# Free families

1. May 29, 2015

### geoffrey159

1. The problem statement, all variables and given/known data
Prove that the following families are free in the vector space of continuous mappings from $\mathbb{R} \rightarrow \mathbb{R}$, with real scalars :
1 - $(f_\lambda)_{\lambda \in \mathbb{R}^+} : f_\lambda(x) = \cos(\lambda x)$
2 - $(f_\lambda)_{\lambda \in \mathbb{R}} : f_\lambda(x) = |x-\lambda|$

2. Relevant equations

3. The attempt at a solution

Ok, I'm not used to that kind of exercise so it may be all wrong, and I've worked this problem long enough so that it may be even worse than that.

My approach will be to show that any finite sub-family of $(f_\lambda)_{\lambda\in \mathbb{R}}$ is free.
Let $\lambda_1 < ... < \lambda_N$ be distinct lambdas, and $(x_1,...,x_N)$ be real scalars such that $\sum_{i = 1}^N x_i f_{\lambda_i}(x) = 0$, for any real $x$. I want to show that $x_1 = ... = x_N = 0$

1 -
In this case, $0\le \lambda_1 < ... < \lambda_N$. Since one can derivate infinitely many times the function $f_\lambda$, then for any $k\ge 0$,

$0 = \frac{d^{2k}}{dx^{2k}}(\sum_{i = 1}^N x_i f_{\lambda_i}) = \sum_{i = 1}^N x_i \frac{d^{2k}}{dx^{2k}}(f_{\lambda_i}) = (-1)^k \sum_{i = 1}^N x_i \lambda_i^{2k} f_{\lambda_i}$.

The $(-1)^k$ can be overlooked since the whole term is equal to zero. Taking $x = 0$, then $0 =\sum_{i = 1}^N x_i \lambda_i^{2k}$.

So $x_N = - \sum_{i = 1}^N x_i (\frac{\lambda_i}{\lambda_N})^{2k}$. Since $0 \le (\frac{\lambda_i}{\lambda_N})^{2k} < 1$ for any $k$, then $x_N = 0$ by taking the limit as $k$ tends to infinity.

Repeating this process, I find $x_N = ... = x_2 = 0$. Now if $\lambda_1 = 0$, then
$0 = \sum_{i = 1}^N x_i f_{\lambda_i}(x) = x_1 f_{\lambda_1}(x) = x_1$. If $\lambda_1 \neq 0$, then $0 = \sum_{i = 1}^N x_i \lambda_i^{2k} = x_1 \lambda_1^{2k}$, so $x_1 = 0$ too !

2 - I haven't reached to the conclusion yet.
What I can show is that for any $x > \max(0,\lambda_N)$, then $x \sum_{i = 1}^N x_i = \sum_{i = 1}^N x_i \lambda_i$.
Dividing by $x$ on both side (legal because x > 0) and taking the limit as $x$ goes to infinity,
$0 = \sum_{i = 1}^N x_i = \sum_{i = 1}^N x_i \lambda_i$. I need help for this one

2. May 29, 2015

### wabbit

Regarding 1, your proof looks fine to me - the idea is quite elegant actually.

For 2, you may try differentiating too - just once. Even though it is not differentiable everywhere, you might want to look at the behaviour of the derivative of the function $f=\sum x_i f_{\lambda_i}$.

Last edited by a moderator: May 29, 2015
3. May 29, 2015

### geoffrey159

You gave me the idea, thank you !

So I keep one piece of what I found for 2 : $\sum_{i=1}^N x_i = 0$

For any x such that $\lambda_{N-1} < x < \lambda_N$, then $0 = \sum_{i=1}^N x_i |x-\lambda_i| = x_N (\lambda_N - x) + \sum_{i=1}^{N-1} x_i (x-\lambda_i)$
If you take the derivative with respect to $x$, then you get $x_N = \sum_{i=1}^{N-1} x_i = -x_N$. So $x_N = 0$.

Repeat the process for $\lambda_{k-1} < x < \lambda_k$

4. May 30, 2015

### wabbit

Yep that works - and you don't even actually need to repeat the process : if the family is not free then there exist a combination $\sum_{i=1}^N x_i f_{\lambda_i}=0$ with non-zero coefficients, i.e. such that $\forall i, x_i\neq 0$. By showing that $x_N=0$ you have already exhibited a contradiction.

5. May 30, 2015

### geoffrey159

Thank you, that was quality help, it solved my problem in minutes !

However, I don't quite understand your last post: if the family is not free, it seems to me that there exist at least one $x_i \neq 0$ rather than all $x_i \neq 0$. I don't see how you can skip the looping process ?

6. May 30, 2015

### wabbit

No problem, nice to know it was helpful.

If you start with a set of coefficients which are not all zero, first extract the subset of non-zero coefficients : voilà, a combination with only non-zero coefficients !

It might sound like a shady trick, but if you think about it, you'll see its perfectly honest : )

7. Jun 3, 2015

### micromass

Just some aside information which might or might not be helpful: if you consider distributions, then $|x|$ (just like any continuous function) is differentiable everywhere, so you can simply apply the same method for every one of these.

Also, your method in 1 was nice, and this has actually be generalized: http://en.wikipedia.org/wiki/Wronskian

8. Jun 3, 2015

### geoffrey159

Thank you for the info !