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I, just for fun, want to find a contradiction to the assumption that $C([0,1],\Bbb R)$ is finite-dimensional by constructing a continuous function that cannot be a linear combination of the proposed finite Hamel basis $\{f_1, ..., f_N \}$.

To be clear we have the assumed properties that $\forall f \in C([0,1], \Bbb R)$

$$ f = \sum_{k = 1}^N\alpha_kf_k$$

and

$$ \sum_{k = 1}^N\alpha_kf_k = 0 \implies \alpha_k = 0 \space \space\forall k \leq N$$

Now it is clear that if we take some countable linearly independent sequence of functions like $\sin (nx)$ or $x^n$, the assumption that they can all be written as a finite sum of our basis vectors can lead us to a contradiction, but I want instead to construct a single continuous function as a counterexample.

My construction, which is wrong:

we must have that each $f_k$ is nonzero in some neighbourhood contained in some $(\frac jN,\frac{j+1}N)$, then either:

  • we can ensure that these neighbourhoods lie in distinct such intervals,

or

  • otherwise there would be some $k$ intervals where at most $k-1$ of the proposed basis functions are nonzero (by Hall's marriage theorem), and we will continue the construction only using those $k-1$ functions, and setting $f$ to be zero outside of the $k$ intervals.

Now $f$ defined to be equal to $1$ on $\{\frac jN : j \leq N\}$, $0$ at the point in the middle of the distinct neighbourhoods where each $f_k$ is nonzero, and linear in between.

I wanted to conclude that all $\alpha_k$ must be zero, but this isn't true necessarily.

I'm asking really for an elegant construction; a nice function that evades linear combinations of out $\{f_1,...f_N\}$

Christian Fieldhouse
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  • "we must have that each $f_k$ is nonzero in some neighbourhood contained in some $(\frac jN,\frac{j+1}N)$, then either" I'm not sure this makes much sense. Do you mean that for each $k$ there is some $j_k$ and an open set $U_k\subset \left(\frac {j_k}N,\frac{j_k+1}N\right)$ such that $f_k\neq 0$ in $U_k$? – Fimpellizzeri Apr 14 '18 at 17:12
  • Here is an interpretation of your question: "Given $f_1,\dots,f_N\in C[0,1]$, construct a function $f\in C[0,1]$ which is not a linear combination of the $f_j$." Is this what you want? – Jan Bohr Apr 14 '18 at 17:14
  • @Jan Bohr, yes, sorry if that was unclear – Christian Fieldhouse Apr 14 '18 at 17:14
  • @Fimpellizieri, I mean what you write, and that the $U_k$ are contained in different $(\frac jN, \frac{j+1}N )$ – Christian Fieldhouse Apr 14 '18 at 17:17

1 Answers1

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Let $F=\{f_1,\dots,f_N\}\subset C[0,1]$ arbitrary. We will construct a function $f\in C[0,1]$ which is linearly independent of $F$.

Take $N+1$ distinct points $x_1,\dots,x_{N+1}\in [0,1]$ and define $$\Phi\colon C[0,1]\rightarrow\mathbb{R}^{N+1}, g\mapsto (g(x_1),\dots,g(x_{N+1})).$$ $\Phi$ is linear and $$\mathrm{rank}(\Phi\vert_{\mathrm{span}F})\le N,$$ because $\mathrm{span}F$ has dimension at most $N$. Hence there is a point $y=(b_1,\dots,b_{N+1})\in\mathbb{R}^{N+1}$ which does not lie in the range of $\Phi\vert_{\mathrm{span}F}$. (You can actually compute one!) Now let $f$ any continuous function on $[0,1]$ with $f(x_i)=b_i$ (e.g. a piecewise linear one).

If $f$ was not linearly independent of $F$, it would lie in the span of $F$ and hence $$(b_1,\dots,b_N)=\Phi(f)\in \mathrm{range}(\Phi\vert_{\mathrm{span}F}),$$ which is wrong by construction.

Jan Bohr
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