2

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.

sequence
  • 9,638
  • It seems to me that it would be easier to just construct an infinite linearly independent set. – carmichael561 Nov 25 '16 at 05:41
  • How can it can be done? – sequence Nov 25 '16 at 05:43
  • 2
    How about ${1,x,x^2,x^3,\dots}$? – carmichael561 Nov 25 '16 at 05:43
  • Do you mean just to argue that, since there exists an infinite basis for a subspace of $C[0,1]$, $C[0,1]$ must be infinite dimensional? – sequence Nov 25 '16 at 05:45
  • 1
    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. – carmichael561 Nov 25 '16 at 05:46
  • 2
    It's WRONG that in any vector space compactness is equivalent to boundedness and closedness. This is only a property of finite dimensional spaces! – Vim Nov 25 '16 at 05:47
  • Exactly, but this is a proof by contradiction (I'm assuming that $C[0,1]$ is finite dimensional). – sequence Nov 25 '16 at 05:50
  • 1
    @sequence you should prove it's not compact, not that it's not closed. – Vim Nov 25 '16 at 05:51
  • @carmichael561 oh sorry I misinterpretted your comment. – Vim Nov 25 '16 at 06:02
  • 1
    @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 – Mariano Suárez-Álvarez Nov 25 '16 at 06:08
  • @sequence, your approach is unnecessarily complicated. Just exhibit an infinite linearly independent set in your space, like the one carmicheal suggested in a comment above. – Mariano Suárez-Álvarez Nov 25 '16 at 06:11
  • @MarianoSuárez-Álvarez fair point. Thanks for the bug spotting. – Vim Nov 25 '16 at 06:13
  • But still, would it be logically correct to assume, for contradiction, that $C[0,1]$ is finite dimensional, and thus $S$ is compact, hence $S$ is closed and bounded. Then show that $S$ is bounded, but not closed? – sequence Nov 25 '16 at 06:24
  • @sequence. No. Because $S$ is bounded and closed. Again, once you've shown $S$ isn't compact it's done, but you do not need to, nor are able to show $S$ isn't bounded and closed because it is! – Vim Nov 25 '16 at 06:27
  • @Vim, but then, if convergence for some $x$ is not an issue, then $f_n=e^{-2\pi nx}$ will converge to $0$, correct? – sequence Nov 25 '16 at 06:31
  • 1
    @Vim, it is relevant to my proof, where I show that $\exists f_0$, a limit point of $S$, not in $S$. Showing that $S$ is not closed with the stated assumption that $C[0,1]$ is finite dimensional. – sequence Nov 25 '16 at 06:48
  • If you define $S$ as the unit sphere of $C[0,1]$ then $S$ is automatically closed, whether it's finite dimensional or not. – Vim Nov 25 '16 at 06:53
  • @sequence wait a minute, your $f_n = e^{-2\pi n t}$ does not converge to any function in $C[0,1]$. Since convergence in the uniform norm implies pointwise convergence, which means if there were some function $f$ to which $f_n$ converges to, $f$ would be zero on $[0,1)$ but attains $1$ at $1$, and would hence not possibly be continuous. – Vim Nov 25 '16 at 09:32

3 Answers3

2

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.
Vim
  • 13,640
  • What if I pick $\log(nex)$? It does not have a convergent subsequence and its sup-norm is $1$ for all n and some $x\in [0,1]$. – sequence Nov 25 '16 at 06:02
  • Actually, ${ e^{in2\pi t} }$ is a convergent sequence for $t=0$. – sequence Nov 25 '16 at 06:04
  • 1
    @sequence sorry I mistook it for $L^2$. I didn't check that in $|\cdot|_\infty$. – Vim Nov 25 '16 at 06:07
  • @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) – Vim Nov 25 '16 at 06:11
  • 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 $t$. – sequence Nov 25 '16 at 06:22
  • @sequence 1. For reals you take $\sin 2\pi nt$ instead. 2. Convergence for some t isn't a problem at all. What is the definition of the $|\cdot|_\infty$ norm? – Vim Nov 25 '16 at 06:23
  • $\lVert f_n \rVert_\infty = \max\limits_{t\in[0,1]} {f_n(t)}$. $\sin2\pi nt$ may converge if $t$ happens to satisfy the norm and simultaneously make the sequence constant for $n>N$. – sequence Nov 25 '16 at 06:28
2

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.

Community
  • 1
Najib Idrissi
  • 54,185
  • Caveat: lots of people might take "countable" as "at most countable". – Vim Nov 25 '16 at 09:27
  • @Vim It's fixed. – Najib Idrissi Nov 25 '16 at 09:41
  • @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?
     – sequence Nov 30 '16 at 03:25
  • 1
    @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 ${x^k}$. – Vim Nov 30 '16 at 05:29
  • @najib In fact, as per sequence's question, I don't think your last paragraph is necessary since you have already shown that $f=0$ implies zero coefficients of the linear span. Am I getting the point? – Vim Nov 30 '16 at 05:42
  • @NajibIdrissi, so does what you said under 2. mean that the number of finitely many $a_k\ne 0$ in this case is zero? – sequence Nov 30 '16 at 06:54
  • 1
    @Vim The last paragraph is just an interesting tidbit I thought OP would appreciate, it's not necessary for the question. – Najib Idrissi Nov 30 '16 at 08:18
  • @sequence Is it clearer now? It's just linear algebra... – Najib Idrissi Nov 30 '16 at 08:18
  • @NajibIdrissi No. I mean the last paragraph of your original answer that says $(p_1,\cdots)$ must be infinitely nonzero. Not the addendum about uncountable Hamel basis. (But it looks you have modified it.) – Vim Nov 30 '16 at 08:35
  • @Vim Ah, well I'm just summarizing what has been proven so far, hoping that my answer is clearer with it... – Najib Idrissi Nov 30 '16 at 08:36
  • @NajibIdrissi, so does what you said under 2. mean that the number of finitely many ak≠0 in this case is zero? – sequence Dec 01 '16 at 20:36
1

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.

copper.hat
  • 172,524