a) Let X={f in C[0,1]:f(0)=0}X=\{f \in C[0,1]: f(0)=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\}
Prove that Y\mathrm{Y} is a proper subspace of X\mathrm{X}. Is Y\mathrm{Y} a closed subspace of X\mathrm{X} ? Justify your answer.
Answer:
To prove that YY is a proper subspace of XX and determine whether YY is a closed subspace of XX, we need to examine the definitions and properties of these sets within the context of functional analysis.
XX and YY Defined
X={f in C[0,1]:f(0)=0}X = \{f \in C[0,1]: f(0) = 0\} is the set of all continuous functions on the interval [0,1][0,1] that vanish at 00.
Y={g in X:int_(0)^(1)g(t)dt=0}Y = \left\{g \in X: \int_0^1 g(t) dt = 0\right\} is the set of all functions in XX whose integral over [0,1][0,1] is 00.
Proving YY is a Proper Subspace of XX
To show that YY is a proper subspace of XX, we must verify that YY satisfies the following criteria for being a subspace:
Non-emptiness: YY contains the zero function, which is the function g(t)=0g(t) = 0 for all t in[0,1]t \in [0,1]. This function clearly belongs to XX and satisfies the integral condition, so YY is non-empty.
Closed under addition: If g_(1),g_(2)in Yg_1, g_2 \in Y, then g_(1)+g_(2)in Yg_1 + g_2 \in Y. This is because the integral of the sum is the sum of the integrals, each of which is 00, so their sum is also 00.
Closed under scalar multiplication: If g in Yg \in Y and alpha\alpha is a scalar, then alpha g in Y\alpha g \in Y. This follows because 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 = 0.
Since YY satisfies these criteria, it is a subspace of XX. It is a proper subspace because there exist functions in XX that do not satisfy the integral condition, such as f(t)=tf(t) = t, which is in XX but not in YY since int_(0)^(1)tdt=1//2!=0\int_0^1 t dt = 1/2 \neq 0.
Is YY a Closed Subspace of XX?
A subspace YY is closed in XX if it contains all its limit points; that is, if a sequence of functions {g_(n)}\{g_n\} in YY converges uniformly to a function gg, then gg must also be in YY.
To show YY is closed, consider a sequence {g_(n)}sub Y\{g_n\} \subset Y that converges uniformly to g in Xg \in X. We need to show that g in Yg \in Y, meaning int_(0)^(1)g(t)dt=0\int_0^1 g(t) dt = 0.
Uniform convergence of {g_(n)}\{g_n\} to gg implies that for every epsilon > 0\epsilon > 0, there exists an NN such that for all n >= Nn \geq N, we have |g_(n)(t)-g(t)| < epsilon|g_n(t) – g(t)| < \epsilon for all t in[0,1]t \in [0,1]. By the properties of integrals and the limit of a sequence of functions,
Since each g_(n)in Yg_n \in Y, we have int_(0)^(1)g_(n)(t)dt=0\int_0^1 g_n(t) dt = 0. Thus, the limit of these integrals as n rarr oon \to \infty is also 00, which means int_(0)^(1)g(t)dt=0\int_0^1 g(t) dt = 0, and hence g in Yg \in Y.
Therefore, YY is a closed subspace of XX because it contains all its limit points under uniform convergence.
b) Let X=L^(p)[0,1]X=L^p[0,1] and x=x(t)=t^(2)x=x(t)=t^2. Find ||x||_(p)\|x\|_p for p=4p=4 and oo\infty.
Answer:
To find the L^(p)L^p norm of x(t)=t^(2)x(t) = t^2 on the interval [0,1][0,1] for p=4p=4 and p=oop=\infty, we’ll use the definitions of the L^(p)L^p norms.
