Mixing
Topological Spaces and Continuous Functions
$begingroup$
I have no idea where to even start with this problem. We were going over metric spaces in class and this never showed up in the chapter. If someone could help me out it would be much appreciated!
Let $f:[0,infty) to [0,infty)$ be a continuously twice-differentiable, strictly increasing, and
concave (also called concave down; i.e. $f''< 0)$ function such that $f(0) = 0$.
$\$
A: Show that the following function $ϕ: [0, ∞) → R$ is decreasing for any fixed $t > 0$:
$ϕ(x) = { f(x + t) − f(x)over t }$
B: Prove for $x ≥ 0$ and $t > 0$ that:
$f(x + t) ≤ f(x) + f(t)$.
C: Show that the rational function ${xover 1+x}$
satisfies the inequality in part (b).
general-topology metric-spaces
edited Feb 1 at 3:02
Noah Schweber
128k10152294
asked Jan 31 at 22:35
DataD96DataD96
295
$endgroup$
add a comment |
$begingroup$
I have no idea where to even start with this problem. We were going over metric spaces in class and this never showed up in the chapter. If someone could help me out it would be much appreciated!
Let $f:[0,infty) to [0,infty)$ be a continuously twice-differentiable, strictly increasing, and
concave (also called concave down; i.e. $f''< 0)$ function such that $f(0) = 0$.
$\$
A: Show that the following function $ϕ: [0, ∞) → R$ is decreasing for any fixed $t > 0$:
$ϕ(x) = { f(x + t) − f(x)over t }$
B: Prove for $x ≥ 0$ and $t > 0$ that:
$f(x + t) ≤ f(x) + f(t)$.
C: Show that the rational function ${xover 1+x}$
satisfies the inequality in part (b).
general-topology metric-spaces
edited Feb 1 at 3:02
Noah Schweber
128k10152294
asked Jan 31 at 22:35
DataD96DataD96
295
$endgroup$
$begingroup$
This has nothing to do with set theory.
$endgroup$
– Noah Schweber
Feb 1 at 3:02
add a comment |
$begingroup$
I have no idea where to even start with this problem. We were going over metric spaces in class and this never showed up in the chapter. If someone could help me out it would be much appreciated!
Let $f:[0,infty) to [0,infty)$ be a continuously twice-differentiable, strictly increasing, and
concave (also called concave down; i.e. $f''< 0)$ function such that $f(0) = 0$.
$\$
A: Show that the following function $ϕ: [0, ∞) → R$ is decreasing for any fixed $t > 0$:
$ϕ(x) = { f(x + t) − f(x)over t }$
B: Prove for $x ≥ 0$ and $t > 0$ that:
$f(x + t) ≤ f(x) + f(t)$.
C: Show that the rational function ${xover 1+x}$
satisfies the inequality in part (b).
general-topology metric-spaces
edited Feb 1 at 3:02
Noah Schweber
128k10152294
asked Jan 31 at 22:35
DataD96DataD96
295
$endgroup$
I have no idea where to even start with this problem. We were going over metric spaces in class and this never showed up in the chapter. If someone could help me out it would be much appreciated!
Let $f:[0,infty) to [0,infty)$ be a continuously twice-differentiable, strictly increasing, and
concave (also called concave down; i.e. $f''< 0)$ function such that $f(0) = 0$.
$\$
A: Show that the following function $ϕ: [0, ∞) → R$ is decreasing for any fixed $t > 0$:
$ϕ(x) = { f(x + t) − f(x)over t }$
B: Prove for $x ≥ 0$ and $t > 0$ that:
$f(x + t) ≤ f(x) + f(t)$.
C: Show that the rational function ${xover 1+x}$
satisfies the inequality in part (b).
general-topology metric-spaces
general-topology metric-spaces
edited Feb 1 at 3:02
Noah Schweber
128k10152294
asked Jan 31 at 22:35
DataD96DataD96
295
edited Feb 1 at 3:02
Noah Schweber
128k10152294
asked Jan 31 at 22:35
DataD96DataD96
295
edited Feb 1 at 3:02
Noah Schweber
128k10152294
edited Feb 1 at 3:02
Noah Schweber
128k10152294
edited Feb 1 at 3:02
Noah Schweber
128k10152294
128k10152294
asked Jan 31 at 22:35
DataD96DataD96
295
asked Jan 31 at 22:35
DataD96DataD96
295
asked Jan 31 at 22:35
DataD96DataD96
295
295
$begingroup$
This has nothing to do with set theory.
