$x,|x|,x^2,x|x|$ linearly dependent or independent?

Question

Let $V$ be a vector space of continuous functions on $[-1,1]$ over $\Bbb R$. Let $u_1,u_2,u_3,u_4\in V$ defined as $u_1(x)=x,u_2(x)=|x|,u_3(x)=x^2,u_4(x)=x|x|$ then

1. $\{u_1,u_2\}$ is linearly dependent.
2. $\{u_1,u_3,u_4\}$ is linearly dependent.
3. $\{u_1,u_2,u_4\}$ is linearly dependent.
4. None of the above.

I think none of the above is correct: Plugging $x=1$ in $ax=|x|$ gives us $a=1$, but then for $x=-1$ we don't get $ax=|x|$. So, (1) fails.

Plugging $x=1,-1$ in $x=ax^2+bx|x|$ gives $a=0,b=1$, but for $x=\frac12$, $x=ax^2+bx|x|$ fails.

Plugging $x=1,\frac12$ in $x=a|x|+bx|x|$ gives $a=1,b=0$, but for $x=-\frac12$, $x=a|x|+bx|x|$ fails.

so, none of the above given set is linearly dependent, right?

Answer 1

Correct. More generally, the four functions form a linearly independent set, because if: $$ ax + b|x| + cx^2 + dx|x| = 0 $$

then it can be shown that $a = b = c = d = 0$. Just evaluate the equation at $x = \pm 1, \pm \frac{1}{2}$ and solve the resulting system of four equations in four unknowns.

Answer 2

Suppose $f(x) = \sum_k \alpha_k u_k(x) = 0$.

Note that $f$ is smooth for $x \neq 0$.

Note that $f''(x) = 2 \alpha_3 u_3''(x) + 2 \alpha_4 u_4''(x)$. Evaluating at $x=\pm 1$ shows that $\alpha_3 = \alpha_4 = 0$.

Then $f'(x) = \alpha_1 u_1'(x) + \alpha_2 u_2'(x)$, evaluating at $x= \pm 1$ shows that $\alpha_1 = \alpha_2 = 0$.

Hence $u_1,...,u_4$ are linearly independent.