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Re: monotonic function
Posted:
Nov 29, 2012 5:01 AM
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On 11/28/2012 09:27 AM, Timothy Murphy wrote:
> quasi wrote:
>
>> Any increasing function is monotonic.
>>
>> Any decreasing function is monotonic.
> ...
>
> Also, maybe: a continuous function f: R->R is strictly monotonic
> if and only if it is injective?
>
My deductions show that:
If a continuous function f: R -> R isn't strictly monotonic,
it might or might not be injective.
If f is not injective, we're done.
--
Then the remaining case is where f is injective:
f not strictly monotonic and yet f injective over an
interval [a, b] with a< b would give one of two graph shapes:
(f continuous)
x = a/\
\x = b
or:
/x = b
x=a\/
There's a strict maximum at some point between a and b (1st case),
and a strict minimum of f at some point between a and b (2nd case).
This is because f is continuous and [a, b] is a compact interval.
Then, by the intermediate value theorem (whether it be the 1st or
2nd case), it's easy to see that some value in the range of
f on [a, b] is attained at at least two distinct points
in [a, b]. So f is not injective on [a, b].
So f: R -> R isn't injective.
This contradicts the assumption that f is injective.
--
So, assuming f: R->R is continuous,
f not strictly monotonic ==> f not injective.
By the contrapositive, under the same assumptions on f,
f injective ===> f strictly monotonic.
On the other hand, assuming f: R->R is continuous,
if f is strictly monotonic, a<b ==> f(a) < f(b) without
loss of generality. That is seen to rule out f being
non-injective.
So on the other hand, assuming f: R->R is continuous,
f strictly monotonic ==> f is injective.
Taken together, we get:
Under the assumption that f: R -> R is continuous, then:
f is injective <==> f is strictly monotonic.
David Bernier
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