$endgroup$
– Noah Schweber
Feb 1 at 3:02
add a comment |
$begingroup$
This has nothing to do with set theory.
$endgroup$
– Noah Schweber
Feb 1 at 3:02
$begingroup$
This has nothing to do with set theory.
$endgroup$
– Noah Schweber
Feb 1 at 3:02
$begingroup$
This has nothing to do with set theory.
$endgroup$
– Noah Schweber
Feb 1 at 3:02
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
$phi (x)=frac {f(x+t)-f(x)} t =frac 1 tint_x^{x+t} f'(u)du=int_0^{1} f'(x +tv)dv$ by the substitution $v=frac {u-x} t$. Since $f'$ is decreasing it is obvious from this expression that $phi$ is decreasing.
For b) use the fact that $phi (x) leq phi (0)$. c) is by direct verification.
answered Jan 31 at 23:32
Kavi Rama MurthyKavi Rama Murthy
73.4k53170
$endgroup$
add a comment |
$begingroup$
(1).$quad phi'(x)=frac {f'(x+t)-f'(x)}{t},$ which by the MVT is equal to $f''(x+s)$ for some $sin (0,t).$ So $phi'(x)<0.$
So if $y>0$ then $$phi(x+y)-phi(x)=ycdot frac {phi(x+y)-phi(y)}{y}$$ which by the MVT is equal to $yphi'(x+z)$ for some $zin (0,y).$ So $phi(x+y)-phi(x)<0.$
(2).$quad$For a fixed $t>0$ let $g(x) =f(x+t)-f(x)-f(t).$ Then $g(0)=-f(0)le 0.$
By (1) we have $g'(y)=tphi'(y)<0.$ For $x>0$ we have $$g(x)=g(0)+xcdotfrac {g(x)-g(0)}{x}$$ which by the MVT is equal to $g(0)+xg'(y)$ for some $yin (0,x).$ So $g(x)=g(0)+xg'(y)<g(0)le 0.$
answered Feb 1 at 2:59
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
add a comment |
$begingroup$
This is for part A.
Show that
is decreasing.
is nothing more than the subtraction of two functions.
The function is just a function being shifted to the left t units. Because t is always positive the shift will always be a horizontal shift to the left graphically.
The function is the same function only without the horizontally shift.
Because essentially these are the same functions the one with the shift has more time to grow or increase than the nonshifted one. So we can conclude that:
The shifted function is always greater than the non-shifted function. In other words:
&space;frac{f(x))}{t}" target="_blank">&space;frac{f(x))}{t}" title="frac{f(x+t)}{t}> frac{f(x))}{t}" />
Thus we can conclude that:
&space;0" target="_blank">&space;0" title="phi (x)> 0" />
Which means that between these two functions there must be some vertical distance between them that exists for any given particular x value.
Now we have three cases. These two functions can either diverge away from each other, converge on each other, or have a constant distance between them.
Consider the convergent case. If the two functions are converging then the distance values between them keep geting smaller and smaller.
Because this distance is getting smaller and smaller as x increases this is what allows us to conclude that is actually decreasing.
answered Feb 1 at 4:24
Erock BroxErock Brox
24128
$endgroup$
add a comment |
$begingroup$
Check the derivative of $phi$, and use the 2 conditions provided.
Use that $phi$ is decreasing to conclude that $phi(0) <= phi(x)$. And rearrange to get the answer. Keep in mind that $t$ is positive so you can cancel that from the denominator.
Show that $f$ satisfies the conditions described by the exercise.
edited Feb 1 at 5:43
feynhat
506414
answered Jan 31 at 23:18
AlexandrosAlexandros
1,0151413
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$phi (x)=frac {f(x+t)-f(x)} t =frac 1 tint_x^{x+t} f'(u)du=int_0^{1} f'(x +tv)dv$ by the substitution $v=frac {u-x} t$. Since $f'$ is decreasing it is obvious from this expression that $phi$ is decreasing.
For b) use the fact that $phi (x) leq phi (0)$. c) is by direct verification.
answered Jan 31 at 23:32
Kavi Rama MurthyKavi Rama Murthy
73.4k53170
$endgroup$
add a comment |
$begingroup$
$phi (x)=frac {f(x+t)-f(x)} t =frac 1 tint_x^{x+t} f'(u)du=int_0^{1} f'(x +tv)dv$ by the substitution $v=frac {u-x} t$. Since $f'$ is decreasing it is obvious from this expression that $phi$ is decreasing.
