Show that if f is differentiable and f'(x) ≥ 0 on (a, b), then f is strictly increasing
$begingroup$
Show that if f is differentiable and f'(x) ≥ $0$ on (a, b), then f is strictly increasing provided there is no sub interval (c, d) with с < d on which f' is identically zero.
So so far I'm trying to do this by contradiction:
Suppose not, that is suppose we have function $f$ where f(x)$geq$$0$ on (a,b) where f ' is not identically $0$ for a sub interval of (a,b) and f is not strictly increasing. Since f is not strictly increasing this implies there exists $x_1$ and $x_2$ where a <$x_1$ < $x_2$ < b and f($x_1$) = f($x_2$). Then for all y $in$ [$x_1$, $x_2$], f(y) = f($x_2$) which means that f is constant and f '(y) = 0.
Since f'(y)=$0$ for all y $in$ [$x_1$, $x_2$] this means f ' is identically $0$ which is a contradiction. Thus f is strictly increasing. $square$
I'm not sure if there is a better way to do this but any help or comments would be appreciated!
real-analysis
asked Dec 3 '18 at 10:54
NolandoNolando
111
$endgroup$
add a comment |
$begingroup$
Show that if f is differentiable and f'(x) ≥ $0$ on (a, b), then f is strictly increasing provided there is no sub interval (c, d) with с < d on which f' is identically zero.
So so far I'm trying to do this by contradiction:
Suppose not, that is suppose we have function $f$ where f(x)$geq$$0$ on (a,b) where f ' is not identically $0$ for a sub interval of (a,b) and f is not strictly increasing. Since f is not strictly increasing this implies there exists $x_1$ and $x_2$ where a <$x_1$ < $x_2$ < b and f($x_1$) = f($x_2$). Then for all y $in$ [$x_1$, $x_2$], f(y) = f($x_2$) which means that f is constant and f '(y) = 0.
Since f'(y)=$0$ for all y $in$ [$x_1$, $x_2$] this means f ' is identically $0$ which is a contradiction. Thus f is strictly increasing. $square$
I'm not sure if there is a better way to do this but any help or comments would be appreciated!
real-analysis
asked Dec 3 '18 at 10:54
NolandoNolando
111
$endgroup$
$begingroup$
I'm not sure how you conclude that $f$ is constant on a sub interval. That doesn't follow from the fact that $f$ is not strictly increasing (it may as well be decreasing or not monotonic at all).
$endgroup$
– freakish
Dec 3 '18 at 11:00
add a comment |
$begingroup$
Show that if f is differentiable and f'(x) ≥ $0$ on (a, b), then f is strictly increasing provided there is no sub interval (c, d) with с < d on which f' is identically zero.
So so far I'm trying to do this by contradiction:
Suppose not, that is suppose we have function $f$ where f(x)$geq$$0$ on (a,b) where f ' is not identically $0$ for a sub interval of (a,b) and f is not strictly increasing. Since f is not strictly increasing this implies there exists $x_1$ and $x_2$ where a <$x_1$ < $x_2$ < b and f($x_1$) = f($x_2$). Then for all y $in$ [$x_1$, $x_2$], f(y) = f($x_2$) which means that f is constant and f '(y) = 0.
Since f'(y)=$0$ for all y $in$ [$x_1$, $x_2$] this means f ' is identically $0$ which is a contradiction. Thus f is strictly increasing. $square$
I'm not sure if there is a better way to do this but any help or comments would be appreciated!
real-analysis
asked Dec 3 '18 at 10:54
NolandoNolando
111
$endgroup$
Show that if f is differentiable and f'(x) ≥ $0$ on (a, b), then f is strictly increasing provided there is no sub interval (c, d) with с < d on which f' is identically zero.
So so far I'm trying to do this by contradiction:
Suppose not, that is suppose we have function $f$ where f(x)$geq$$0$ on (a,b) where f ' is not identically $0$ for a sub interval of (a,b) and f is not strictly increasing. Since f is not strictly increasing this implies there exists $x_1$ and $x_2$ where a <$x_1$ < $x_2$ < b and f($x_1$) = f($x_2$). Then for all y $in$ [$x_1$, $x_2$], f(y) = f($x_2$) which means that f is constant and f '(y) = 0.
