Behaviour of a continuous function that is always smaller than another

Question

Let there be two continuous functions $f,g$ on $[0,1]$ such that $0<g(x)<f(x) \forall x $, why is it true that we can always find a $\lambda$ such that $\lambda g(x) \leq f(x) \forall x \in [0,1]$? More importantly, why is this fact important? Since my professor stressed this fact so many times.

Answer 1

It's because of compactness of $[0,1]$. Consider $f-g$. It is continuous and positive everywhere. Since its domain is compact, it attains its minimum value $m$. In other words, $f(x)-g(x)\geq m$ for all $x$ in $[0,1]$. You can use that to obtain $\lambda$.

Answer 2

I think you actually want $\lambda > 1$, else it's trivial if we can take $\lambda=1.$

Consider the function $h=f/g$. It is clear that $h$ is also continuous and that $h(x)>1$ for all $x\in[0,1]$. Since $[0,1]$ is a compact set, $h$ attains its minimum i.e. there is $x^* \in [0,1]$ such that $$h(x^*)=\min_{x\in[0,1]} h(x).$$

Since $h>1$ we must have $\min_{x\in[0,1]} \left(f(x)/g(x)\right) = h(x^*)>1$.

Finally, it is easy to see that $\lambda=\min_{x\in[0,1]} \left(f(x)/g(x)\right)$ satisfies the condition you want.

PS. As to why this fact is important to your professor one can only guess. What I can say is this "fact" doesn't hold if our domain is an unbounded interval or an open interval.