Linear independence of function u(x), u(-x)

Question

Hopefully someone can give me a bit of a hint here.

Given function $u$ on [-1,1], which is neither even nor odd, I need to show that u(x) and u(-x) are linearly independent.

Thanks!

Answer 1

What does it mean for two functions $f,g$ to be linearly independent? It means that $\alpha f+\beta g\equiv 0$ implies $\alpha=\beta=0$.

Suppose that $\alpha u(x)+\beta u(-x)=0$ for all $x\in[-1,1]$, for some $(\alpha,\beta)\neq(0,0)$.

Note that there must exist $x_0\neq0$ such that $u(x_0)\neq 0$; otherwise, we have $u(x)=0=u(-x)$ for all $x\in[-1,1]$, contradicting our assumption that $u$ is not even or odd.

Further, it must be the case that $\beta\neq 0$: otherwise, we have $\alpha u(x)=0$, $\alpha\neq 0$, meaning that $u(x)=0$ for all $x$, and is therefore even. Similarly, $\alpha\neq 0$.

Now, plugging in $x=x_0$, we find that $\alpha u(x_0)+\beta u(-x_0)=0$; equivalently, $$ \frac{\alpha}{\beta}=\frac{u(-x_0)}{u(x_0)}, $$ since $u(x_0)\neq 0$ and $\beta\neq0$. Plugging in $x=-x_0$ yields, similarly, $$ \frac{\alpha}{\beta}=\frac{u(x_0)}{u(-x_0)}. $$ So, we find that $$ \frac{\alpha}{\beta}=\frac{u(-x_0)}{u(x_0)}=\frac{\beta}{\alpha}, $$ so that $\alpha=\pm\beta$. But, $\alpha\neq\beta$, as this would imply that $u$ is odd; also $\alpha\neq-\beta$, as this would imply that $u$ is even.