If $f (x)  0$ such that $f + δ ≤ g$. Question:
Prove that if $f, g: [a, b] → \mathbb{R}$ are continuous and $f (x) <g (x) $ for  every $x \in [a, b]$, then there exists $δ> 0$ such that $f + δ ≤ g$.
I have tried to take the $f-g$ function and apply something like intermediate value theorem, but couldn't conclude anything.  I've seen a solution that uses covers, but we haven't seen that yet and would like to know if there is an easier way to fix it.
 A: let $h=g-f>0$ on $[a,b]$. Since $h$ is continuous on a closed interval, it has a global minimum (extreme value theorem).
then let $\delta=\min_{x\in[a,b]}{h(x)}$
Then, by definition $\delta>0$ and $\forall x\in[a,b], f(x)+\delta\leq g(x) $
A: Suppose not. Then for every $\delta>0$ there is a point $x\in[a,b]$ where $g(x)-f(x)<\delta$. In particular, there is a sequence $x_n\in[a,b]$ with $g(x_n)-f(x_n)<1/n$.
The sequence $x_n$ doesn't necessarily have a limit, but by Bolzano-Weierstrass there is a convergent subsequence $x_{n_i}\to x\in[a,b]$. Now $g(x_{n_i})-f(x_{n_i})\to 0$, so $g(x)=f(x)$.
A: The idea is that you want to prove that there exists some constant $\delta$ such that $g(x)-f(x) \geqslant \delta >0$ for all $x \in [a,b]$. Your only hypothesis are the continuity of $f$ and $g$, with $g >f$, over the set $[a,b]$.
But there is a well-known property of continuous functions over closed bounded interval : if $h : [a,b] \to \mathbb{R}$ is continuous, then $h$ is bounded and its infimum and supremum are maximum and minimum.
Thus, if $h = g-f$, then $h$ is continuous over $[a,b]$ and has a minimum :
\begin{align}
\exists c \in [a,b],~ \forall x \in [a,b],~ h(x) \geqslant h(c)
\end{align}
Consequently,
\begin{align}
\exists c \in [a,b],~ \forall x \in [a,b],~ g(x) \geqslant h(c) + f(x)
\end{align}
By assumption, $h(c) = g(c)-f(c) >0$, and so the result follows by defining $\delta = h(c)$.
