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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.


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When both are negative, your $\alpha>1$. You must focus on case when both are positive like I did in my answer. But you had the right idea :) – genepeer Apr 27 '13 at 21:36
up vote 1 down vote accepted

Suppose by contradiction that $$\forall \alpha<1\quad \exists x_\alpha\in[a,b]\quad|\quad f(x_\alpha)>\alpha g(x_\alpha)$$ so if we take $\alpha=1-\frac{1}{n}$ we have $$\forall n\in\mathbb{N},\quad\exists x_n\in[a,b]\quad|\quad g(x_n)>f(x_n)>\left(1-\frac{1}{n}\right)g(x_n)\tag{1}$$

and since $[a,b]$ is compact there's a subsequence $(x_{n_k})$ convergent to say $\ell\in[a,b]$ then we pass to limit in $(1)$ and by continuity we find $$g(\ell)\geq f(\ell)\geq g(\ell)$$ hence $$f(\ell)=g(\ell)$$ which is absurd.

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That is kinda neat. – genepeer Apr 27 '13 at 21:40

Hint: $f\le \alpha g\iff (1-\alpha)f\le \alpha(g-f)$.

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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 $[a,b]$.

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  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.
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You almost have it already. If $f(x)<0$ then for any $\alpha<1$ we have that $f(x) < \alpha f(x) < \alpha g(x)$. Now let $S=\{x\in [a,b]\;|\;g(x)\leq 0\}$. Since $f$ is continuous and strictly less than $g$, it must be negative on some open set $S'\supseteq S$ and the condition is trivially satisfied for all $x\in S'$. Then just proceed on the set $[a,b]\setminus S'$ where $g$ has no zeroes - just pick $$\alpha = \sup_{x \in [a,b]\setminus S'}\left(\frac{f(x)}{g(x)}\right)$$.

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