Javascript required
Skip to content Skip to sidebar Skip to footer
Home / Prove That a Continuous Function on an Interval is Injective if and Only if It is Strictly Monotone

Prove That a Continuous Function on an Interval is Injective if and Only if It is Strictly Monotone

If a function is continuous and injective on a closed interval is it always strictly monotonic

To prove it you note that if the function is not strictly monotone you have a set of arguments where the value "turns"(*), that is $a<b<c$, but $f(b)$ is strictly greater (or less) than both $f(a)$ and $f(c)$.

Now select $y$ such that $y<f(b)$ but larger than both $f(a)$ and $f(c)$. By the intermediate value theorem you have that $f$ takes the value $y$ both somewhere in $(a,b)$ and $(b,c)$ which contradicts the assumption that $f$ is injective.

Note that the fact that the interval is closed was never used.

Of course the opposite is true: if a function is strictly monotone it's injective - which means that for continuous functions on an interval strict monotonicity and injectivity is equivalent.

(*) This is "obvious", but as often these can be proven (with some effort). You assume that the function is non-monotone which means that there is $b>a$ with $f(b)>f(a)$ and $d>c$ with $f(d)<f(c)$. Now it's basically to consider the possibilities of the relation between these arguments and the corresponding values - it's a couple of cases one need to cover.

For example if $b>d>c>a$ we have that $0<f(b)-f(a) = (f(b)-f(d)) + (f(d)-f(c)) + (f(c)-f(a))$, but as $f(d)-f(c)$ is negative at least one of the other terms must be positive. Which means that we have the claimed "turn" (either $b$-$d$-$c$ or $d$-$c$-$a$)

Related videos on Youtube

Comments

  • (B)I note that if a function is continuous and injective, then very often the function is monotonic, and thus strictly monotonic, this is nearly always true?

    That is long as the domain and range are connected, and/or are real valued, closed and bound intervals, that this is correct, and that the counter-examples often pertain, to disjoint intervals and sets,such as: $$[0,3)\cup{4}\to R$$.

    Are these counter-examples only related to non-connected degenerate cases such as these?. I presume, they don not apply to apply when the domain and range are singular continuous, closed and bounded real intervals, or connected sets, such as $$[0,2]\to[0,3]$$?

    • The intermediate value theorem can be used to prove that if $f:[a,b] \rightarrow \mathbb{R}$ is injective and continuous, then it is monotone. To go about proving this, try to prove the contrapositive.

    • Thanks, I was just trying to be careful as a few counter-examples were mentioned in some of the other questions of this nature. I presume these depend on the domain being dis-connected (not a closed and bonded real interval) which you have specified to be case. Thanks for this again

  • Do you know if such functions; that is strictly monotonic increasing functions are always Strictly quasi convex and strictly quasi concave. Or is this only when continuous. I presume then that a strictly monotonic increasing function can be as a biconditional: x>y iff F(x)>F(y), x<y iff F(x)<F(y);x=y iff F(x)=F(y). As an order imbedding. It seems obvious but I want to be sure

  • If you mean the "turn" then it's guaranteed regardless of the continuity of $f$, only the non-monotonicity is needed for this to happen.

  • yeah I mean x<=y implies F(x)<=F(y) and F(x)<=F(y) implies x<=y

  • generally one requires strict monotonicity for this double inference

  • yeah I mean(A) $[x\leq y]$ implies $[F(x)\leqF(y)]$ and(B)$ [F(x)\leqF(y) ]$implies $[x\leq y]$; Functions that are monotonic increasing,but not strictly monotonic increasing may not satisfy latter (B), A constant function for example, where if $x>y$, then $x \geq y$, and so$ F(x) \geq F(y)$ and $F(x)=F(y)$ would satisfy this. But then $[F(x)=F(y)] \rightarrow F(x) \leq F(y) $ is satisfied, buts not the case that $x \leq y$ as $x>y,$

  • I presume that 'injectivity plus monotone increasing' generally implies strict monotone increasing? Simply in virtue of the fact that when $x>y$ $x \neq y$, $x \geq y$ and hence by monotonicity $F(x) \geq F(y)$ but by injectivity $F(x)\neq F(y)$ as $x\neq y$, so $F(x)>F(y)$,

  • But yes I agree with your argument. I was just now inquiring about the fact that strict monotonicity allows you to invert the inference (but not monotonic increasing alone); if the function is continuous and injective one canbut that because its both' injective and monotonic' and thus strictly monotonic). And if an injective continuous function is monotonic, and injectivity given monotonicity entails strict monotonicity (which seems relatively obvious) then its strictly monotonic

Recents

What is the matrix and directed graph corresponding to the relation $\{(1, 1), (2, 2), (3, 3), (4, 4), (4, 3), (4, 1), (3, 2), (3, 1)\}$?

Related

skirvingaboughtters.blogspot.com

Source: https://9to5science.com/if-a-function-is-continuous-and-injective-on-a-closed-interval-is-it-always-strictly-monotonic