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**: $f\le \alpha g\iff (1-\alpha)f\le \alpha(g-f)$.
1. Prove, using continuity, that for each $x\in [a,b]$ there is a number $h\_x<1$ and a nbhd $O(x,h\_x)$ of $x$
such that $f(y)\le h\_x g(y)$ for all $y\in O(x,h\_x)$.
2. By compactness of $[a,b]$, finitely many of these nbhd's cover $[a,b]$.
3. Now just let your $\alpha$ be the largest of these finitely many $h\_x$'s.