0 2 ( 4 x x 3 ) d x + 2 1 ( x 3 4 x ) d x d(f,g)=int_(0)^(2)(4x-x^(3))dx+int_(2)^(1)(x^(3)-4x)dxd(f, g) = \int_0^{2} (4x – x^3) \, dx + \int_{2}^1 (x^3 – 4x)\, dxd(f,g)=02(4xx3)dx+21(x34x)dx
Now, we can calculate these integrals:
d ( f , g ) = [ 2 x 2 1 4 x 4 ] 0 2 + [ 1 4 x 4 2 x 2 ] 2 1 = ( 2 ( 2 ) 2 1 4 ( 2 ) 4 ) ( 0 0 ) + ( 1 4 ( 1 ) 4 2 ( 1 ) 2 ) ( 1 4 ( 2 ) 4 2 ( 2 ) 2 ) = ( 8 4 ) + ( 1 4 2 ) ( 4 8 ) = 4 7 4 ( 4 ) = 16 4 7 4 + 16 4 = 25 4 = 6.25 d ( f , g ) = 2 x 2 1 4 x 4 0 2 + 1 4 x 4 2 x 2 2 1 = 2 ( 2 ) 2 1 4 ( 2 ) 4 0 0 + 1 4 ( 1 ) 4 2 ( 1 ) 2 1 4 ( 2 ) 4 2 ( 2 ) 2 = 8 4 + 1 4 2 4 8 = 4 7 4 ( 4 ) = 16 4 7 4 + 16 4 = 25 4 = 6.25 {:[d(f”,”g)=[2x^(2)-(1)/(4)x^(4)]_(0)^(2)+[(1)/(4)x^(4)-2x^(2)]_(2)^(1)],[=(2(2)^(2)-(1)/(4)(2)^(4))-(0-0)+((1)/(4)(1)^(4)-2(1)^(2))-((1)/(4)(2)^(4)-2(2)^(2))],[=(8-4)+((1)/(4)-2)-(4-8)],[=4-(7)/(4)-(-4)],[=(16)/(4)-(7)/(4)+(16)/(4)],[=(25)/(4)],[=6.25]:}\begin{align*} d(f, g) &= \left[2x^2 – \frac{1}{4}x^4\right]_0^{2} + \left[\frac{1}{4}x^4 – 2x^2\right]_{2}^1 \\ &= \left(2(2)^2 – \frac{1}{4}(2)^4\right) – \left(0 – 0\right) + \left(\frac{1}{4}(1)^4 – 2(1)^2\right) – \left(\frac{1}{4}(2)^4 – 2(2)^2\right) \\ &= \left(8 – 4\right) + \left(\frac{1}{4} – 2\right) – \left(4 – 8\right) \\ &= 4 – \frac{7}{4} – (-4) \\ &= \frac{16}{4} – \frac{7}{4} + \frac{16}{4} \\ &= \frac{25}{4} \\ &= 6.25 \end{align*}d(f,g)=[2x214x4]02+[14x42x2]21=(2(2)214(2)4)(00)+(14(1)42(1)2)(14(2)42(2)2)=(84)+(142)(48)=474(4)=16474+164=254=6.25
Therefore, d ( f , g ) = 6.25 d ( f , g ) = 6.25 d(f,g)=6.25d(f, g) = 6.25d(f,g)=6.25.
b) Let ( X , d ) X , d ) X,d)X, d)X,d) be a metric space and a X a X a in Xa \in XaX be a fixed point of X X XXX. Show that the function f a : X R f a : X R f_(a):X rarrRf_a: X \rightarrow \mathbf{R}fa:XR given by f a ( x ) = d ( x , a ) f a ( x ) = d ( x , a ) f_(a)(x)=d(x,a)\mathrm{f}_{\mathrm{a}}(\mathrm{x})=\mathrm{d}(\mathrm{x}, \mathrm{a})fa(x)=d(x,a) is continuous. Is it uniformly continuous? Justify you answer.
