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Let $C([0,1])$ equipped with $\lVert \cdot \rVert_\infty$ be set of all functions continuous on $[0,1]$. Prove that $C([0,1])$ is not finite dimensional.

There is a theorem which states that a normed vector space is finite dimensional if and only if its closed unit ball $\{v\in V: \lVert v \rVert =1 \}$ is compact. In our case, $S=\{f\in C([0,1]): \lVert f \rVert_\infty =1 \}$. We know that a subset of a vector space is compact if and only if it is closed and bounded. Clearly, $S$ is bounded. So all is needed to be proved here is that $S$ cannot be closed to establish that $(C([0,1]),\lVert \cdot \rVert_\infty)$ is not finite dimensional.

One of the approaches that I've tried is this:

If $f_0$ is a limit point of $S$ then $\exists \{f_n\}\subset S$ such that $\{f_n\}$ converges to $f$. In this case, one would have to prove that somehow $\lVert f_0 \rVert_\infty \ne 1$. This can be done by providing a counterexample that $\exists f_n\subset S$ converging to $f_0$ where $\lVert f_0 \rVert_\infty \ne 1$. I came up with the sequence of functions $f_n=e^{-nx}$, which supposedly converges to the zero function. The only problem is: what if $x=0$? Then $f_n$ converges to $1$.

I would appreciate some advice.

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    $\begingroup$ How about $\{1,x,x^2,x^3,\dots\}$? $\endgroup$ – carmichael561 Nov 25 '16 at 5:43
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    $\begingroup$ My argument is that if a vector space $V$ has finite dimension $n$, then any linearly independent subset of $V$ has cardinality at most $n$. Therefore a vector space containing an infinite linearly independent set must be infinite-dimensional. $\endgroup$ – carmichael561 Nov 25 '16 at 5:46
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    $\begingroup$ It's WRONG that in any vector space compactness is equivalent to boundedness and closedness. This is only a property of finite dimensional spaces! $\endgroup$ – Vim Nov 25 '16 at 5:47
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    $\begingroup$ @sequence you should prove it's not compact, not that it's not closed. $\endgroup$ – Vim Nov 25 '16 at 5:51
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    $\begingroup$ @vim, there are other topological vector spaces in which a subset is compact iff it is bounded and closed, apart from finite dimensional ones. If you assume the topological vector space is a Banach space, though, then that is true.ç $\endgroup$ – Mariano Suárez-Álvarez Nov 25 '16 at 6:08
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Your argument is not gonna work because $S$ is closed in any normed space. The point is not to show it's not closed, but to show it's not compact: as addressed in my comment, the equivalence between compactness and closedness-boundedness is only a property of finite dimensional vector spaces. (Edit: as spotted by @Mariano Suárez-Álvarez, we need to assume first that the vector space is complete.)

To prove non-compactness you might want to use the following (equivalent) characterisation of compactness in any metric space: if $K$ is compact, then any sequence in it has a subsequential limit in it. Now pick $\{e^{in2\pi t}\}\subset S$, and you should be able to show it has no subsequential limit at all.


Or, for a much more straightforward approach, just see @carmichael561's comment: $\{1,x,x^2,\cdots\}$ are linear independent, but no finite-dimensional vector space admits an infinite set of linearly independent vectors.

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  • $\begingroup$ What if I pick $\log(nex)$? It does not have a convergent subsequence and its sup-norm is $1$ for all n and some $x\in [0,1]$. $\endgroup$ – sequence Nov 25 '16 at 6:02
  • $\begingroup$ Actually, $\{ e^{in2\pi t} \}$ is a convergent sequence for $t=0$. $\endgroup$ – sequence Nov 25 '16 at 6:04
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    $\begingroup$ @sequence sorry I mistook it for $L^2$. I didn't check that in $\|\cdot\|_\infty$. $\endgroup$ – Vim Nov 25 '16 at 6:07
  • $\begingroup$ @sequence Actually it still holds in $\|\|_\infty$. Try to show that any distinct pair of elements in this sequence has distance $2$. (Imagine the functions as rotations along $S^1$ for visual aid) $\endgroup$ – Vim Nov 25 '16 at 6:11
  • $\begingroup$ Yes, we can find such elements, but $C[0,1]$ is a set of continuous real functions, not complex. Also, the sequence you gave still converges for some $t$. $\endgroup$ – sequence Nov 25 '16 at 6:22
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The norm is completely irrelevant for this exercise, it's just linear algebra.

Let $p_n(x) = x^n$, so that $p_n \in C([0,1])$. Suppose that $f := \sum a_k p_k = 0$ for some coefficients $(a_0, a_1, \dots)$ (where only a finite number are nonzero). Then $a_k = k! \cdot f^{(k)}(0)$, simply by differentiating the polynomial and evaluating at $x = 0$. Since $f$ is actually the zero function, all its derivatives vanish, therefore $f^{(k)}(0) = 0$ and thus $a_k = 0$.

In other words, as soon as a we pick a linear combination of the $p_k$ which vanish, then all the coefficients must vanish. By definition, this means that the family $(p_0, p_1, \dots)$ is linearly independent. It is also infinite, thus $C([0,1])$ is infinite-dimensional, as otherwise any infinite family would be linearly dependent.


Now, something more interesting is that the dimension of $C([0,1])$ is not even countable. Indeed, it is a Banach space, and infinite-dimensional Banach spaces never have a countable basis (by the Baire category theorem). For this the norm is really relevant.

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  • $\begingroup$ Caveat: lots of people might take "countable" as "at most countable". $\endgroup$ – Vim Nov 25 '16 at 9:27
  • $\begingroup$ @Vim It's fixed. $\endgroup$ – Najib Idrissi Nov 25 '16 at 9:41
  • $\begingroup$ @NajibIdrissi, can you please clarify the following? 1. Why is $a_k = k! f^{(k)}(0)$? I think it's related to Maclaurin series, but I don't see how you derived this. 2. Why does $a_k$ have to be zero if some of the coefficients (finite number thereof) are nonzero? 3. How exactly does it follow that $(p_0, p_1,...)$ is infinite? $\endgroup$ – sequence Nov 30 '16 at 3:25
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    $\begingroup$ @sequence 1. You don't have to think of Maculaurin series or power series because as hypothesised $f$ is in the (finite) span of $\{x^k\}$ so all you have to do is to differentiate a polynomial $1,2,3,\cdot n$ times respectively at $x=0$, from which you get $a_k=\dfrac1{k!}f^{(k)}(0)$. 2. Showinh linear independence is equivalent to showing if $f=\sum_{i=1}^n a_k p_k=0$ then $a_k=0$ for each $k$. This directly from the previous analysis. 3. In 1,2 we have shown the finitely many coefficient case is unlikely to admit independent $\{x^k\}$. $\endgroup$ – Vim Nov 30 '16 at 5:29
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    $\begingroup$ @Vim The last paragraph is just an interesting tidbit I thought OP would appreciate, it's not necessary for the question. $\endgroup$ – Najib Idrissi Nov 30 '16 at 8:18
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Let $f_n$ linearly interpolate the points $(0,0), ({1 \over n+1},0), ({1\over 2}({1 \over n}+ { 1\over n+1}),1), ({1 \over n},0),(1,0)$.

Suppose $\sum_n \alpha_n f_n = 0$, then be evaluating at ${1\over 2}({1 \over k}+ { 1\over k+1} )$, we see that $\alpha_k = 0$, hence the $f_n$ are linearly independent.

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