# If $f: \Bbb R\to \Bbb R$, $g: \Bbb R\to \Bbb R$, where $f$ and $g$ are continuous and $f(x_0) >g(x_0)$?

You can take for granted that the difference of continuous functions is continuous and you can use $\epsilon$ - $\delta$ definition of continuity.

i) If $f \colon \Bbb R \rightarrow \Bbb R$, $g \colon \Bbb R \rightarrow \Bbb R$, where $f$ and $g$ are continuous and $f(x_0) >g(x_0)$ for some $x_0$ element in $\Bbb R$, then there exists a $\delta$ such that $f(x)$-$g(x)$ > 1/2 $(f(x_0)$ - $g(x_0))$ for all $x$ element in ($x_0$ - $\delta$, $x_0$ + $\delta$).

ii) use i) to show by contradiction that if $f \colon \Bbb R \rightarrow \Bbb R$, $g \colon \Bbb R \rightarrow \Bbb R$, where $f$ and $g$ are continuous, and $f(x)$=$g(x)$ a.e. then $f$=$g$.

My TA told me to not look at this problem again after I asked him because we will not be covering this topic. I was still wondering what the proof will look like though. It seems like a good proof to look at for something I might learn later on in another self study class.

OK. I understand part i) of the question. How would be prove part ii)?

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did you chose random tags? I mean real analysis ok, but where is that a bit of algebraic topology – Dominic Michaelis Apr 15 '13 at 6:19
Sorry. It was by mistake. – 9959 Apr 15 '13 at 6:23
Is it possible you can help me with this proof? – 9959 Apr 15 '13 at 6:24
Here it just says to use i) to show by contradiction that f(x)=g(x) a.e. then f=g. – 9959 Apr 15 '13 at 7:04
It is still no group theory nor measure theory – Dominic Michaelis Apr 15 '13 at 7:07

Let $h(x)=f(x)-g(x)$. We are told that $h(x_0)\gt 0$. Let $\epsilon=\frac{h(x_0)}{2}$. By the definition of continuity, there is a $\delta$ such that if $|x-x_0|\lt \delta$, then $|h(x)-h(x_0)|\lt \epsilon$.

The statement "if $|x-x_0|\lt \delta$ just says "if $x_0-\delta\lt x\lt x_0+\delta$."

The statement $|h(x)-h(x_0)|\lt \epsilon$ implies that $-\frac{h(x_0)}{2}\lt h(x)-h(x_0)\lt \frac{h(x_0)}{2}$. The left inequality is equivalent to $h(x)\gt \frac{h(x_0)}{2}$. That is exactly what we wanted to show.

Remark: If we view things geometrically, the result is obvious. Let $a=h(x_0)$. Note that $a$ is positive. Mark the point $a$ on the positive $x$-axis, fairly close to $0$. Let $b=h(x)$. We picked $\epsilon=\frac{a}{2}$. For suitable $\delta$, the distance from $b$ to $a$ is less than $\frac{a}{2}$. Put down a dot $b$ which is at distance less than $\frac{a}{2}$ from $a$. Then the distance from $b$ to $0$ is greater than $\frac{a}{2}$. That's all there is to it.

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Andre thanks for helping me figure out that part. How do we prove by contradiction that f(x)=g(x) then f=g for part ii) – 9959 Apr 15 '13 at 6:53
If they are not equal then for some $x_0$ the difference is not $0$. If the difference is positive, be happy. If it is negative, replace $f$ by $-f$ and $g$ by $-g$, or intercange the roles of $f$ and $g$. So by (i) $f(x)-g(x)\gt 0$ in some interval. But then the functions are not equal a.e.. – André Nicolas Apr 15 '13 at 7:13
I get it now. Thanks. This helps. – 9959 Apr 15 '13 at 7:23
@9959: You are welcome. Many of these results are geometric intuition made formal. – André Nicolas Apr 15 '13 at 7:31