Let $f,g$ be analytic functions on $(0,1)$. Show that they are linearly dependent over $\mathbb R$ if and only if $$W:=\bigg(\matrix {f & g\\ f' & g'}\bigg)$$ is singular everywhere in $(0,1)$.
Let now $C$ be the curve in $\mathbb R^3$ parametrized by $(x,f(x),g(x)$, $x\in(0,1)$. Consider the lines through two distinct points of $C$, corresponding to $x_1\neq x_2\in(0,1)$. Show that if the union on these lines doesn't include open sets of $\mathbb R^3$, then $C$ lies in a plane, and conversely.
The first point is easy, I think: for the not obvious implication, one writes the identity $fg'=gf'$ as the product of series and obtains $$\sum_{h+k=n}f_h(k+1)g_{k+1}=\sum_{h+k=n}g_h(k+1)f_{k+1}$$ for every $n$ (where $f=\sum f_n x^n$ and $g=g+n x^n$). From that one should obtain many identities which, recursively, give the linear dependance, but I'm not sure whether this is the simplest way. Do you have any suggestions?
For the second point, there is an easy implication (if $C$ lies in a plane, so do all the lines). But for the converse? The point should be that if the union of all lines doesn't lie in a plane, one gets a contradiction since fo any two distinct lines there exists two distinct points where they meet $C$ respectively, and these points must be connected by another line: and this idea should lead to 'fill' the union of the lines so that it happens to include an open set of $\mathbb R^3$.
But how to so say this correctly and completely? And are there simpler solutions?
Thank you in advance.
For the first question: Suppose $f'g-fg'\equiv 0.$ If $g\equiv 0,$ then $f,g$ are linearly dependent. If $g\not \equiv 0,$ then $g$ is nonzero in some open subinterval $I$ of $(0,1)$ of positive length. On $I$, by the quotient rule, we see $(f/g)' \equiv 0.$ Thus on $I,$ $f/g \equiv c$ for some constant $c,$ which implies $f=cg$ on $I.$ By the idenitity principle, $f=cg$ on $(0,1),$ which implies $f,g$ are linearly dependent.