Since f'(y)=$0$ for all y $in$ [$x_1$, $x_2$] this means f ' is identically $0$ which is a contradiction. Thus f is strictly increasing. $square$
I'm not sure if there is a better way to do this but any help or comments would be appreciated!
real-analysis
real-analysis
asked Dec 3 '18 at 10:54
NolandoNolando
111
asked Dec 3 '18 at 10:54
NolandoNolando
111
asked Dec 3 '18 at 10:54
NolandoNolando
111
asked Dec 3 '18 at 10:54
NolandoNolando
111
asked Dec 3 '18 at 10:54
NolandoNolando
111
111
$begingroup$
I'm not sure how you conclude that $f$ is constant on a sub interval. That doesn't follow from the fact that $f$ is not strictly increasing (it may as well be decreasing or not monotonic at all).
$endgroup$
– freakish
Dec 3 '18 at 11:00
add a comment |
$begingroup$
I'm not sure how you conclude that $f$ is constant on a sub interval. That doesn't follow from the fact that $f$ is not strictly increasing (it may as well be decreasing or not monotonic at all).
$endgroup$
– freakish
Dec 3 '18 at 11:00
$begingroup$
I'm not sure how you conclude that $f$ is constant on a sub interval. That doesn't follow from the fact that $f$ is not strictly increasing (it may as well be decreasing or not monotonic at all).
$endgroup$
– freakish
Dec 3 '18 at 11:00
$begingroup$
I'm not sure how you conclude that $f$ is constant on a sub interval. That doesn't follow from the fact that $f$ is not strictly increasing (it may as well be decreasing or not monotonic at all).
$endgroup$
– freakish
Dec 3 '18 at 11:00
add a comment |
2 Answers
2
active
oldest
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$begingroup$
Your idea of going by contradiction is a good one, but it is not performed all that well. In particular, your argument that there exist $x_1<x_2$ such that $f(x_1)=f(x_2)$ is ok, but it is not clear how to conclude from that the fact that $f$ is constant of $[x_1,x_2]$.
My advice is to either use Rolle's theorem on $x_1,x_2$, or to restart your proof, write exactly what "not strictly increasing" means, and go with the classics and employ Lagrange's mean value theorem.
answered Dec 3 '18 at 10:58
5xum5xum
90.2k393161
$endgroup$
$begingroup$
I was wondering how I would define not strictly increasing, would I negate the formal definition or find the contrapositive of the definition? Or is there another way?
$endgroup$
– Nolando
Dec 3 '18 at 12:02
$begingroup$
@Nolando That's exactly it. A function is strictly increasing if for all $x_1,x_2$ such that $x_1<x_2$, we have $f(x_1)<f(x_2)$. Symbolically, this can be written as $$forall x_1, x_2: x_1<x_2implies f(x_1)<f(x_2)$$ By definition, a function is not strictly increasing if the negation of that holds, i.e. $$exists x_1,x_2:x_1<x_2land f(x_1)geq f(x_2)$$ meaning there exists some pair $x_1,x_2$ such that $x_1<x_2$ and $f(x_1)geq f(x_2)$.
$endgroup$
– 5xum
Dec 3 '18 at 12:41
$begingroup$
Thanks! I was thinking wondering if i should go along the lines of $dfrac{f(x2)-f(x1)}{x2-x1}$ $geq$ 0 since f '(x) $geq$ 0 then f(x2)-f(x1) $geq$ 0 but from the fact f is not strictly increasing and f(x2) -f(x1) $leq$ 0. So f(x2)-f(x1) = 0. I'm not sure if this is a good way to go about it either.
$endgroup$
– Nolando
Dec 3 '18 at 13:27
add a comment |
$begingroup$
To prove that by contradiction, firstly we can prove that $f'(x)>0 implies f(x)$ strictly increasing.
Suppose indeed that exist $x_1<x_2$ such that $f(x_1)>f(x_2)$ then by MVT
$$f(x_2)-f(x_1)=(x_2-x_1)f'(c)>0$$
which is a contradiction.
Form here we can now extend the result to $f'(x)ge 0$, with $f'(x)=0$ only for a set of isolated points, $implies f(x)$ strictly increasing, indeed assuming again by contradiction that $f(x)$ is not strictly increasing then it would exist an interval with $f'(x)=0$ which is against the hypotesis.
answered Dec 3 '18 at 11:24
gimusigimusi
92.9k94494
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Your idea of going by contradiction is a good one, but it is not performed all that well. In particular, your argument that there exist $x_1<x_2$ such that $f(x_1)=f(x_2)$ is ok, but it is not clear how to conclude from that the fact that $f$ is constant of $[x_1,x_2]$.
My advice is to either use Rolle's theorem on $x_1,x_2$, or to restart your proof, write exactly what "not strictly increasing" means, and go with the classics and employ Lagrange's mean value theorem.
answered Dec 3 '18 at 10:58
5xum5xum
90.2k393161
$endgroup$
$begingroup$
I was wondering how I would define not strictly increasing, would I negate the formal definition or find the contrapositive of the definition? Or is there another way?
