I'm working on the following question from a past Masters' exam in Analysis:
Prove the following:
(a) If $f:\mathbb R\to\mathbb R$ is continuous and differentiable everywhere and $f'$ has at most $n$ distinct zeros, then $f$ has at most $n+1$ distinct zeros.
(b) If $f:\mathbb R\to\mathbb R$ is defined by $f(x)=a_0+\sum_{i=1}^n a_i e^{b_i x}$ (where $a_i$ and $b_i$ are real numbers), then either $f$ is identically zero or it has at most $n$ distinct zeros.
Part (a) has already been answered here: Proving that if $f'$ has at most $n-1$ zeros, then $f$ has at most $n$ zeros. I'm trying to use induction to prove part (b) using part (a). Here's what I tried so far:
In the base case $n=1$ we have that $f(x)=a_0+a_1e^{b_1x}$ for $a_0,a_1,b_1\in\mathbb R$. Suppose $f$ is not identically zero. The only case where $f$ is nonconstant is when $a_1\neq 0$ and $b_1\neq 0$, and in this case it's easy to show $f$ is injective so that it has at most one zero.
So suppose the statement holds for $k$. I want to show it holds for $k+1$, so I let $f(x)=a_0+\sum_{i=1}^{k+1}a_ie^{b_ix}$. Then we have that $f'(x)=\sum_{i=1}^{k+1}a_ib_ie^{b_ix}$. I get stuck trying to show that $f'$ has at most $k$ zeros. I know by our assumption that $\sum_{i=1}^k a_ib_ie^{b_ix}$ has at most $k$ zeros, but I'm not sure how that helps. Any hints here?