Mixing
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Let $alpha >1.$ Then $forall xgt 0: psi(alpha x)leq alpha psi( x);.$ True or False?
$begingroup$
Let $ psi$ be a function satisfying :
$psi: mathbb{R}^+rightarrow mathbb{R}^+$ .
$psi $ is non-decreasing.
$psi (x)< x, forall x> 0$.
I want to know if the following statement is true:
$$text{Let } alpha >1. text{ Then }, forall x> 0: psi(alpha x)leq alpha psi( x);.$$
If not, can you give me a counter example please.
real-analysis functional-analysis analysis
edited Jan 17 at 16:40
jordan_glen
1
asked Jan 17 at 15:49
MotakaMotaka
238111
$endgroup$
add a comment |
$begingroup$
Let $ psi$ be a function satisfying :
$psi: mathbb{R}^+rightarrow mathbb{R}^+$ .
$psi $ is non-decreasing.
$psi (x)< x, forall x> 0$.
I want to know if the following statement is true:
$$text{Let } alpha >1. text{ Then }, forall x> 0: psi(alpha x)leq alpha psi( x);.$$
If not, can you give me a counter example please.
real-analysis functional-analysis analysis
edited Jan 17 at 16:40
jordan_glen
1
asked Jan 17 at 15:49
MotakaMotaka
238111
$endgroup$
$begingroup$
Not true. The function can increase dramatically over a short interval.
$endgroup$
– Don Thousand
Jan 17 at 15:54
add a comment |
$begingroup$
Let $ psi$ be a function satisfying :
$psi: mathbb{R}^+rightarrow mathbb{R}^+$ .
$psi $ is non-decreasing.
$psi (x)< x, forall x> 0$.
I want to know if the following statement is true:
$$text{Let } alpha >1. text{ Then }, forall x> 0: psi(alpha x)leq alpha psi( x);.$$
If not, can you give me a counter example please.
real-analysis functional-analysis analysis
edited Jan 17 at 16:40
jordan_glen
1
asked Jan 17 at 15:49
MotakaMotaka
238111
$endgroup$
Let $ psi$ be a function satisfying :
$psi: mathbb{R}^+rightarrow mathbb{R}^+$ .
$psi $ is non-decreasing.
$psi (x)< x, forall x> 0$.
I want to know if the following statement is true:
$$text{Let } alpha >1. text{ Then }, forall x> 0: psi(alpha x)leq alpha psi( x);.$$
If not, can you give me a counter example please.
real-analysis functional-analysis analysis
real-analysis functional-analysis analysis
edited Jan 17 at 16:40
jordan_glen
1
asked Jan 17 at 15:49
MotakaMotaka
238111
edited Jan 17 at 16:40
jordan_glen
1
asked Jan 17 at 15:49
MotakaMotaka
238111
edited Jan 17 at 16:40
jordan_glen
1
edited Jan 17 at 16:40
jordan_glen
1
edited Jan 17 at 16:40
jordan_glen
1
1
asked Jan 17 at 15:49
MotakaMotaka
238111
asked Jan 17 at 15:49
MotakaMotaka
238111
asked Jan 17 at 15:49
MotakaMotaka
238111
238111
$begingroup$
Not true. The function can increase dramatically over a short interval.
$endgroup$
– Don Thousand
Jan 17 at 15:54
add a comment |
$begingroup$
Not true. The function can increase dramatically over a short interval.
$endgroup$
– Don Thousand
Jan 17 at 15:54
$begingroup$
Not true. The function can increase dramatically over a short interval.
$endgroup$
– Don Thousand
Jan 17 at 15:54
$begingroup$
Not true. The function can increase dramatically over a short interval.
$endgroup$
– Don Thousand
Jan 17 at 15:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider
$$
f(x) = begin{cases} x, & x > 1 \ x^2 & x in [0,1] end{cases}
$$
around $x approx 0.9$.
answered Jan 17 at 15:59
gt6989bgt6989b
34.4k22456
$endgroup$
add a comment |
$begingroup$
There is an easy counterexample: imagine the function $psi(x)$ defined by parts as
$$psi(x)=begin{cases}dfrac{x}{10} &xleq 1000\ dfrac{2x}{5}+60 &x>1000end{cases}$$
Clearly $psi(x)leq x$ for all $x$, it is non-decreasing, but for $x=1000$ and $alpha=2>1$ we have $psi(2cdot 1000)=500>200=2cdot psi(1000)$.
answered Jan 17 at 16:02
Darío GDarío G
4,062613
$endgroup$
1
$begingroup$
$psi(2.2000)$ is not 500,+ and has two values at the same time
$endgroup$
– Motaka
Jan 17 at 16:27
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider
$$
f(x) = begin{cases} x, & x > 1 \ x^2 & x in [0,1] end{cases}
$$
around $x approx 0.9$.
