Sample Solution

MMT-006 Solved Assignment 2024 Sample Solution

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

a) Let $X = {f \in C [0, 1] : f (0) = 0}$

Y = {g \in x : \int_{0}^{1} g (t) d t = 0}

Prove that

Y

is a proper subspace of

X

. Is

Y

a closed subspace of

X

? Justify your answer.

Answer:

To prove that

Y

is a proper subspace of

X

and determine whether

Y

is a closed subspace of

X

, we need to examine the definitions and properties of these sets within the context of functional analysis.

$X$ and $Y$ Defined

$X = {f \in C [0, 1] : f (0) = 0}$ is the set of all continuous functions on the interval $[0, 1]$ that vanish at $0$ .
$Y = {g \in X : \int_{0}^{1} g (t) d t = 0}$ is the set of all functions in $X$ whose integral over $[0, 1]$ is $0$ .

Proving $Y$ is a Proper Subspace of $X$

To show that

Y

is a proper subspace of

X

, we must verify that

Y

satisfies the following criteria for being a subspace:

Non-emptiness: $Y$ contains the zero function, which is the function $g (t) = 0$ for all $t \in [0, 1]$ . This function clearly belongs to $X$ and satisfies the integral condition, so $Y$ is non-empty.
Closed under addition: If $g_{1}, g_{2} \in Y$ , then $g_{1} + g_{2} \in Y$ . This is because the integral of the sum is the sum of the integrals, each of which is $0$ , so their sum is also $0$ .
Closed under scalar multiplication: If $g \in Y$ and $α$ is a scalar, then $α g \in Y$ . This follows because $\int_{0}^{1} α g (t) d t = α \int_{0}^{1} g (t) d t = α \cdot 0 = 0$ .

Since

Y

satisfies these criteria, it is a subspace of

X

. It is a proper subspace because there exist functions in

X

that do not satisfy the integral condition, such as

f (t) = t

, which is in

X

but not in

Y

since

\int_{0}^{1} t d t = 1 / 2 \neq 0

Is $Y$ a Closed Subspace of $X$ ?

A subspace

Y

is closed in

X

if it contains all its limit points; that is, if a sequence of functions

{g_{n}}

Y

converges uniformly to a function

g

, then

g

must also be in

Y

To show

Y

is closed, consider a sequence

{g_{n}} \subset Y

that converges uniformly to

g \in X

. We need to show that

g \in Y

, meaning

\int_{0}^{1} g (t) d t = 0

Uniform convergence of

{g_{n}}

g

implies that for every

ϵ > 0

, there exists an

N

such that for all

n \geq N

, we have

| g_{n} (t) - g (t) | < ϵ

for all

t \in [0, 1]

. By the properties of integrals and the limit of a sequence of functions,

lim_{n \to \infty} \int_{0}^{1} g_{n} (t) d t = \int_{0}^{1} lim_{n \to \infty} g_{n} (t) d t = \int_{0}^{1} g (t) d t .

Since each

g_{n} \in Y

, we have

\int_{0}^{1} g_{n} (t) d t = 0

. Thus, the limit of these integrals as

n \to \infty

is also

0

, which means

\int_{0}^{1} g (t) d t = 0

, and hence

g \in Y

Therefore,

Y

is a closed subspace of

X

because it contains all its limit points under uniform convergence.

b) Let

X = L^{p} [0, 1]

and

x = x (t) = t^{2}

. Find

∥ x ∥_{p}

for

p = 4

and

\infty

Answer:

To find the

L^{p}

norm of

x (t) = t^{2}

on the interval

[0, 1]

for

p = 4

and

p = \infty

, we’ll use the definitions of the

L^{p}

norms.

For $p = 4$

The

L^{p}

norm for

p = 4

is defined as

∥ x ∥_{4} = {(\int_{0}^{1} | t^{2} |^{4} d t)}^{1 / 4}

This simplifies to

∥ x ∥_{4} = {(\int_{0}^{1} t^{8} d t)}^{1 / 4}

To compute this integral, we use the formula for the integral of

t^{n}

, which is

\frac{t^{n + 1}}{n + 1}

for

n \neq - 1

, from

0

1

\int_{0}^{1} t^{8} d t = \frac{t^{9}}{9} |_{0}^{1} = \frac{1}{9}

Therefore,

∥ x ∥_{4} = {(\frac{1}{9})}^{1 / 4}

For $p = \infty$

The

L^{\infty}

norm, or the supremum norm, is defined as the essential supremum of

| x (t) |

over the interval. For

x (t) = t^{2}

[0, 1]

