Sample Solution

mmt-006-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

mmt-006-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

  1. a) Let X = { f C [ 0 , 1 ] : f ( 0 ) = 0 } X = { f C [ 0 , 1 ] : f ( 0 ) = 0 } X={f in C[0,1]:f(0)=0}X=\{f \in C[0,1]: f(0)=0\}X={fC[0,1]:f(0)=0}
Y = { g x : 0 1 g ( t ) d t = 0 } Y = g x : 0 1 g ( t ) d t = 0 Y={g in x:int_(0)^(1)g(t)dt=0}Y=\left\{g \in x: \int_0^1 g(t) d t=0\right\}Y={gx:01g(t)dt=0}
Prove that Y Y Y\mathrm{Y}Y is a proper subspace of X X X\mathrm{X}X. Is Y Y Y\mathrm{Y}Y a closed subspace of X X X\mathrm{X}X ? Justify your answer.
Answer:
To prove that Y Y YYY is a proper subspace of X X XXX and determine whether Y Y YYY is a closed subspace of X X XXX, we need to examine the definitions and properties of these sets within the context of functional analysis.

X X XXX and Y Y YYY Defined

  • X = { f C [ 0 , 1 ] : f ( 0 ) = 0 } X = { f C [ 0 , 1 ] : f ( 0 ) = 0 } X={f in C[0,1]:f(0)=0}X = \{f \in C[0,1]: f(0) = 0\}X={fC[0,1]:f(0)=0} is the set of all continuous functions on the interval [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1] that vanish at 0 0 000.
  • Y = { g X : 0 1 g ( t ) d t = 0 } Y = g X : 0 1 g ( t ) d t = 0 Y={g in X:int_(0)^(1)g(t)dt=0}Y = \left\{g \in X: \int_0^1 g(t) dt = 0\right\}Y={gX:01g(t)dt=0} is the set of all functions in X X XXX whose integral over [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1] is 0 0 000.

Proving Y Y YYY is a Proper Subspace of X X XXX

To show that Y Y YYY is a proper subspace of X X XXX, we must verify that Y Y YYY satisfies the following criteria for being a subspace:
  1. Non-emptiness: Y Y YYY contains the zero function, which is the function g ( t ) = 0 g ( t ) = 0 g(t)=0g(t) = 0g(t)=0 for all t [ 0 , 1 ] t [ 0 , 1 ] t in[0,1]t \in [0,1]t[0,1]. This function clearly belongs to X X XXX and satisfies the integral condition, so Y Y YYY is non-empty.
  2. Closed under addition: If g 1 , g 2 Y g 1 , g 2 Y g_(1),g_(2)in Yg_1, g_2 \in Yg1,g2Y, then g 1 + g 2 Y g 1 + g 2 Y g_(1)+g_(2)in Yg_1 + g_2 \in Yg1+g2Y. This is because the integral of the sum is the sum of the integrals, each of which is 0 0 000, so their sum is also 0 0 000.
  3. Closed under scalar multiplication: If g Y g Y g in Yg \in YgY and α α alpha\alphaα is a scalar, then α g Y α g Y alpha g in Y\alpha g \in YαgY. This follows because 0 1 α g ( t ) d t = α 0 1 g ( t ) d t = α 0 = 0 0 1 α g ( t ) d t = α 0 1 g ( t ) d t = α 0 = 0 int_(0)^(1)alpha g(t)dt=alphaint_(0)^(1)g(t)dt=alpha*0=0\int_0^1 \alpha g(t) dt = \alpha \int_0^1 g(t) dt = \alpha \cdot 0 = 001αg(t)dt=α01g(t)dt=α0=0.
Since Y Y YYY satisfies these criteria, it is a subspace of X X XXX. It is a proper subspace because there exist functions in X X XXX that do not satisfy the integral condition, such as f ( t ) = t f ( t ) = t f(t)=tf(t) = tf(t)=t, which is in X X XXX but not in Y Y YYY since 0 1 t d t = 1 / 2 0 0 1 t d t = 1 / 2 0 int_(0)^(1)tdt=1//2!=0\int_0^1 t dt = 1/2 \neq 001tdt=1/20.

Is Y Y YYY a Closed Subspace of X X XXX?

