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- Homework Statement
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- Relevant Equations
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Let ##X=\{(x_i)\in \ell^\infty | x_i=1 \text{ for } i\leq n, x_i=0 \text{ for } i>n, n\in \mathbb{N}\}##.
For every sequence ##(x_i)## in ##X##, ##(x_i)## has the property that the first ##n## entries are 1s and the rest are 0s. So, every sequence in ##X## trivially converges to ##X## and hence ##X## is a Banach space.
Define ##f:X\to \mathbb{R}## given by ##f((x_i))=(x_i)\cdot v## where ##v=(\frac{1}{2},\frac{1}{4},\frac{1}{8},...)##. ##f## is a linear functional by construction.
$$\displaystyle{\lim_{n \to \infty}} \sum_{j=1}^n \frac{1}{2^j}=1.$$
Hence ##f## is bounded and the supremum of ##f## over ##X## is ##1##.
##||(x_i)||_\infty\leq 1##, so ##f(x_i)_{||(x_i)||\leq 1}=1##.
Lastly, ##n## is finite, so there does not exist ##(x_i)\in X## such that ##f((x_i))= 1##.
I want to make a similar argument for a space of sequences over the closed 1-D unit ball that converge to 0, using the same linear functional.
For every sequence ##(x_i)## in ##X##, ##(x_i)## has the property that the first ##n## entries are 1s and the rest are 0s. So, every sequence in ##X## trivially converges to ##X## and hence ##X## is a Banach space.
Define ##f:X\to \mathbb{R}## given by ##f((x_i))=(x_i)\cdot v## where ##v=(\frac{1}{2},\frac{1}{4},\frac{1}{8},...)##. ##f## is a linear functional by construction.
$$\displaystyle{\lim_{n \to \infty}} \sum_{j=1}^n \frac{1}{2^j}=1.$$
Hence ##f## is bounded and the supremum of ##f## over ##X## is ##1##.
##||(x_i)||_\infty\leq 1##, so ##f(x_i)_{||(x_i)||\leq 1}=1##.
Lastly, ##n## is finite, so there does not exist ##(x_i)\in X## such that ##f((x_i))= 1##.
I want to make a similar argument for a space of sequences over the closed 1-D unit ball that converge to 0, using the same linear functional.
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