Suppose that $f$ and $g$ are continuous on $[a,b]$ and for each $x$, it holds that $f(x)<g(x)$. Prove that there exists $\alpha>0$ such that for each $x$, it holds that $f(x) + \alpha <g(x)$ (Notice that $\alpha$ may not depend on $x$).

We have done a similar problem with $f(x)>0$ and proving $\alpha>0$ exists such that $f(x)>\alpha$, but I'm not exactly sure how to use supremum and infimum to solve the above version.

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    $\begingroup$ Hint: $f(x)+\alpha<g(x)$ iff $(g-h)(x)=g(x)-f(x)>\alpha$. $\endgroup$ – Julien May 3 '13 at 1:46
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    $\begingroup$ Consider the function $h(x)=g(x)-f(x)>0$. $\endgroup$ – Milind May 3 '13 at 1:47
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    $\begingroup$ Hint: The continuous function $g-f$ attains a minimum value on $[a,b]$. $\endgroup$ – David Mitra May 3 '13 at 1:47

The comments answer the question, while the answer does not (it attempts to prove a statement that the OP already knows how to prove). So, I'm going to record here that the statement about $f<g$ follows by applying the statement about $f>0$ to the function $g-f$.


It only needs to use finite covering theorem for compact interval $[a, b]$. This question ,namely :suppose $f(x)>0 ,\forall x\in [a, b]$ ,we need to proof $\exists \alpha>0 , f(x)>\alpha , \forall x\in [a, b].$\ Finite covering theorem tells us that $\exists finite\ x_i and\ \delta>0,such that\ x\in (x_i,\delta)for\ some\ x_i,x\in[a,b].$\ by continuity of $f(x)$,$$f(x)>\frac{f(x_i)}{2}>0,x\in (x_i,\delta)$$ when we select the minimum value of $\frac{f(x_i)}{2},here\ index \ (i) finite.$ thus ,we have solved this question.

  • $\begingroup$ If fact, strictly speaking,we should use continuity of $f(x)$ firstly. then use finite covering theorem for compact interval $[a, b]$.maybe having finite $\delta_i$.but that doesn't matter. $\endgroup$ – user63788 May 3 '13 at 2:37

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