# Continuity and differentiability answer check.

• May 9th 2009, 03:27 AM
Showcase_22
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Let $\displaystyle f:[-1,1] \rightarrow \mathbb{R}$ be continuous with $\displaystyle f(-1)=2$ and $\displaystyle f(1)=1.$
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i). If also f is injective prove that f is decreasing.
Assume f is injective but not decreasing. This implies f is either an oscillating function, strictly increasing, increasing or constant.

f cannot be an oscillating function because if f was an oscillating function over $\displaystyle [-1,1]$ then if $\displaystyle f(a)=b$ where $\displaystyle a \in [-1,1]$ and $\displaystyle b \in Im(f)$ then by the definition of an oscillating function $\displaystyle \exists c \in [-1,1]$ such that $\displaystyle f(a+c)=b$. This would contradict the fact that f is injective.

So we now know that f is either strictly increasing, increasing or constant.

Once again, f cannot be a constant function. If f was a constant function (similar to before) if $\displaystyle f(a)=b$ then $\displaystyle f(a+ \delta)=b$ $\displaystyle \forall \ \delta>0$.

f is either a strictly increasing or an increasing function.

If f is strictly increasing then if $\displaystyle [a,b] \subset [-1,1]$ then $\displaystyle f(b)>f(a)$.

This gives a contradiction since $\displaystyle f(1)<f(-1)$ so f cannot be strictly increasing.

A similar argument holds for an increasing function:
If f is increasing then if $\displaystyle [a,b] \subset [-1,1]$ then $\displaystyle f(b) \geq f(a)$.
This contradicts the definition of injectivity since it is possible for $\displaystyle f(a)=f(b)$.

Hence we are forced to conclude that f is a decreasing function.

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ii) if f takes only positive values show that there exists a positive real number $\displaystyle \delta$ with $\displaystyle f(x)> \delta \ \forall x \in[-1,1]$, carefully stating any general resullts you use.
If f only takes positive values then $\displaystyle f(x) \neq 0$. Since f is also continous, f must be bounded since it's domain is bounded (general result). f is bounded above by $\displaystyle \max \{ f(x)|x \in[-1,1] \}$ and below by $\displaystyle \min \{f(x)| x \in [-1,1] \}$.

Take $\displaystyle \delta= \min \{ f(x)| x \in[-1,1] \}- \epsilon$ for some $\displaystyle \epsilon>0$.

This gives $\displaystyle f(x) \geq \min \{f(x)| x \in[-1,1] \}> \min \{f(x)| x \in[-1,1] \}- \epsilon= \delta$.

Hence $\displaystyle f(x)> \delta$ as required.

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b). Suppose $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}$ is increasing and satisfies the conclusion of the intermediate value theorem. Prove that f is left continuous (ie. prove that $\displaystyle \forall \ c \in \mathbb{R}$ the restriction of f to $\displaystyle (- \infty, c]$ is continuous at c).
This is the part i'm really having trouble with (provided my previous two proofs are okay! (Worried)).

I need to find $\displaystyle |f(x)-f(c)|< \epsilon$ when $\displaystyle |x-c|< \delta$ where $\displaystyle x <c$.

Let $\displaystyle x=c-k \Rightarrow |c-k-c|=|k|<\delta \ \forall k>0$.

We now need to find $\displaystyle |f(c-k)-f(c)|< \epsilon$ when $\displaystyle |k|< \delta.$

...and this is where I get stuck (Crying)

Can someone please tell me whether my two above proofs are fine and show me how to do this last part?

I think there's a way of doing parts i) and ii) using the intermediate value theorem, but I couldn't see that way of doing it.

Any help is appreciated. Thankyou!
• May 9th 2009, 08:21 AM
running-gag
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Originally Posted by Showcase_22
If f only takes positive values then $\displaystyle f(x) \neq 0$. Since f is also continous, f must be bounded since it's domain is bounded (general result). f is bounded above by $\displaystyle \max \{ f(x)|x \in[-1,1] \}$ and below by $\displaystyle \min \{f(x)| x \in [-1,1] \}$.

Take $\displaystyle \delta= \min \{ f(x)| x \in[-1,1] \}- \epsilon$ for some $\displaystyle \epsilon>0$.

This gives $\displaystyle f(x) \geq \min \{f(x)| x \in[-1,1] \}> \min \{f(x)| x \in[-1,1] \}- \epsilon= \delta$.

Hence $\displaystyle f(x)> \delta$ as required.

Hum ...

It is important to notice that $\displaystyle \min \{f(x)| x \in [-1,1] \} > 0$

Of course due to the fact that f is positive $\displaystyle \min \{f(x)| x \in [-1,1] \} \geq 0$

And $\displaystyle \min \{f(x)| x \in [-1,1] \} \neq 0$ because if it was 0 then there would exist $\displaystyle c \in [-1,1] \mbox{ such that} f(c) = 0$ (this is due to the fact that [-1,1] is a closed interval. If you take g(x) = x over )0,1], you have $\displaystyle \forall x\in )0,1]~g(x) > 0$ but g has no minimum, only a lower bound which is 0)

Take $\displaystyle \delta = \frac{\min \{f(x)| x \in [-1,1] \}}{2}$

$\displaystyle \forall x\in[-1,1]~f(x) \geq \min \{f(x)| x \in [-1,1] \} > \delta > 0$

A drawing is better than a long discussion ...

http://nsa07.casimages.com/img/2009/...6283675757.jpg
• May 10th 2009, 01:02 AM
Showcase_22
I understand what you're saying, and the picture was super!

I also think that your $\displaystyle \delta$ was much better than mine!(Cool)

Do you think my proof for the injectivity was fine and do you have any ideas for the last part? I looked at it again today and it was still highly confusing! (Worried)