$endgroup$
– Nolando
Dec 3 '18 at 12:02
$begingroup$
@Nolando That's exactly it. A function is strictly increasing if for all $x_1,x_2$ such that $x_1<x_2$, we have $f(x_1)<f(x_2)$. Symbolically, this can be written as $$forall x_1, x_2: x_1<x_2implies f(x_1)<f(x_2)$$ By definition, a function is not strictly increasing if the negation of that holds, i.e. $$exists x_1,x_2:x_1<x_2land f(x_1)geq f(x_2)$$ meaning there exists some pair $x_1,x_2$ such that $x_1<x_2$ and $f(x_1)geq f(x_2)$.
$endgroup$
– 5xum
Dec 3 '18 at 12:41
$begingroup$
Thanks! I was thinking wondering if i should go along the lines of $dfrac{f(x2)-f(x1)}{x2-x1}$ $geq$ 0 since f '(x) $geq$ 0 then f(x2)-f(x1) $geq$ 0 but from the fact f is not strictly increasing and f(x2) -f(x1) $leq$ 0. So f(x2)-f(x1) = 0. I'm not sure if this is a good way to go about it either.
$endgroup$
– Nolando
Dec 3 '18 at 13:27
add a comment |
$begingroup$
Your idea of going by contradiction is a good one, but it is not performed all that well. In particular, your argument that there exist $x_1<x_2$ such that $f(x_1)=f(x_2)$ is ok, but it is not clear how to conclude from that the fact that $f$ is constant of $[x_1,x_2]$.
My advice is to either use Rolle's theorem on $x_1,x_2$, or to restart your proof, write exactly what "not strictly increasing" means, and go with the classics and employ Lagrange's mean value theorem.
answered Dec 3 '18 at 10:58
5xum5xum
90.2k393161
$endgroup$
$begingroup$
I was wondering how I would define not strictly increasing, would I negate the formal definition or find the contrapositive of the definition? Or is there another way?
$endgroup$
– Nolando
Dec 3 '18 at 12:02
$begingroup$
@Nolando That's exactly it. A function is strictly increasing if for all $x_1,x_2$ such that $x_1<x_2$, we have $f(x_1)<f(x_2)$. Symbolically, this can be written as $$forall x_1, x_2: x_1<x_2implies f(x_1)<f(x_2)$$ By definition, a function is not strictly increasing if the negation of that holds, i.e. $$exists x_1,x_2:x_1<x_2land f(x_1)geq f(x_2)$$ meaning there exists some pair $x_1,x_2$ such that $x_1<x_2$ and $f(x_1)geq f(x_2)$.
$endgroup$
– 5xum
Dec 3 '18 at 12:41
$begingroup$
Thanks! I was thinking wondering if i should go along the lines of $dfrac{f(x2)-f(x1)}{x2-x1}$ $geq$ 0 since f '(x) $geq$ 0 then f(x2)-f(x1) $geq$ 0 but from the fact f is not strictly increasing and f(x2) -f(x1) $leq$ 0. So f(x2)-f(x1) = 0. I'm not sure if this is a good way to go about it either.
$endgroup$
– Nolando
Dec 3 '18 at 13:27
add a comment |
$begingroup$
Your idea of going by contradiction is a good one, but it is not performed all that well. In particular, your argument that there exist $x_1<x_2$ such that $f(x_1)=f(x_2)$ is ok, but it is not clear how to conclude from that the fact that $f$ is constant of $[x_1,x_2]$.
My advice is to either use Rolle's theorem on $x_1,x_2$, or to restart your proof, write exactly what "not strictly increasing" means, and go with the classics and employ Lagrange's mean value theorem.
answered Dec 3 '18 at 10:58
5xum5xum
90.2k393161
$endgroup$
Your idea of going by contradiction is a good one, but it is not performed all that well. In particular, your argument that there exist $x_1<x_2$ such that $f(x_1)=f(x_2)$ is ok, but it is not clear how to conclude from that the fact that $f$ is constant of $[x_1,x_2]$.
My advice is to either use Rolle's theorem on $x_1,x_2$, or to restart your proof, write exactly what "not strictly increasing" means, and go with the classics and employ Lagrange's mean value theorem.
answered Dec 3 '18 at 10:58
5xum5xum
90.2k393161
answered Dec 3 '18 at 10:58
5xum5xum
90.2k393161
answered Dec 3 '18 at 10:58
5xum5xum
90.2k393161
answered Dec 3 '18 at 10:58
5xum5xum
90.2k393161
90.2k393161
$begingroup$
I was wondering how I would define not strictly increasing, would I negate the formal definition or find the contrapositive of the definition? Or is there another way?