answered Jan 17 at 15:59
gt6989bgt6989b
34.4k22456
$endgroup$
add a comment |
$begingroup$
Consider
$$
f(x) = begin{cases} x, & x > 1 \ x^2 & x in [0,1] end{cases}
$$
around $x approx 0.9$.
answered Jan 17 at 15:59
gt6989bgt6989b
34.4k22456
$endgroup$
add a comment |
$begingroup$
Consider
$$
f(x) = begin{cases} x, & x > 1 \ x^2 & x in [0,1] end{cases}
$$
around $x approx 0.9$.
answered Jan 17 at 15:59
gt6989bgt6989b
34.4k22456
$endgroup$
Consider
$$
f(x) = begin{cases} x, & x > 1 \ x^2 & x in [0,1] end{cases}
$$
around $x approx 0.9$.
answered Jan 17 at 15:59
gt6989bgt6989b
34.4k22456
answered Jan 17 at 15:59
gt6989bgt6989b
34.4k22456
answered Jan 17 at 15:59
gt6989bgt6989b
34.4k22456
answered Jan 17 at 15:59
gt6989bgt6989b
34.4k22456
34.4k22456
add a comment |
add a comment |
$begingroup$
There is an easy counterexample: imagine the function $psi(x)$ defined by parts as
$$psi(x)=begin{cases}dfrac{x}{10} &xleq 1000\ dfrac{2x}{5}+60 &x>1000end{cases}$$
Clearly $psi(x)leq x$ for all $x$, it is non-decreasing, but for $x=1000$ and $alpha=2>1$ we have $psi(2cdot 1000)=500>200=2cdot psi(1000)$.
answered Jan 17 at 16:02
Darío GDarío G
4,062613
$endgroup$
1
$begingroup$
$psi(2.2000)$ is not 500,+ and has two values at the same time
$endgroup$
– Motaka
Jan 17 at 16:27
add a comment |
$begingroup$
There is an easy counterexample: imagine the function $psi(x)$ defined by parts as
$$psi(x)=begin{cases}dfrac{x}{10} &xleq 1000\ dfrac{2x}{5}+60 &x>1000end{cases}$$
Clearly $psi(x)leq x$ for all $x$, it is non-decreasing, but for $x=1000$ and $alpha=2>1$ we have $psi(2cdot 1000)=500>200=2cdot psi(1000)$.
answered Jan 17 at 16:02
Darío GDarío G
4,062613
$endgroup$
1
$begingroup$
$psi(2.2000)$ is not 500,+ and has two values at the same time
$endgroup$
– Motaka
Jan 17 at 16:27
add a comment |
$begingroup$
There is an easy counterexample: imagine the function $psi(x)$ defined by parts as
$$psi(x)=begin{cases}dfrac{x}{10} &xleq 1000\ dfrac{2x}{5}+60 &x>1000end{cases}$$
Clearly $psi(x)leq x$ for all $x$, it is non-decreasing, but for $x=1000$ and $alpha=2>1$ we have $psi(2cdot 1000)=500>200=2cdot psi(1000)$.
answered Jan 17 at 16:02
Darío GDarío G
4,062613
$endgroup$
There is an easy counterexample: imagine the function $psi(x)$ defined by parts as
$$psi(x)=begin{cases}dfrac{x}{10} &xleq 1000\ dfrac{2x}{5}+60 &x>1000end{cases}$$
Clearly $psi(x)leq x$ for all $x$, it is non-decreasing, but for $x=1000$ and $alpha=2>1$ we have $psi(2cdot 1000)=500>200=2cdot psi(1000)$.
answered Jan 17 at 16:02
Darío GDarío G
4,062613
edited Jan 17 at 16:31
answered Jan 17 at 16:02
Darío GDarío G
4,062613
answered Jan 17 at 16:02
Darío GDarío G
4,062613
answered Jan 17 at 16:02
Darío GDarío G
4,062613
4,062613
1
$begingroup$
$psi(2.2000)$ is not 500,+ and has two values at the same time
$endgroup$
– Motaka
Jan 17 at 16:27
add a comment |
1
$begingroup$
$psi(2.2000)$ is not 500,+ and has two values at the same time
$endgroup$
– Motaka
Jan 17 at 16:27
1
1
$begingroup$
$psi(2.2000)$ is not 500,+ and has two values at the same time
$endgroup$
– Motaka
Jan 17 at 16:27
$begingroup$
$psi(2.2000)$ is not 500,+ and has two values at the same time
$endgroup$
– Motaka
Jan 17 at 16:27
add a comment |
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$begingroup$
Not true. The function can increase dramatically over a short interval.
$endgroup$
– Don Thousand
Jan 17 at 15:54