For b) use the fact that $phi (x) leq phi (0)$. c) is by direct verification.
answered Jan 31 at 23:32
Kavi Rama MurthyKavi Rama Murthy
73.4k53170
$endgroup$
add a comment |
$begingroup$
$phi (x)=frac {f(x+t)-f(x)} t =frac 1 tint_x^{x+t} f'(u)du=int_0^{1} f'(x +tv)dv$ by the substitution $v=frac {u-x} t$. Since $f'$ is decreasing it is obvious from this expression that $phi$ is decreasing.
For b) use the fact that $phi (x) leq phi (0)$. c) is by direct verification.
answered Jan 31 at 23:32
Kavi Rama MurthyKavi Rama Murthy
73.4k53170
$endgroup$
$phi (x)=frac {f(x+t)-f(x)} t =frac 1 tint_x^{x+t} f'(u)du=int_0^{1} f'(x +tv)dv$ by the substitution $v=frac {u-x} t$. Since $f'$ is decreasing it is obvious from this expression that $phi$ is decreasing.
For b) use the fact that $phi (x) leq phi (0)$. c) is by direct verification.
answered Jan 31 at 23:32
Kavi Rama MurthyKavi Rama Murthy
73.4k53170
answered Jan 31 at 23:32
Kavi Rama MurthyKavi Rama Murthy
73.4k53170
answered Jan 31 at 23:32
Kavi Rama MurthyKavi Rama Murthy
73.4k53170
answered Jan 31 at 23:32
Kavi Rama MurthyKavi Rama Murthy
73.4k53170
73.4k53170
add a comment |
add a comment |
$begingroup$
(1).$quad phi'(x)=frac {f'(x+t)-f'(x)}{t},$ which by the MVT is equal to $f''(x+s)$ for some $sin (0,t).$ So $phi'(x)<0.$
So if $y>0$ then $$phi(x+y)-phi(x)=ycdot frac {phi(x+y)-phi(y)}{y}$$ which by the MVT is equal to $yphi'(x+z)$ for some $zin (0,y).$ So $phi(x+y)-phi(x)<0.$
(2).$quad$For a fixed $t>0$ let $g(x) =f(x+t)-f(x)-f(t).$ Then $g(0)=-f(0)le 0.$
By (1) we have $g'(y)=tphi'(y)<0.$ For $x>0$ we have $$g(x)=g(0)+xcdotfrac {g(x)-g(0)}{x}$$ which by the MVT is equal to $g(0)+xg'(y)$ for some $yin (0,x).$ So $g(x)=g(0)+xg'(y)<g(0)le 0.$
answered Feb 1 at 2:59
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
add a comment |
$begingroup$
(1).$quad phi'(x)=frac {f'(x+t)-f'(x)}{t},$ which by the MVT is equal to $f''(x+s)$ for some $sin (0,t).$ So $phi'(x)<0.$
So if $y>0$ then $$phi(x+y)-phi(x)=ycdot frac {phi(x+y)-phi(y)}{y}$$ which by the MVT is equal to $yphi'(x+z)$ for some $zin (0,y).$ So $phi(x+y)-phi(x)<0.$
(2).$quad$For a fixed $t>0$ let $g(x) =f(x+t)-f(x)-f(t).$ Then $g(0)=-f(0)le 0.$
By (1) we have $g'(y)=tphi'(y)<0.$ For $x>0$ we have $$g(x)=g(0)+xcdotfrac {g(x)-g(0)}{x}$$ which by the MVT is equal to $g(0)+xg'(y)$ for some $yin (0,x).$ So $g(x)=g(0)+xg'(y)<g(0)le 0.$
answered Feb 1 at 2:59
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
add a comment |
$begingroup$
(1).$quad phi'(x)=frac {f'(x+t)-f'(x)}{t},$ which by the MVT is equal to $f''(x+s)$ for some $sin (0,t).$ So $phi'(x)<0.$
So if $y>0$ then $$phi(x+y)-phi(x)=ycdot frac {phi(x+y)-phi(y)}{y}$$ which by the MVT is equal to $yphi'(x+z)$ for some $zin (0,y).$ So $phi(x+y)-phi(x)<0.$
(2).$quad$For a fixed $t>0$ let $g(x) =f(x+t)-f(x)-f(t).$ Then $g(0)=-f(0)le 0.$
By (1) we have $g'(y)=tphi'(y)<0.$ For $x>0$ we have $$g(x)=g(0)+xcdotfrac {g(x)-g(0)}{x}$$ which by the MVT is equal to $g(0)+xg'(y)$ for some $yin (0,x).$ So $g(x)=g(0)+xg'(y)<g(0)le 0.$
answered Feb 1 at 2:59
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
(1).$quad phi'(x)=frac {f'(x+t)-f'(x)}{t},$ which by the MVT is equal to $f''(x+s)$ for some $sin (0,t).$ So $phi'(x)<0.$
So if $y>0$ then $$phi(x+y)-phi(x)=ycdot frac {phi(x+y)-phi(y)}{y}$$ which by the MVT is equal to $yphi'(x+z)$ for some $zin (0,y).$ So $phi(x+y)-phi(x)<0.$
(2).$quad$For a fixed $t>0$ let $g(x) =f(x+t)-f(x)-f(t).$ Then $g(0)=-f(0)le 0.$
By (1) we have $g'(y)=tphi'(y)<0.$ For $x>0$ we have $$g(x)=g(0)+xcdotfrac {g(x)-g(0)}{x}$$ which by the MVT is equal to $g(0)+xg'(y)$ for some $yin (0,x).$ So $g(x)=g(0)+xg'(y)<g(0)le 0.$
answered Feb 1 at 2:59
DanielWainfleetDanielWainfleet
35.8k31648
answered Feb 1 at 2:59
DanielWainfleetDanielWainfleet
35.8k31648
answered Feb 1 at 2:59
DanielWainfleetDanielWainfleet
35.8k31648