To compute this integral, we use the formula for the integral of t^(n)t^n, which is (t^(n+1))/(n+1)\frac{t^{n+1}}{n+1} for n!=-1n \neq -1, from 00 to 11:
The L^( oo)L^\infty norm, or the supremum norm, is defined as the essential supremum of |x(t)||x(t)| over the interval. For x(t)=t^(2)x(t) = t^2 on [0,1][0,1], this is simply the maximum value of t^(2)t^2 on the interval, which occurs at t=1t=1. Therefore,
For p=4p=4, ||x||_(4)=(1)/(sqrt3)\|x\|_4 = \frac{1}{\sqrt{3}}.
For p=oop=\infty, ||x||_(oo)=1\|x\|_\infty = 1.
c) Let E be a subset of a normed space X,Y=span EX, Y=\operatorname{span} E and a in Xa \in X. Show that a in bar(Y)a \in \bar{Y} if and only if f(a)=0f(a)=0 whenever f inX^(‘)f \in X^{\prime} and f=0f=0 everywhere on EE.
Answer:
To show that a in bar(Y)a \in \bar{Y} if and only if f(a)=0f(a) = 0 for every f inX^(‘)f \in X’ (the dual space of XX) where f=0f = 0 on EE, we’ll use some fundamental properties of normed spaces, their duals, and the concept of closure.
Definitions:
XX is a normed space, and X^(‘)X’ is its dual space, consisting of all continuous linear functionals on XX.
EE is a subset of XX, and Y=span EY = \operatorname{span}E is the subspace spanned by EE.
bar(Y)\bar{Y} denotes the closure of YY in XX, which includes all limits of convergent sequences in YY.
(=>\Rightarrow) If a in bar(Y)a \in \bar{Y}, then f(a)=0f(a) = 0 for every f inX^(‘)f \in X’ where f=0f = 0 on EE:
Assume a in bar(Y)a \in \bar{Y}. This means that for every epsilon > 0\epsilon > 0, there exists y in Yy \in Y such that ||a-y|| < epsilon\|a – y\| < \epsilon, since aa can be approximated arbitrarily closely by elements of YY.
Let f inX^(‘)f \in X’ be such that f=0f = 0 on EE. Since Y=span EY = \operatorname{span}E, ff also vanishes on YY because any linear combination of elements in EE (which ff maps to 00) will also be mapped to 00 by ff.
Given a in bar(Y)a \in \bar{Y}, and considering ff is continuous, we have f(a)=lim_(y rarr a,y in Y)f(y)f(a) = \lim_{y \to a, y \in Y} f(y). Since f(y)=0f(y) = 0 for all y in Yy \in Y, it follows that f(a)=0f(a) = 0.
(lArr\Leftarrow) If f(a)=0f(a) = 0 for every f inX^(‘)f \in X’ where f=0f = 0 on EE, then a in bar(Y)a \in \bar{Y}:
Assume f(a)=0f(a) = 0 for every f inX^(‘)f \in X’ where f=0f = 0 on EE. Suppose, for the sake of contradiction, that a!in bar(Y)a \notin \bar{Y}. By the Hahn-Banach theorem, there exists a continuous linear functional f inX^(‘)f \in X’ such that f(a)!=0f(a) \neq 0 and f=0f = 0 on bar(Y)\bar{Y} (since aa and bar(Y)\bar{Y} can be separated).
However, since f=0f = 0 on bar(Y)\bar{Y}, and E sube Y sube bar(Y)E \subseteq Y \subseteq \bar{Y}, it follows that f=0f = 0 on EE. But we assumed that f(a)=0f(a) = 0 for all such ff, contradicting the existence of such an ff that separates aa from bar(Y)\bar{Y}.
Therefore, our assumption that a!in bar(Y)a \notin \bar{Y} must be false, implying a in bar(Y)a \in \bar{Y}.
Conclusion:
We’ve shown that a in bar(Y)a \in \bar{Y} if and only if f(a)=0f(a) = 0 for every f inX^(‘)f \in X’ where f=0f = 0 on EE, using the properties of normed spaces, their duals, and the Hahn-Banach theorem for the separation argument.