Show that if $f neq 0$ and $x < y Rightarrow f(x) le frac {f(y) - f(x)}{y - x } le f(y)$ then $f$ is...
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I have the following problem :
$f neq 0$ satisfies $x < y Rightarrow f(x) le frac {f(y) - f(x)}{y - x } le f(y)$ then $f$ is differentiable
Try
For $h > 0, f(x) le frac {f(x +h) - f(x)}{h} le f(x + h)$.
By squeeze theorem, $frac {f(x +h) - f(x)}{h} le f(x + h) to f(x)$ as $h to 0^+$, but this assumes continuity of $f$. I'm stuck at showing $f$ is continuous.
calculus real-analysis
edited Nov 17 at 8:54
DanielV
17.7k42753
asked Nov 17 at 8:21
Moreblue
770216
add a comment |
up vote
3
down vote
favorite
I have the following problem :
$f neq 0$ satisfies $x < y Rightarrow f(x) le frac {f(y) - f(x)}{y - x } le f(y)$ then $f$ is differentiable
Try
For $h > 0, f(x) le frac {f(x +h) - f(x)}{h} le f(x + h)$.
By squeeze theorem, $frac {f(x +h) - f(x)}{h} le f(x + h) to f(x)$ as $h to 0^+$, but this assumes continuity of $f$. I'm stuck at showing $f$ is continuous.
calculus real-analysis
edited Nov 17 at 8:54
DanielV
17.7k42753
asked Nov 17 at 8:21
Moreblue
770216
1
How exactly did you make this so unreadable?
– Lord Shark the Unknown
Nov 17 at 8:22
@LordSharktheUnknown Edited.
– Moreblue
Nov 17 at 8:24
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I have the following problem :
$f neq 0$ satisfies $x < y Rightarrow f(x) le frac {f(y) - f(x)}{y - x } le f(y)$ then $f$ is differentiable
Try
For $h > 0, f(x) le frac {f(x +h) - f(x)}{h} le f(x + h)$.
By squeeze theorem, $frac {f(x +h) - f(x)}{h} le f(x + h) to f(x)$ as $h to 0^+$, but this assumes continuity of $f$. I'm stuck at showing $f$ is continuous.
calculus real-analysis
edited Nov 17 at 8:54
DanielV
17.7k42753
asked Nov 17 at 8:21
Moreblue
770216
I have the following problem :
$f neq 0$ satisfies $x < y Rightarrow f(x) le frac {f(y) - f(x)}{y - x } le f(y)$ then $f$ is differentiable
Try
For $h > 0, f(x) le frac {f(x +h) - f(x)}{h} le f(x + h)$.
By squeeze theorem, $frac {f(x +h) - f(x)}{h} le f(x + h) to f(x)$ as $h to 0^+$, but this assumes continuity of $f$. I'm stuck at showing $f$ is continuous.
calculus real-analysis
calculus real-analysis
edited Nov 17 at 8:54
DanielV
17.7k42753
asked Nov 17 at 8:21
Moreblue
770216
edited Nov 17 at 8:54
DanielV
17.7k42753
asked Nov 17 at 8:21
Moreblue
770216
edited Nov 17 at 8:54
DanielV
17.7k42753
edited Nov 17 at 8:54
DanielV
17.7k42753
edited Nov 17 at 8:54
DanielV
17.7k42753
17.7k42753
asked Nov 17 at 8:21
Moreblue
770216
asked Nov 17 at 8:21
Moreblue
770216
asked Nov 17 at 8:21
Moreblue
770216
770216
1
How exactly did you make this so unreadable?
– Lord Shark the Unknown
Nov 17 at 8:22
@LordSharktheUnknown Edited.
– Moreblue
Nov 17 at 8:24
add a comment |
1
How exactly did you make this so unreadable?
– Lord Shark the Unknown
Nov 17 at 8:22
@LordSharktheUnknown Edited.
– Moreblue
Nov 17 at 8:24
1
1
How exactly did you make this so unreadable?
– Lord Shark the Unknown
Nov 17 at 8:22
How exactly did you make this so unreadable?
– Lord Shark the Unknown
Nov 17 at 8:22
@LordSharktheUnknown Edited.
– Moreblue
Nov 17 at 8:24
@LordSharktheUnknown Edited.
– Moreblue
Nov 17 at 8:24
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
$f(x)(y - x) le f(y) - f(x) le f(y)(y - x)$.
Thus as $x to y, f(x) to f(y).$
From such considerations and that $f$ is ascending,
continuity should be forth coming.
edited Nov 17 at 9:39
Moreblue
770216
answered Nov 17 at 8:47
William Elliot
6,8702518
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$f(x)(y - x) le f(y) - f(x) le f(y)(y - x)$.
Thus as $x to y, f(x) to f(y).$
From such considerations and that $f$ is ascending,
continuity should be forth coming.
edited Nov 17 at 9:39
Moreblue
770216
answered Nov 17 at 8:47
William Elliot
6,8702518
add a comment |
up vote
2
down vote
accepted
$f(x)(y - x) le f(y) - f(x) le f(y)(y - x)$.
Thus as $x to y, f(x) to f(y).$
From such considerations and that $f$ is ascending,
continuity should be forth coming.
edited Nov 17 at 9:39
Moreblue
770216
answered Nov 17 at 8:47
William Elliot
6,8702518
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$f(x)(y - x) le f(y) - f(x) le f(y)(y - x)$.
Thus as $x to y, f(x) to f(y).$
From such considerations and that $f$ is ascending,
continuity should be forth coming.
edited Nov 17 at 9:39
Moreblue
770216
answered Nov 17 at 8:47
William Elliot
6,8702518
$f(x)(y - x) le f(y) - f(x) le f(y)(y - x)$.
Thus as $x to y, f(x) to f(y).$
From such considerations and that $f$ is ascending,
continuity should be forth coming.
edited Nov 17 at 9:39
Moreblue
770216
answered Nov 17 at 8:47
William Elliot
6,8702518
edited Nov 17 at 9:39
Moreblue
770216
edited Nov 17 at 9:39
Moreblue
770216
edited Nov 17 at 9:39
Moreblue
770216
770216
answered Nov 17 at 8:47
William Elliot
6,8702518
answered Nov 17 at 8:47
William Elliot
6,8702518
answered Nov 17 at 8:47
William Elliot
6,8702518
6,8702518
add a comment |
add a comment |
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1
How exactly did you make this so unreadable?
– Lord Shark the Unknown
Nov 17 at 8:22
@LordSharktheUnknown Edited.
– Moreblue
Nov 17 at 8:24