$endgroup$
– Nolando
Dec 3 '18 at 12:02
$begingroup$
@Nolando That's exactly it. A function is strictly increasing if for all $x_1,x_2$ such that $x_1<x_2$, we have $f(x_1)<f(x_2)$. Symbolically, this can be written as $$forall x_1, x_2: x_1<x_2implies f(x_1)<f(x_2)$$ By definition, a function is not strictly increasing if the negation of that holds, i.e. $$exists x_1,x_2:x_1<x_2land f(x_1)geq f(x_2)$$ meaning there exists some pair $x_1,x_2$ such that $x_1<x_2$ and $f(x_1)geq f(x_2)$.
$endgroup$
– 5xum
Dec 3 '18 at 12:41
$begingroup$
Thanks! I was thinking wondering if i should go along the lines of $dfrac{f(x2)-f(x1)}{x2-x1}$ $geq$ 0 since f '(x) $geq$ 0 then f(x2)-f(x1) $geq$ 0 but from the fact f is not strictly increasing and f(x2) -f(x1) $leq$ 0. So f(x2)-f(x1) = 0. I'm not sure if this is a good way to go about it either.
$endgroup$
– Nolando
Dec 3 '18 at 13:27
add a comment |
$begingroup$
I was wondering how I would define not strictly increasing, would I negate the formal definition or find the contrapositive of the definition? Or is there another way?
$endgroup$
– Nolando
Dec 3 '18 at 12:02
$begingroup$
@Nolando That's exactly it. A function is strictly increasing if for all $x_1,x_2$ such that $x_1<x_2$, we have $f(x_1)<f(x_2)$. Symbolically, this can be written as $$forall x_1, x_2: x_1<x_2implies f(x_1)<f(x_2)$$ By definition, a function is not strictly increasing if the negation of that holds, i.e. $$exists x_1,x_2:x_1<x_2land f(x_1)geq f(x_2)$$ meaning there exists some pair $x_1,x_2$ such that $x_1<x_2$ and $f(x_1)geq f(x_2)$.
$endgroup$
– 5xum
Dec 3 '18 at 12:41
$begingroup$
Thanks! I was thinking wondering if i should go along the lines of $dfrac{f(x2)-f(x1)}{x2-x1}$ $geq$ 0 since f '(x) $geq$ 0 then f(x2)-f(x1) $geq$ 0 but from the fact f is not strictly increasing and f(x2) -f(x1) $leq$ 0. So f(x2)-f(x1) = 0. I'm not sure if this is a good way to go about it either.
$endgroup$
– Nolando
Dec 3 '18 at 13:27
$begingroup$
I was wondering how I would define not strictly increasing, would I negate the formal definition or find the contrapositive of the definition? Or is there another way?
$endgroup$
– Nolando
Dec 3 '18 at 12:02
$begingroup$
I was wondering how I would define not strictly increasing, would I negate the formal definition or find the contrapositive of the definition? Or is there another way?
$endgroup$
– Nolando
Dec 3 '18 at 12:02
$begingroup$
@Nolando That's exactly it. A function is strictly increasing if for all $x_1,x_2$ such that $x_1<x_2$, we have $f(x_1)<f(x_2)$. Symbolically, this can be written as $$forall x_1, x_2: x_1<x_2implies f(x_1)<f(x_2)$$ By definition, a function is not strictly increasing if the negation of that holds, i.e. $$exists x_1,x_2:x_1<x_2land f(x_1)geq f(x_2)$$ meaning there exists some pair $x_1,x_2$ such that $x_1<x_2$ and $f(x_1)geq f(x_2)$.
$endgroup$
– 5xum
Dec 3 '18 at 12:41
$begingroup$
@Nolando That's exactly it. A function is strictly increasing if for all $x_1,x_2$ such that $x_1<x_2$, we have $f(x_1)<f(x_2)$. Symbolically, this can be written as $$forall x_1, x_2: x_1<x_2implies f(x_1)<f(x_2)$$ By definition, a function is not strictly increasing if the negation of that holds, i.e. $$exists x_1,x_2:x_1<x_2land f(x_1)geq f(x_2)$$ meaning there exists some pair $x_1,x_2$ such that $x_1<x_2$ and $f(x_1)geq f(x_2)$.