A subspace Y Y YYY is closed in X X XXX if it contains all its limit points; that is, if a sequence of functions { g n } { g n } {g_(n)}\{g_n\}{gn} in Y Y YYY converges uniformly to a function g g ggg, then g g ggg must also be in Y Y YYY.
To show Y Y YYY is closed, consider a sequence { g n } Y { g n } Y {g_(n)}sub Y\{g_n\} \subset Y{gn}Y that converges uniformly to g X g X g in Xg \in XgX. We need to show that g Y g Y g in Yg \in YgY, meaning 0 1 g ( t ) d t = 0 0 1 g ( t ) d t = 0 int_(0)^(1)g(t)dt=0\int_0^1 g(t) dt = 001g(t)dt=0.
Uniform convergence of { g n } { g n } {g_(n)}\{g_n\}{gn} to g g ggg implies that for every ϵ > 0 ϵ > 0 epsilon > 0\epsilon > 0ϵ>0, there exists an N N NNN such that for all n N n N n >= Nn \geq NnN, we have | g n ( t ) g ( t ) | < ϵ | g n ( t ) g ( t ) | < ϵ |g_(n)(t)-g(t)| < epsilon|g_n(t) – g(t)| < \epsilon|gn(t)g(t)|<ϵ for all t [ 0 , 1 ] t [ 0 , 1 ] t in[0,1]t \in [0,1]t[0,1]. By the properties of integrals and the limit of a sequence of functions,
lim n 0 1 g n ( t ) d t = 0 1 lim n g n ( t ) d t = 0 1 g ( t ) d t . lim n 0 1 g n ( t ) d t = 0 1 lim n g n ( t ) d t = 0 1 g ( t ) d t . lim_(n rarr oo)int_(0)^(1)g_(n)(t)dt=int_(0)^(1)lim_(n rarr oo)g_(n)(t)dt=int_(0)^(1)g(t)dt.\lim_{n \to \infty} \int_0^1 g_n(t) dt = \int_0^1 \lim_{n \to \infty} g_n(t) dt = \int_0^1 g(t) dt.limn01gn(t)dt=01limngn(t)dt=01g(t)dt.
Since each g n Y g n Y g_(n)in Yg_n \in YgnY, we have 0 1 g n ( t ) d t = 0 0 1 g n ( t ) d t = 0 int_(0)^(1)g_(n)(t)dt=0\int_0^1 g_n(t) dt = 001gn(t)dt=0. Thus, the limit of these integrals as n n n rarr oon \to \inftyn is also 0 0 000, which means 0 1 g ( t ) d t = 0 0 1 g ( t ) d t = 0 int_(0)^(1)g(t)dt=0\int_0^1 g(t) dt = 001g(t)dt=0, and hence g Y g Y g in Yg \in YgY.
Therefore, Y Y YYY is a closed subspace of X X XXX because it contains all its limit points under uniform convergence.
b) Let X = L p [ 0 , 1 ] X = L p [ 0 , 1 ] X=L^(p)[0,1]X=L^p[0,1]X=Lp[0,1] and x = x ( t ) = t 2 x = x ( t ) = t 2 x=x(t)=t^(2)x=x(t)=t^2x=x(t)=t2. Find x p x p ||x||_(p)\|x\|_pxp for p = 4 p = 4 p=4p=4p=4 and oo\infty.
Answer:
To find the L p L p L^(p)L^pLp norm of x ( t ) = t 2 x ( t ) = t 2 x(t)=t^(2)x(t) = t^2x(t)=t2 on the interval [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1] for p = 4 p = 4 p=4p=4p=4 and p = p = p=oop=\inftyp=, we’ll use the definitions of the L p L p L^(p)L^pLp norms.

For p = 4 p = 4 p=4p=4p=4

The L p L p L^(p)L^pLp norm for p = 4 p = 4 p=4p=4p=4 is defined as
x 4 = ( 0 1 | t 2 | 4 d t ) 1 / 4 x 4 = 0 1 | t 2 | 4 d t 1 / 4 ||x||_(4)=(int_(0)^(1)|t^(2)|^(4)dt)^(1//4)\|x\|_4 = \left( \int_0^1 |t^2|^4 dt \right)^{1/4}x4=(01|t2|4dt)1/4
This simplifies to
x 4 = ( 0 1 t 8 d t ) 1 / 4 x 4 = 0 1 t 8 d t 1 / 4 ||x||_(4)=(int_(0)^(1)t^(8)dt)^(1//4)\|x\|_4 = \left( \int_0^1 t^8 dt \right)^{1/4}x4=(01t8dt)1/4
To compute this integral, we use the formula for the integral of t n t n t^(n)t^ntn, which is t n + 1 n + 1 t n + 1 n + 1 (t^(n+1))/(n+1)\frac{t^{n+1}}{n+1}tn+1n+1 for n 1 n 1 n!=-1n \neq -1n1, from 0 0 000 to 1 1 111:
0 1 t 8 d t = t 9 9 | 0 1 = 1 9 0 1 t 8 d t = t 9 9 | 0 1 = 1 9 int_(0)^(1)t^(8)dt=(t^(9))/(9)|_(0)^(1)=(1)/(9)\int_0^1 t^8 dt = \frac{t^9}{9} \Big|_0^1 = \frac{1}{9}01t8dt=t99|01=19
Therefore,
x 4 = ( 1 9 ) 1 / 4 x 4 = 1 9 1 / 4 ||x||_(4)=((1)/(9))^(1//4)\|x\|_4 = \left( \frac{1}{9} \right)^{1/4}x4=(19)1/4

