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The problem:

Let $f$ and $g$ be continuous functions defined on an open interval $I$. Let $a\in I$ such that $f(a)<g(a)$. Show that there exists an open interval $J\subset I$ with $a\in J$ such that $f(x)<g(x)$ for all $x\in J$

I'm honestly struggling to even conceptualize what this problem is asking me to prove let alone to how start proving it. Can anyone guide me or help me through?

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  • $\begingroup$ You are being told that $f\lt g$ for one value, i.e. $f(a)\lt g(a)$. Given that both functions are continuous show that you can find an interval around $a$ such that $f(x)\lt g(x)$ for all $x$ in the interval. $\endgroup$ – John Douma Dec 20 '18 at 3:51
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    $\begingroup$ That's the wrong question. Now that you know what is being asked you should spend some time trying to solve this problem for yourself. $\endgroup$ – John Douma Dec 20 '18 at 3:54
  • $\begingroup$ Do you know what continuity of $f$ at $a$ means in informal sense? If yes then the answer to your question is immediate. $\endgroup$ – Paramanand Singh Dec 20 '18 at 5:22
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Let $\epsilon =\frac {g(a)-f(a)} 2$. There exists $\delta >0$ such that $|x-a| <\delta$ implies $|f(x)-f(a)| <\epsilon$ and $|g(x)-g(a)| <\epsilon$. Verify that for $x \in (a-\delta, a+\delta) $ we have $f(x) <g(x)$.

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$h(x)=g(x)-f(x)$. $h$ is continuous on $I$ and $h(a)>0$. Since $h$ is continuous, in particular at $a$, we have $$\lim_{k\to 0}h(a-k)=\lim_{x\to a^-}h(x)=h(a)=\lim_{x\to a^{+}}h(x)=\lim_{k\to 0}h(a+k)$$

We see that for $k>0$, $$\quad h(a-k)< h(a)< h(a+k)\implies h(x)>0\;\text{for}\;x\in(a,a+k)\subset I$$

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