answered Feb 1 at 2:59
DanielWainfleetDanielWainfleet
35.8k31648
35.8k31648
add a comment |
add a comment |
$begingroup$
This is for part A.
Show that
is decreasing.
is nothing more than the subtraction of two functions.
The function is just a function being shifted to the left t units. Because t is always positive the shift will always be a horizontal shift to the left graphically.
The function is the same function only without the horizontally shift.
Because essentially these are the same functions the one with the shift has more time to grow or increase than the nonshifted one. So we can conclude that:
The shifted function is always greater than the non-shifted function. In other words:
&space;frac{f(x))}{t}" target="_blank">&space;frac{f(x))}{t}" title="frac{f(x+t)}{t}> frac{f(x))}{t}" />
Thus we can conclude that:
&space;0" target="_blank">&space;0" title="phi (x)> 0" />
Which means that between these two functions there must be some vertical distance between them that exists for any given particular x value.
Now we have three cases. These two functions can either diverge away from each other, converge on each other, or have a constant distance between them.
Consider the convergent case. If the two functions are converging then the distance values between them keep geting smaller and smaller.
Because this distance is getting smaller and smaller as x increases this is what allows us to conclude that is actually decreasing.
answered Feb 1 at 4:24
Erock BroxErock Brox
24128
$endgroup$
add a comment |
$begingroup$
This is for part A.
Show that
is decreasing.
is nothing more than the subtraction of two functions.
The function is just a function being shifted to the left t units. Because t is always positive the shift will always be a horizontal shift to the left graphically.
The function is the same function only without the horizontally shift.
Because essentially these are the same functions the one with the shift has more time to grow or increase than the nonshifted one. So we can conclude that:
The shifted function is always greater than the non-shifted function. In other words:
&space;frac{f(x))}{t}" target="_blank">&space;frac{f(x))}{t}" title="frac{f(x+t)}{t}> frac{f(x))}{t}" />
Thus we can conclude that:
&space;0" target="_blank">&space;0" title="phi (x)> 0" />
Which means that between these two functions there must be some vertical distance between them that exists for any given particular x value.
Now we have three cases. These two functions can either diverge away from each other, converge on each other, or have a constant distance between them.
Consider the convergent case. If the two functions are converging then the distance values between them keep geting smaller and smaller.
Because this distance is getting smaller and smaller as x increases this is what allows us to conclude that is actually decreasing.
answered Feb 1 at 4:24
Erock BroxErock Brox
24128
$endgroup$
add a comment |
$begingroup$
This is for part A.
Show that
is decreasing.
is nothing more than the subtraction of two functions.
The function is just a function being shifted to the left t units. Because t is always positive the shift will always be a horizontal shift to the left graphically.
The function is the same function only without the horizontally shift.
Because essentially these are the same functions the one with the shift has more time to grow or increase than the nonshifted one. So we can conclude that:
The shifted function is always greater than the non-shifted function. In other words:
&space;frac{f(x))}{t}" target="_blank">&space;frac{f(x))}{t}" title="frac{f(x+t)}{t}> frac{f(x))}{t}" />
Thus we can conclude that:
&space;0" target="_blank">&space;0" title="phi (x)> 0" />
Which means that between these two functions there must be some vertical distance between them that exists for any given particular x value.