For p = p = p=oop=\inftyp=

The L L L^( oo)L^\inftyL norm, or the supremum norm, is defined as the essential supremum of | x ( t ) | | x ( t ) | |x(t)||x(t)||x(t)| over the interval. For x ( t ) = t 2 x ( t ) = t 2 x(t)=t^(2)x(t) = t^2x(t)=t2 on [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1], this is simply the maximum value of t 2 t 2 t^(2)t^2t2 on the interval, which occurs at t = 1 t = 1 t=1t=1t=1. Therefore,
x = max t [ 0 , 1 ] | t 2 | = 1 2 = 1 x = max t [ 0 , 1 ] | t 2 | = 1 2 = 1 ||x||_(oo)=max_(t in[0,1])|t^(2)|=1^(2)=1\|x\|_\infty = \max_{t \in [0,1]} |t^2| = 1^2 = 1x=maxt[0,1]|t2|=12=1

Calculation

Let’s calculate x 4 x 4 ||x||_(4)\|x\|_4x4 explicitly:
x 4 = ( 1 9 ) 1 / 4 x 4 = 1 9 1 / 4 ||x||_(4)=((1)/(9))^(1//4)\|x\|_4 = \left( \frac{1}{9} \right)^{1/4}x4=(19)1/4
This calculation yields:
x 4 = ( 1 9 ) 1 / 4 = 1 3 1 / 2 = 1 3 x 4 = 1 9 1 / 4 = 1 3 1 / 2 = 1 3 ||x||_(4)=((1)/(9))^(1//4)=(1)/(3^(1//2))=(1)/(sqrt3)\|x\|_4 = \left( \frac{1}{9} \right)^{1/4} = \frac{1}{3^{1/2}} = \frac{1}{\sqrt{3}}x4=(19)1/4=131/2=13
So, summarizing:
  • For p = 4 p = 4 p=4p=4p=4, x 4 = 1 3 x 4 = 1 3 ||x||_(4)=(1)/(sqrt3)\|x\|_4 = \frac{1}{\sqrt{3}}x4=13.
  • For p = p = p=oop=\inftyp=, x = 1 x = 1 ||x||_(oo)=1\|x\|_\infty = 1x=1.
c) Let E be a subset of a normed space X , Y = span E X , Y = span E X,Y=span EX, Y=\operatorname{span} EX,Y=spanE and a X a X a in Xa \in XaX. Show that a Y ¯ a Y ¯ a in bar(Y)a \in \bar{Y}aY¯ if and only if f ( a ) = 0 f ( a ) = 0 f(a)=0f(a)=0f(a)=0 whenever f X f X f inX^(‘)f \in X^{\prime}fX and f = 0 f = 0 f=0f=0f=0 everywhere on E E EEE.
Answer:
To show that a Y ¯ a Y ¯ a in bar(Y)a \in \bar{Y}aY¯ if and only if f ( a ) = 0 f ( a ) = 0 f(a)=0f(a) = 0f(a)=0 for every f X f X f inX^(‘)f \in X’fX (the dual space of X X XXX) where f = 0 f = 0 f=0f = 0f=0 on E E EEE, we’ll use some fundamental properties of normed spaces, their duals, and the concept of closure.

Definitions:

  • X X XXX is a normed space, and X X X^(‘)X’X is its dual space, consisting of all continuous linear functionals on X X XXX.
  • E E EEE is a subset of X X XXX, and Y = span E Y = span E Y=span EY = \operatorname{span}EY=spanE is the subspace spanned by E E EEE.
  • Y ¯ Y ¯ bar(Y)\bar{Y}Y¯ denotes the closure of Y Y YYY in X X XXX, which includes all limits of convergent sequences in Y Y YYY.

