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**: $f\le \alpha g\iff (1-\alpha)f\le \alpha(g-f)$.
This is true for any $\alpha \in (0,1)$ and closed interval in which $g(x) \le 0$, because: $f(x)<g(x)\le \alpha g(x) \le 0$. Focus on the set $I$ in $[a, b]$ for which $0< f(x) < g(x)$. If $I=\emptyset$ then pick any $\alpha$. Otherwise, pick $\displaystyle \alpha = \sup\_{x \in I} \frac{f(x)}{g(x)}$. This $\alpha \ne 1$ since $f(x) < g(x)$ on $[a,b]$.