Quote:

Let $\displaystyle f:[-1,1] \rightarrow \mathbb{R}$ be continuous with $\displaystyle f(-1)=2$ and $\displaystyle f(1)=1.$

Assume f is injective but not decreasing. This implies f is either an oscillating function, strictly increasing, increasing or constant.Quote:

i). If also f is injective prove that f is decreasing.

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.

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 $\displaystyle \min \{f(x)| x \in [-1,1] \}$.Quote:

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.

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.

This is the part i'm really having trouble with (provided my previous two proofs are okay! (Worried)).Quote:

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).

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!