Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)?

share|improve this question
    
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

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.