# linear functionals, closed sets, convergence

• Jun 21st 2008, 12:16 AM
marianne
linear functionals, closed sets, convergence
This is the last round of questions, I promise.

(1) Identify the space of all continuous linear functionals on the space $\displaystyle c_0$.

(2) $\displaystyle X=(C([0, 2], \mathbb{R}), || . ||_2)$,
$\displaystyle Y=\{f \in C([0, 2], \mathbb{R}: f(1-x)=f(1+x), x \in [0,1]\}$
Is Y closed vector space in X? What is the complement of Y?

(3) Prove that the series
$\displaystyle \sum_{k=1}^{\infty}x_k b_k$ converge for every vector
$\displaystyle x=(x_k) \in l^p, 1<p<\infty$ if and only if $\displaystyle b=(b_k) \in l^q, 1/p +1/q=1$.

These are the last ones I don't know what to do with, and if anyone has the time and is willing to help, I would be really grateful.
• Jun 21st 2008, 01:31 PM
Opalg
Quote:

Originally Posted by marianne
(1) Identify the space of all continuous linear functionals on the space $\displaystyle c_0$.

The dual space of $\displaystyle c_0$ is (isometrically isomorphic to) $\displaystyle l^1$. If $\displaystyle x=(x_n)\in c_0$ and $\displaystyle y=(y_n)\in l^1$ then the value of the functional y at the point x is $\displaystyle \sum x_ny_n$. This is a very standard result which you should be able to find in any relevant textbook.

Quote:

Originally Posted by marianne
(2) $\displaystyle X=(C([0, 2], \mathbb{R}), || . ||_2)$,
$\displaystyle Y=\{f \in C([0, 2], \mathbb{R}): f(1-x)=f(1+x), x \in [0,1]\}$
Is Y closed vector space in X? What is the complement of Y?

This is more interesting. Take the second question first. If g(x) is in the orthogonal complement of Y, then $\displaystyle \int_0^2f(x)g(x)\,dx=0$ for all f in Y. But $\displaystyle \int_0^2f(x)g(x)\,dx = \int_0^1f(x)g(x)\,dx + \int_1^2f(x)g(x)\,dx$. If you change variables in the two integrals, putting y=1-x in the first and y=1+x in the second, then you find that $\displaystyle \int_0^1f(1-y)\bigl(g(1-y)+g(1+y)\bigr)dy=0$ (remembering that f(1+y)=f(1-y)). This holds for all continuous functions on [0,1], so you conclude that g(1+y)=–g(1-y) for all y in [0,1].

Thus the orthogonal complement of Y is $\displaystyle Z=\{g \in C([0, 2], \mathbb{R}): g(1-x)=-g(1+x), x \in [0,1]\}.$ If you repeat the same argument on the subspace Z, you will find that its orthogonal complement is Y. Therefore Y is equal to its second orthogonal complement, which is closed in X. Hence Y is closed.

Quote:

Originally Posted by marianne
(3) Prove that the series
$\displaystyle \sum_{k=1}^{\infty}x_k b_k$ converge for every vector
$\displaystyle x=(x_k) \in l^p, 1<p<\infty$ if and only if $\displaystyle b=(b_k) \in l^q, 1/p +1/q=1$.

This is another standard piece of bookwork. Look it up in any good textbook. The proof uses Hölder's inequality.