# Continuity in open intervals

The problem:

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

• 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. – John Douma Dec 20 '18 at 3:51
• 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. – John Douma Dec 20 '18 at 3:54
• Do you know what continuity of $f$ at $a$ means in informal sense? If yes then the answer to your question is immediate. – Paramanand Singh Dec 20 '18 at 5:22

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 $$f(x) .
$$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$$