( =>\Rightarrow) If a Y ¯ a Y ¯ a in bar(Y)a \in \bar{Y}aY¯, then f ( a ) = 0 f ( a ) = 0 f(a)=0f(a) = 0f(a)=0 for every f X f X f inX^(‘)f \in X’fX where f = 0 f = 0 f=0f = 0f=0 on E E EEE:

Assume a Y ¯ a Y ¯ a in bar(Y)a \in \bar{Y}aY¯. This means that for every ϵ > 0 ϵ > 0 epsilon > 0\epsilon > 0ϵ>0, there exists y Y y Y y in Yy \in YyY such that a y < ϵ a y < ϵ ||a-y|| < epsilon\|a – y\| < \epsilonay<ϵ, since a a aaa can be approximated arbitrarily closely by elements of Y Y YYY.
Let f X f X f inX^(‘)f \in X’fX be such that f = 0 f = 0 f=0f = 0f=0 on E E EEE. Since Y = span E Y = span E Y=span EY = \operatorname{span}EY=spanE, f f fff also vanishes on Y Y YYY because any linear combination of elements in E E EEE (which f f fff maps to 0 0 000) will also be mapped to 0 0 000 by f f fff.
Given a Y ¯ a Y ¯ a in bar(Y)a \in \bar{Y}aY¯, and considering f f fff is continuous, we have f ( a ) = lim y a , y Y f ( y ) f ( a ) = lim y a , y Y f ( y ) f(a)=lim_(y rarr a,y in Y)f(y)f(a) = \lim_{y \to a, y \in Y} f(y)f(a)=limya,yYf(y). Since f ( y ) = 0 f ( y ) = 0 f(y)=0f(y) = 0f(y)=0 for all y Y y Y y in Yy \in YyY, it follows that f ( a ) = 0 f ( a ) = 0 f(a)=0f(a) = 0f(a)=0.

( lArr\Leftarrow) If f ( a ) = 0 f ( a ) = 0 f(a)=0f(a) = 0f(a)=0 for every f X f X f inX^(‘)f \in X’fX where f = 0 f = 0 f=0f = 0f=0 on E E EEE, then a Y ¯ a Y ¯ a in bar(Y)a \in \bar{Y}aY¯:

Assume f ( a ) = 0 f ( a ) = 0 f(a)=0f(a) = 0f(a)=0 for every f X f X f inX^(‘)f \in X’fX where f = 0 f = 0 f=0f = 0f=0 on E E EEE. Suppose, for the sake of contradiction, that a Y ¯ a Y ¯ a!in bar(Y)a \notin \bar{Y}aY¯. By the Hahn-Banach theorem, there exists a continuous linear functional f X f X f inX^(‘)f \in X’fX such that f ( a ) 0 f ( a ) 0 f(a)!=0f(a) \neq 0f(a)0 and f = 0 f = 0 f=0f = 0f=0 on Y ¯ Y ¯ bar(Y)\bar{Y}Y¯ (since a a aaa and Y ¯ Y ¯ bar(Y)\bar{Y}Y¯ can be separated).
However, since f = 0 f = 0 f=0f = 0f=0 on Y ¯ Y ¯ bar(Y)\bar{Y}Y¯, and E Y Y ¯ E Y Y ¯ E sube Y sube bar(Y)E \subseteq Y \subseteq \bar{Y}EYY¯, it follows that f = 0 f = 0 f=0f = 0f=0 on E E EEE. But we assumed that f ( a ) = 0 f ( a ) = 0 f(a)=0f(a) = 0f(a)=0 for all such f f fff, contradicting the existence of such an f f fff that separates a a aaa from Y ¯ Y ¯ bar(Y)\bar{Y}Y¯.
Therefore, our assumption that a Y ¯ a Y ¯ a!in bar(Y)a \notin \bar{Y}aY¯ must be false, implying a Y ¯ a Y ¯ a in bar(Y)a \in \bar{Y}aY¯.

Conclusion:

We’ve shown that a Y ¯ a Y ¯ a in bar(Y)a \in \bar{Y}aY¯ if and only if f ( a ) = 0 f ( a ) = 0 f(a)=0f(a) = 0f(a)=0 for every f X f X f inX^(‘)f \in X’fX where f = 0 f = 0 f=0f = 0f=0 on E E EEE, using the properties of normed spaces, their duals, and the Hahn-Banach theorem for the separation argument.
Verified Answer
5/5
Scroll to Top
Scroll to Top