Show that there exists some constant c>0c > 0 such that f(x)>g(x)+cf(x) > g(x) + c

Let f,g:[a,b]→Rf,g : [a,b] \rightarrow \mathbb{R} be continuous, with f(x)>g(x)f(x) > g(x) for all xx in [a,b][a,b]. Show that there exists some constant c>0c > 0 such that f(x)>g(x)+cf(x) > g(x) + c

I need to show there is a gap between the two graphs. Is the idea to show that for every point in g(x), there is always an f(x) > g(x)?

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Hint: h(x)=f(x)−g(x)>0h(x)=f(x)-g(x) \gt 0 must have a minimum on [a,b][a,b].
– dxiv
2 days ago

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2 Answers
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Since [a,b][a,b] is compact, the continuous map x↦f(x)−g(x)x\mapsto f(x)-g(x) attains a minimum. Let c=12min{f(x)−g(x):x∈[a,b]},c = \frac12\min\{f(x)-g(x):x\in[a,b]\}, then f(x)>g(x)+cf(x)>g(x)+c for all x\in[0,1]x\in[0,1] as desired.

  

 

@DaanMichiels Indeed.
– Math1000
2 days ago

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Of course you mean [a,b][a,b] instead of [0,1][0,1].
– Abdallah Hammam
2 days ago

  

 

@AbdallahHammam When I think “closed interval” it is [0,1][0,1] that comes to mind 🙂
– Math1000
2 days ago

  

 

What about my answer below ?
– Abdallah Hammam
2 days ago

Assume that

\forall c>0 \exists x\in [a,b] : f(x)\leq g(x)+c\forall c>0 \exists x\in [a,b] : f(x)\leq g(x)+c

OR ( if we take special cc)

\forall n \in \mathbb N \; \exists x_n\in [a,b]: \; f(x_n)\leq g(x_n)+\frac{1}{n+1}\forall n \in \mathbb N \; \exists x_n\in [a,b]: \; f(x_n)\leq g(x_n)+\frac{1}{n+1}

(x_n)(x_n) is a sequence of the compact [a,b][a,b].

it has a convergent subsequence (y_n)(y_n).

let LL be its limit which is always in the closed [a,b][a,b].

thus we have

\forall n \in \mathbb N f(y_n)\leq g(y_n)+\frac{1}{n+1}\forall n \in \mathbb N f(y_n)\leq g(y_n)+\frac{1}{n+1}

when nn goes to +\infty+\infty

we get by continuity,

f(L)\leq g(L)f(L)\leq g(L)

and the contradiction with the hypotheses.