I'm asked to prove the following: Let $f, g$ be continuous real-valued functions on a compact metric space $M .$ Suppose that $f(x)>g(x)>0$ for all $x \in M .$ Prove that there exists $\lambda>1$ such that $f(x)>\lambda g(x)$ for all $x \in M$.
Intuitively, I think the statement is false. For example, $f(x)=g(x)+1$. Is there something wrong with my understanding? Or are there any mistakes in the problem?
On
Let say that $f,g$ are defined on the compact $K$. As $g$ is supposed to be continuous and not vanishing, it exists $a>0$ such that $g(x)\ge a>0$ for $x \in K$: a continuous function defined on a compact space attains its extreme values.
Then $f/g$ is properly defined on $K$ and by hypothesis, $f/g >1$ on $K$. $f/g$ is continuous as it is the division of two continuous maps, the denominator not vanishing. Therefore $f/g$ attains its minimum $\lambda$ on the compact $K$.
Which means $f(x)/g(x) \ge \lambda >1$ for $x \in K$. We’re done.
Take $\lambda =\frac 1 2 min_x \frac {f(x)} {g(x)}$.
Recall that any continuous positive function on a compact space has a positive minimum.