, this is simply the maximum value of

t^{2}

on the interval, which occurs at

t = 1

. Therefore,

∥ x ∥_{\infty} = max_{t \in [0, 1]} | t^{2} | = 1^{2} = 1

Calculation

Let’s calculate

∥ x ∥_{4}

explicitly:

∥ x ∥_{4} = {(\frac{1}{9})}^{1 / 4}

This calculation yields:

∥ x ∥_{4} = {(\frac{1}{9})}^{1 / 4} = \frac{1}{3^{1 / 2}} = \frac{1}{\sqrt{3}}

So, summarizing:

For $p = 4$ , $∥ x ∥_{4} = \frac{1}{\sqrt{3}}$ .
For $p = \infty$ , $∥ x ∥_{\infty} = 1$ .

c) Let E be a subset of a normed space

X, Y = span E

and

a \in X

. Show that

a \in \bar{Y}

if and only if

f (a) = 0

whenever

f \in X^{'}

and

f = 0

everywhere on

E

Answer:

To show that

a \in \bar{Y}

if and only if

f (a) = 0

for every

f \in X^{'}

(the dual space of

X

) where

f = 0

E

, we’ll use some fundamental properties of normed spaces, their duals, and the concept of closure.

Definitions:

$X$ is a normed space, and $X^{'}$ is its dual space, consisting of all continuous linear functionals on $X$ .
$E$ is a subset of $X$ , and $Y = span E$ is the subspace spanned by $E$ .
$\bar{Y}$ denotes the closure of $Y$ in $X$ , which includes all limits of convergent sequences in $Y$ .

( $\Rightarrow$ ) If $a \in \bar{Y}$ , then $f (a) = 0$ for every $f \in X^{'}$ where $f = 0$ on $E$ :

Assume

a \in \bar{Y}

. This means that for every

ϵ > 0

, there exists

y \in Y

such that

∥ a - y ∥ < ϵ

, since

a

can be approximated arbitrarily closely by elements of

Y

Let

f \in X^{'}

be such that

f = 0

E

. Since

Y = span E

f

also vanishes on

Y

because any linear combination of elements in

E

(which

f

maps to

0

) will also be mapped to

0

f

Given

a \in \bar{Y}

, and considering

f

is continuous, we have

f (a) = lim_{y \to a, y \in Y} f (y)

. Since

f (y) = 0

for all

y \in Y

, it follows that

f (a) = 0

( $\Leftarrow$ ) If $f (a) = 0$ for every $f \in X^{'}$ where $f = 0$ on $E$ , then $a \in \bar{Y}$ :

Assume

f (a) = 0

for every

f \in X^{'}

where

f = 0

E

. Suppose, for the sake of contradiction, that

a \notin \bar{Y}

. By the Hahn-Banach theorem, there exists a continuous linear functional

f \in X^{'}

such that

f (a) \neq 0

and

f = 0

\bar{Y}

(since

a

and

\bar{Y}

can be separated).

However, since

f = 0

\bar{Y}

, and

E \subseteq Y \subseteq \bar{Y}

, it follows that

f = 0

E

. But we assumed that

f (a) = 0

for all such

f

, contradicting the existence of such an

f

that separates

a

from

\bar{Y}

Therefore, our assumption that

a \notin \bar{Y}

must be false, implying

a \in \bar{Y}

Conclusion:

We’ve shown that

a \in \bar{Y}

if and only if

f (a) = 0

for every

f \in X^{'}

where

f = 0

E

, using the properties of normed spaces, their duals, and the Hahn-Banach theorem for the separation argument.

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MMT-006 Solved Assignment 2024

Sample Solution

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

X X XXX and Y Y YYY Defined

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

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

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

For p = ∞ p = ∞ p=oop=\inftyp=∞

Calculation

Definitions:

( ⇒ ⇒ =>\Rightarrow⇒) If a ∈ Y ¯ a ∈ Y ¯ a in bar(Y)a \in \bar{Y}a∈Y¯, 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’f∈X′ where f = 0 f = 0 f=0f = 0f=0 on E E EEE:

( ⇐ ⇐ 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’f∈X′ 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}a∈Y¯:

Conclusion:

$X$ and $Y$ Defined

Proving $Y$ is a Proper Subspace of $X$

Is $Y$ a Closed Subspace of $X$ ?

For $p = 4$

For $p = \infty$

( $\Rightarrow$ ) If $a \in \bar{Y}$ , then $f (a) = 0$ for every $f \in X^{'}$ where $f = 0$ on $E$ :

( $\Leftarrow$ ) If $f (a) = 0$ for every $f \in X^{'}$ where $f = 0$ on $E$ , then $a \in \bar{Y}$ :