$endgroup$
– 5xum
Dec 3 '18 at 12:41
$begingroup$
Thanks! I was thinking wondering if i should go along the lines of $dfrac{f(x2)-f(x1)}{x2-x1}$ $geq$ 0 since f '(x) $geq$ 0 then f(x2)-f(x1) $geq$ 0 but from the fact f is not strictly increasing and f(x2) -f(x1) $leq$ 0. So f(x2)-f(x1) = 0. I'm not sure if this is a good way to go about it either.
$endgroup$
– Nolando
Dec 3 '18 at 13:27
$begingroup$
Thanks! I was thinking wondering if i should go along the lines of $dfrac{f(x2)-f(x1)}{x2-x1}$ $geq$ 0 since f '(x) $geq$ 0 then f(x2)-f(x1) $geq$ 0 but from the fact f is not strictly increasing and f(x2) -f(x1) $leq$ 0. So f(x2)-f(x1) = 0. I'm not sure if this is a good way to go about it either.
$endgroup$
– Nolando
Dec 3 '18 at 13:27
add a comment |
$begingroup$
To prove that by contradiction, firstly we can prove that $f'(x)>0 implies f(x)$ strictly increasing.
Suppose indeed that exist $x_1<x_2$ such that $f(x_1)>f(x_2)$ then by MVT
$$f(x_2)-f(x_1)=(x_2-x_1)f'(c)>0$$
which is a contradiction.
Form here we can now extend the result to $f'(x)ge 0$, with $f'(x)=0$ only for a set of isolated points, $implies f(x)$ strictly increasing, indeed assuming again by contradiction that $f(x)$ is not strictly increasing then it would exist an interval with $f'(x)=0$ which is against the hypotesis.
answered Dec 3 '18 at 11:24
gimusigimusi
92.9k94494
$endgroup$
add a comment |
$begingroup$
To prove that by contradiction, firstly we can prove that $f'(x)>0 implies f(x)$ strictly increasing.
Suppose indeed that exist $x_1<x_2$ such that $f(x_1)>f(x_2)$ then by MVT
$$f(x_2)-f(x_1)=(x_2-x_1)f'(c)>0$$
which is a contradiction.
Form here we can now extend the result to $f'(x)ge 0$, with $f'(x)=0$ only for a set of isolated points, $implies f(x)$ strictly increasing, indeed assuming again by contradiction that $f(x)$ is not strictly increasing then it would exist an interval with $f'(x)=0$ which is against the hypotesis.
answered Dec 3 '18 at 11:24
gimusigimusi
92.9k94494
$endgroup$
add a comment |
$begingroup$
To prove that by contradiction, firstly we can prove that $f'(x)>0 implies f(x)$ strictly increasing.
Suppose indeed that exist $x_1<x_2$ such that $f(x_1)>f(x_2)$ then by MVT
$$f(x_2)-f(x_1)=(x_2-x_1)f'(c)>0$$
which is a contradiction.
Form here we can now extend the result to $f'(x)ge 0$, with $f'(x)=0$ only for a set of isolated points, $implies f(x)$ strictly increasing, indeed assuming again by contradiction that $f(x)$ is not strictly increasing then it would exist an interval with $f'(x)=0$ which is against the hypotesis.
answered Dec 3 '18 at 11:24
gimusigimusi
92.9k94494
$endgroup$
To prove that by contradiction, firstly we can prove that $f'(x)>0 implies f(x)$ strictly increasing.
Suppose indeed that exist $x_1<x_2$ such that $f(x_1)>f(x_2)$ then by MVT
$$f(x_2)-f(x_1)=(x_2-x_1)f'(c)>0$$
which is a contradiction.
Form here we can now extend the result to $f'(x)ge 0$, with $f'(x)=0$ only for a set of isolated points, $implies f(x)$ strictly increasing, indeed assuming again by contradiction that $f(x)$ is not strictly increasing then it would exist an interval with $f'(x)=0$ which is against the hypotesis.
answered Dec 3 '18 at 11:24
gimusigimusi
92.9k94494
edited Dec 3 '18 at 11:34
answered Dec 3 '18 at 11:24
gimusigimusi
92.9k94494
answered Dec 3 '18 at 11:24
gimusigimusi
92.9k94494
answered Dec 3 '18 at 11:24
gimusigimusi
92.9k94494
92.9k94494
add a comment |
add a comment |
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$begingroup$
I'm not sure how you conclude that $f$ is constant on a sub interval. That doesn't follow from the fact that $f$ is not strictly increasing (it may as well be decreasing or not monotonic at all).
$endgroup$
– freakish
Dec 3 '18 at 11:00