Now we have three cases. These two functions can either diverge away from each other, converge on each other, or have a constant distance between them.
Consider the convergent case. If the two functions are converging then the distance values between them keep geting smaller and smaller.
Because this distance is getting smaller and smaller as x increases this is what allows us to conclude that is actually decreasing.
answered Feb 1 at 4:24
Erock BroxErock Brox
24128
$endgroup$
This is for part A.
Show that
is decreasing.
is nothing more than the subtraction of two functions.
The function is just a function being shifted to the left t units. Because t is always positive the shift will always be a horizontal shift to the left graphically.
The function is the same function only without the horizontally shift.
Because essentially these are the same functions the one with the shift has more time to grow or increase than the nonshifted one. So we can conclude that:
The shifted function is always greater than the non-shifted function. In other words:
&space;frac{f(x))}{t}" target="_blank">&space;frac{f(x))}{t}" title="frac{f(x+t)}{t}> frac{f(x))}{t}" />
Thus we can conclude that:
&space;0" target="_blank">&space;0" title="phi (x)> 0" />
Which means that between these two functions there must be some vertical distance between them that exists for any given particular x value.
Now we have three cases. These two functions can either diverge away from each other, converge on each other, or have a constant distance between them.
Consider the convergent case. If the two functions are converging then the distance values between them keep geting smaller and smaller.
Because this distance is getting smaller and smaller as x increases this is what allows us to conclude that is actually decreasing.
answered Feb 1 at 4:24
Erock BroxErock Brox
24128
answered Feb 1 at 4:24
Erock BroxErock Brox
24128
answered Feb 1 at 4:24
Erock BroxErock Brox
24128
answered Feb 1 at 4:24
Erock BroxErock Brox
24128
24128
add a comment |
add a comment |
$begingroup$
Check the derivative of $phi$, and use the 2 conditions provided.
Use that $phi$ is decreasing to conclude that $phi(0) <= phi(x)$. And rearrange to get the answer. Keep in mind that $t$ is positive so you can cancel that from the denominator.
Show that $f$ satisfies the conditions described by the exercise.
edited Feb 1 at 5:43
feynhat
506414
answered Jan 31 at 23:18
AlexandrosAlexandros
1,0151413
$endgroup$
add a comment |
$begingroup$
Check the derivative of $phi$, and use the 2 conditions provided.
Use that $phi$ is decreasing to conclude that $phi(0) <= phi(x)$. And rearrange to get the answer. Keep in mind that $t$ is positive so you can cancel that from the denominator.
Show that $f$ satisfies the conditions described by the exercise.
edited Feb 1 at 5:43
feynhat
506414
answered Jan 31 at 23:18
AlexandrosAlexandros
1,0151413
$endgroup$
add a comment |
$begingroup$
Check the derivative of $phi$, and use the 2 conditions provided.
Use that $phi$ is decreasing to conclude that $phi(0) <= phi(x)$. And rearrange to get the answer. Keep in mind that $t$ is positive so you can cancel that from the denominator.
Show that $f$ satisfies the conditions described by the exercise.
edited Feb 1 at 5:43
feynhat
506414
answered Jan 31 at 23:18
AlexandrosAlexandros
1,0151413
$endgroup$
Check the derivative of $phi$, and use the 2 conditions provided.
Use that $phi$ is decreasing to conclude that $phi(0) <= phi(x)$. And rearrange to get the answer. Keep in mind that $t$ is positive so you can cancel that from the denominator.
Show that $f$ satisfies the conditions described by the exercise.
edited Feb 1 at 5:43
feynhat
506414
answered Jan 31 at 23:18
AlexandrosAlexandros
1,0151413
edited Feb 1 at 5:43
feynhat
506414
edited Feb 1 at 5:43
feynhat
506414
edited Feb 1 at 5:43
feynhat
506414
506414
answered Jan 31 at 23:18
AlexandrosAlexandros
1,0151413
answered Jan 31 at 23:18
AlexandrosAlexandros
1,0151413
answered Jan 31 at 23:18
AlexandrosAlexandros
1,0151413
1,0151413
add a comment |
add a comment |
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$begingroup$
This has nothing to do with set theory.
$endgroup$
– Noah Schweber
Feb 1 at 3:02