Question: I would like to prove the following simple theorem:

Let $f$ and $g$ be continuous functions on the finite interval $[a,b]$. Suppose that $f(x) < g(x)$ for all $x$ in $[a,b]$. Prove that there is an $\alpha < 1$ such that $f(x) \leq \alpha g(x)$ for all $x$ in $[a,b]$.

I can prove this theorem in the case that $g$ does not contain any zeroes in $[a, b]$ by letting $\alpha = \sup\_{x \in [a,b]}(\frac{f(x)}{g(x)})$. However, I'm not sure about the more general case. 

Thanks.

Answer: 
**Hint**: $f\le \alpha g\iff (1-\alpha)f\le \alpha(g-f)$.
You almost have it already. If $f(x)<0$ then for any $\alpha<1$ we have that $f(x) < \alpha f(x) < \alpha g(x)$. Now let $S=\{x\in [a,b]\;|\;g(x)\leq 0\}$. Since $f$ is continuous and strictly less than $g$, it must be negative on some open set $S'\supseteq S$ and the condition is trivially satisfied for all $x\in S'$. 
Then just proceed on the set $[a,b]\setminus S'$ where $g$ has no zeroes - just pick $$\alpha = \sup\_{x \in [a,b]\setminus S'}\left(\frac{f(x)}{g(x)}\right)$$.