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let $f$ be a continuous function in the bounded interval $[0,1]$.

For every $x\in[0,1]$, $f(x)>x$.

Is it true that there exists a constant $c>0$ in a way that for every $x\in [0,1]$,

$f(x)>x+c$

If so, how do I prove it?

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    We have that $c=\inf_{x\in[0,1]} (f(x)-x)$. – 2017-02-28
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    @Masacroso Your construction only ensures $f(x) \geq x + c$, since the inf is attained (and is thus a min). – 2017-02-28
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    Is something missing in the question? You have specified the domain, but not the codomain. For a general real-valued continuous function this statement is false. Take, for example, the function $f(x) = -x$ – 2017-02-28

2 Answers 2

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$g(x) = f(x) - x$ is also a continuous function on $[0,1]$. By the extreme value theorem, a continuous function on $[0,1]$ is bounded and attains its bounds. Thus there exists a $m$ such that $g(x) \geq m$ for all $x \in [0,1]$, and moreover, for some $a \in [0,1]$, $g(a) = m$. But we know $g(x) > 0$ for all $x$, so in particular, $g(a) = m > 0$. Now define $c = m/2$ and note that $f(x) - x \geq m > c > 0$.

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The function $g(x)=f(x)-x$ is defined on the compact interval $[0,1]$ and is continue, there exists $x_0\in [0,1]$ such that $g(x_0)=inf_{x\in [0,1]}g(x)>0$, write $f(x_0)-x_0=d$. Take $c=d/2$. You have $g(x)=f(x)-x\geq f(x_0)-x_0=d>c$ .

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    What if $f(x_0) - x_0 < 0$? – 2017-02-28
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    read the hypothesis – 2017-02-28
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    Whoops, never mind. – 2017-02-28