Let $f$ and $g$ be two continuous functions on the interval $[0,1]$, and suppose that $g(x) < f(x)$ for all $x \in [0,1]$. Show that there exists $\beta > 1$ such that $\beta g(x) \leq f(x)$ for all $x \in [0,1]$.
My attempts: My first attempts 1
Definition of Infimum and Supremum: Let $m = \inf\{f(x) : x \in [0,1]\}$ and $M = \sup\{g(x) : x \in [0,1]\}$. Since $g(x) < f(x)$ for all $x \in [0,1]$, we have $M < m$.
Construction of Function $h$: Consider the function $h(x) = f(x) - \beta g(x)$, where $\beta > 1$. The goal is to find $\beta$ such that $h(x) \geq 0$ for all $x \in [0,1]$.
Assume by Contradiction: Assume, for the sake of contradiction, that there is no such $\beta$, meaning for all $\beta > 1$, there exists $x_0$ such that $h(x_0) = f(x_0) - \beta g(x_0) < 0$.
Exploring $h(x_0)$: For a given $x_0$, we have: $$ h(x_0) = f(x_0) - \beta g(x_0) < 0 $$ Choosing $\beta$ to Create a Contradiction: Since $M < m$, choose $\beta > 1$ such that $\beta g(x_0) > m$. This implies: $$ f(x_0) - \beta g(x_0) > m - m = 0 $$ Contradiction: This contradicts our assumption that $h(x_0) < 0$. Thus, there must exist $\beta > 1$ such that $h(x) \geq 0$ for all $x \in [0,1]$. Therefore, there exists $\beta > 1$ such that $\beta g(x) \leq f(x)$ for all $x \in [0,1]$.
My second attempt:
I would say that it is necessary to show that there exists a $\beta > 1$ such that $\beta \cdot g < f$ on the interval $[0,1]$. If this is indeed the case, then one can proceed by contradiction:
Let's assume that for every $\beta > 1$, there exists $x_{\beta}$ in $[0,1]$ such that $\beta \cdot g(x_{\beta}) \geq f(x_{\beta})$. In particular, for every $n \geq 1$, there would exist $x_n$ in $[0,1]$ such that $\left(1+\frac{1}{n}\right)g(x_n) \geq f(x_n)$ (). Since $x_n$ is a sequence with values in a compact set, we can extract a subsequence $x_{\phi(n)}$ that converges to a certain $x$ in $[0,1]$. Thus, by taking the limit in (), we arrive at $g(x) \geq f(x)$ for a certain $x$ in $[0,1]$, and this is due to the continuity of $f$ and $g$. However, this is contradictory because $f > g$. Therefore, the initial assumption must be false.
We can choose $c>0$ so that $0<g(x)+c<f(x)+c.$ The function $h(x)={f(x)+c\over g(x)+c}$ satisfies $h(x)>1.$ Thus there is $\beta>1$ such that $h(x)\ge \beta.$ Hence $$\beta [g(x)+c]\le f(x)+c$$ We get $$\beta g(x)\le f(x)+c(1-\beta)<f(x)$$