Answer:
To show that the function f a : X R f a : X R f_(a):X rarrRf_a: X \rightarrow \mathbf{R}fa:XR given by f a ( x ) = d ( x , a ) f a ( x ) = d ( x , a ) f_(a)(x)=d(x,a)f_a(x) = d(x, a)fa(x)=d(x,a) is continuous, we need to show that for every ϵ > 0 ϵ > 0 epsilon > 0\epsilon > 0ϵ>0, there exists a δ > 0 δ > 0 delta > 0\delta > 0δ>0 such that for all x , y X x , y X x,y in Xx, y \in Xx,yX, if d ( x , y ) < δ d ( x , y ) < δ d(x,y) < deltad(x, y) < \deltad(x,y)<δ, then | f a ( x ) f a ( y ) | < ϵ | f a ( x ) f a ( y ) | < ϵ |f_(a)(x)-f_(a)(y)| < epsilon|f_a(x) – f_a(y)| < \epsilon|fa(x)fa(y)|<ϵ.
Let ϵ > 0 ϵ > 0 epsilon > 0\epsilon > 0ϵ>0 be given. Set δ = ϵ δ = ϵ delta=epsilon\delta = \epsilonδ=ϵ. Then, for any x , y X x , y X x,y in Xx, y \in Xx,yX such that d ( x , y ) < δ d ( x , y ) < δ d(x,y) < deltad(x, y) < \deltad(x,y)<δ, we have:
| f a ( x ) f a ( y ) | = | d ( x , a ) d ( y , a ) | d ( x , y ) (by the triangle inequality) < δ = ϵ | f a ( x ) f a ( y ) | = | d ( x , a ) d ( y , a ) | d ( x , y ) (by the triangle inequality) < δ = ϵ {:[|f_(a)(x)-f_(a)(y)|=|d(x”,”a)-d(y”,”a)|],[ <= d(x”,”y)quad(by the triangle inequality)],[ < delta],[=epsilon]:}\begin{align*} |f_a(x) – f_a(y)| &= |d(x, a) – d(y, a)| \\ &\leq d(x, y) \quad \text{(by the triangle inequality)} \\ &< \delta \\ &= \epsilon \end{align*}|fa(x)fa(y)|=|d(x,a)d(y,a)|d(x,y)(by the triangle inequality)<δ=ϵ
Hence, f a f a f_(a)f_afa is continuous.
Now, let’s consider whether f a f a f_(a)f_afa is uniformly continuous. A function is uniformly continuous if for every ϵ > 0 ϵ > 0 epsilon > 0\epsilon > 0ϵ>0, there exists a δ > 0 δ > 0 delta > 0\delta > 0δ>0 such that for all x , y X x , y X x,y in Xx, y \in Xx,yX, if d ( x , y ) < δ d ( x , y ) < δ d(x,y) < deltad(x, y) < \deltad(x,y)<δ, then | f a ( x ) f a ( y ) | < ϵ | f a ( x ) f a ( y ) | < ϵ |f_(a)(x)-f_(a)(y)| < epsilon|f_a(x) – f_a(y)| < \epsilon|fa(x)fa(y)|<ϵ, where δ δ delta\deltaδ does not depend on the choice of x x xxx or y y yyy.
In metric spaces, continuity and uniform continuity are not always equivalent. However, if the metric space X X XXX is compact, then every continuous function from X X XXX to R R R\mathbf{R}R is uniformly continuous. This is a consequence of the Heine-Cantor theorem.
If the metric space X X XXX is not compact, we cannot guarantee that f a f a f_(a)f_afa is uniformly continuous. For example, consider the metric space ( R , d ) ( R , d ) (R,d)(\mathbf{R}, d)(R,d) where d ( x , y ) = | x y | d ( x , y ) = | x y | d(x,y)=|x-y|d(x, y) = |x – y|d(x,y)=|xy| is the standard metric on R R R\mathbf{R}R. The function f a ( x ) = | x a | f a ( x ) = | x a | f_(a)(x)=|x-a|f_a(x) = |x – a|fa(x)=|xa| is continuous but not uniformly continuous on R R R\mathbf{R}R.
In conclusion, the function f a f a f_(a)f_afa is continuous, but it is not necessarily uniformly continuous unless the metric space X X XXX has additional properties, such as compactness.

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