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Let $f : [a,b] \to \mathbb{R} $ and $g :[a,b]\to \mathbb{R} $ continuous so that $\forall x \in [a,b], f(x)<g(x)$ .

I want to prove that : $\exists c>0, \forall x \in [a,b], f(x)+c<g(x). $

To do so I studied the continuous function $h=g-f$ to show that $\forall x \in [a,b], h(x)\ge h(c)$ with $h(c)>0$. However I have yet to find a way to get the strict inequality sign. Can anyone help me please? Thanks

Feng
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Pablito
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1 Answers1

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$g-f$ attains its minimum value $m$ at some point $x_0$ and $g(x_0)-f(x_0)>0$ so take $c =\frac 1 2 m$

Kavi Rama Murthy
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