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Let $f,g:[a,b] \to \mathbb{R}$ be two continuous functions, with $f(x)>g(x), \forall x \in [a,b]$. Show that $\exists e > 0, $such that $f(x)>g(x)+e$.

My try is:Let $h(x) = f(x)-g(x), h:[a,b] \to \mathbb{R}$. Then $h$ is continuous in $[a,b]$, and thus continuous $\forall x_o \in [a,b]$. That means that $\forall \epsilon > 0, \exists \delta >0,: |h(x)-h(x_0)|<\epsilon, \forall x \in (x_o - \delta, x_o + \delta)$. Thus:

$$|h(x)| \leq |h(x_0)| + \epsilon$$As $|h(x_0)|> 0 $ and $\epsilon > 0 $, $|h(x_0)|+\epsilon > 0 = e$ and it's proved. Is this correct or am I missing something?

average_discrete_math_enjoyer
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  • Where does your argument use the fact that $[a,b]$ is closed and bounded (i.e., compact). For $(a,b)$ or $[a,\infty)$ etc. the claim is false. – Magdiragdag Apr 15 '21 at 17:19
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    Hint: use that $h(x)$ attains its minimum on $[a,b]$. – Robearz Apr 15 '21 at 17:22
  • Both functions are set in the closed interval $[a,b]$, and since they are continuous they are also bounded with $m \leq f(x) \leq M$ (same goes for g and h). Do you mind elaborating on what you thing I am missing? – average_discrete_math_enjoyer Apr 15 '21 at 17:22
  • Think this. $f$ reachs a minimum and $g$ reachs a maximum, which $e$ could work? (Spoiler: The Average should do) – Marcos Martínez Wagner Apr 15 '21 at 17:23
  • I mean: you argument must be wrong, because you are not using the fact that $[a,b]$ is closed and bounded. So, for example, reread your argument for $(a,b)$, take functions $f$ and $g$ for which your claim is wrong (for $(a,b)$) and see for yourself where it goes wrong. – Magdiragdag Apr 15 '21 at 17:24
  • Be carefull, the way you took the epsilon depends on the $x_0$ you take – Marcos Martínez Wagner Apr 15 '21 at 17:25
  • You argument is faulty in that that is epsilon for that PARTICULAR $x_0$. Doing the analysis at a different point $x_1$ one may get a different $\epsilon_1$. We need to show there is an $\epsilon_{ultimathule}$ so that for any $x$ at all that $f(x) > g(x) + \epsilon_{ultimathule}$. And you will need to us $[a,b]$ is a closed interval. It's not true for an open interval or an unbounded interval. – fleablood Apr 15 '21 at 17:25
  • Another way of looking at your work: what $\epsilon$ did you actually take? – Magdiragdag Apr 15 '21 at 17:28
  • @fleablood Correct, my thinking was to take $\epsilon$ depended on $x_0$, but if this holds for all $x_0 \in [a,b]$, then there will always be a $e$ I can take, and thus it's true. I understand it doesn't hold for closed or unbounded intervals, but I am not asked for those. – average_discrete_math_enjoyer Apr 15 '21 at 17:29
  • Consider $f:(0, 1): \mathbb R$ so $f(x) = x$ and $g:(0,1)$ so that $g= x^2$. for all $x\in (0,1)$ we have $f(x) > g(x)$. Yet there is no $\epsilon$ so that $f(x) > g(x) + \epsilon$ as $\lim_{x\to 0, 1}f(x) = g(x)$ by definition says the exact opposite. – fleablood Apr 15 '21 at 17:31
  • You need the same epsilon to work for every $x_0$ you take. I dont know if you are familiar with the property of compact spaces "Every Open conver has a finite Sub-cover". Maybe, using the compactness of $[a,b]$ will let you partition the interval in finite intervals (with finite epsilons) and just take the minimum epsilon of those – Marcos Martínez Wagner Apr 15 '21 at 17:31
  • If you consider that $\epsilon_w$ is the epsilon you can take for $x=w$ you must choose an $\epsilon_\omega$ so that $\epsilon_\omega \le \epsilon_w$ for all $\epsilon_w$. That's fine. $\epsilon_\omega = \min \epsilon_w$ will do but ... what if $\epsilon_w$ has not minimum. Okay, we use $\epsilon_\omega = \inf \epsilon_w$. That's cool as $\epsilon_w$ are bounded below by $0$ but... what if $\inf \epsilon_w = 0$. Then what do we use? – fleablood Apr 15 '21 at 17:36
  • Echoing @Robearz, generally look for the most powerful applicable tools in your arsenal. Here we have continuous functions on a compact interval, so the powerful tool in question is the Extreme Value Theorem. The result follows directly – Alan Apr 15 '21 at 17:37
  • "Correct, my thinking was to take ϵ depended on x0, but if this holds for all x0∈[a,b], then there will always be a e I can take, and thus it's true" ABSOLUTELY NOT!!!! That would imply that every set bounded below has a minimum. We know that is not true! – fleablood Apr 15 '21 at 17:37
  • @fleablood Yes you are right, I completely wiffed that. As pointed out I will use the extreme value theorem to continue. Thanks =) – average_discrete_math_enjoyer Apr 15 '21 at 17:41
  • That is akin to saying to "there will always be a e I can take" is akin to saying "We can always find a least positive number". Let $x_0 > 0$ then we can take $y =\frac 12 x_0$ and $0 < y$. So we can always a $y$ for every $x_0$ so there is a $y$ that we that is smaller than all $x$. – fleablood Apr 15 '21 at 17:41

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