How to show that the following norms are equivalent?
up vote
0
down vote
favorite
Let $f in C^1[0, 1]$. Then show that the norms
$$|f| =int_0^1|f(t)|, dt+max_{t in [0, 1]}|f'(t)|$$ and $$|f|= max_{t in [0, 1]}|f(t)|+max_{t in [0, 1]}|f'(t)|$$ are equivalent.
I have shown that $|f| leq |f|$. How to show that $|f| leq K |f|$ for some constant K?
norm lp-spaces
edited Nov 23 at 10:55
José Carlos Santos
147k22117218
asked Nov 23 at 10:51
Mittal G
1,192515
add a comment |
up vote
0
down vote
favorite
Let $f in C^1[0, 1]$. Then show that the norms
$$|f| =int_0^1|f(t)|, dt+max_{t in [0, 1]}|f'(t)|$$ and $$|f|= max_{t in [0, 1]}|f(t)|+max_{t in [0, 1]}|f'(t)|$$ are equivalent.
I have shown that $|f| leq |f|$. How to show that $|f| leq K |f|$ for some constant K?
norm lp-spaces
edited Nov 23 at 10:55
José Carlos Santos
147k22117218
asked Nov 23 at 10:51
Mittal G
1,192515
1
The first "norm" is not a norm, please put the modulus inside the integral.
– dan_fulea
Nov 23 at 10:54
The idea is, I think, that if $max|f(t)|$ is much larger than $int |f(t)|dt$, then $max|f'(t)|$ must be large.
– Arthur
Nov 23 at 10:55
Looks like Mean Value Thm
– Richard Martin
Nov 23 at 10:55
what have you tried?
– supinf
Nov 23 at 10:57
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f in C^1[0, 1]$. Then show that the norms
$$|f| =int_0^1|f(t)|, dt+max_{t in [0, 1]}|f'(t)|$$ and $$|f|= max_{t in [0, 1]}|f(t)|+max_{t in [0, 1]}|f'(t)|$$ are equivalent.
I have shown that $|f| leq |f|$. How to show that $|f| leq K |f|$ for some constant K?
norm lp-spaces
edited Nov 23 at 10:55
José Carlos Santos
147k22117218
asked Nov 23 at 10:51
Mittal G
1,192515
Let $f in C^1[0, 1]$. Then show that the norms
$$|f| =int_0^1|f(t)|, dt+max_{t in [0, 1]}|f'(t)|$$ and $$|f|= max_{t in [0, 1]}|f(t)|+max_{t in [0, 1]}|f'(t)|$$ are equivalent.
I have shown that $|f| leq |f|$. How to show that $|f| leq K |f|$ for some constant K?
norm lp-spaces
norm lp-spaces
edited Nov 23 at 10:55
José Carlos Santos
147k22117218
asked Nov 23 at 10:51
Mittal G
1,192515
edited Nov 23 at 10:55
José Carlos Santos
147k22117218
asked Nov 23 at 10:51
Mittal G
1,192515
edited Nov 23 at 10:55
José Carlos Santos
147k22117218
edited Nov 23 at 10:55
José Carlos Santos
147k22117218
edited Nov 23 at 10:55
José Carlos Santos
147k22117218
147k22117218
asked Nov 23 at 10:51
Mittal G
1,192515
asked Nov 23 at 10:51
Mittal G
1,192515
asked Nov 23 at 10:51
Mittal G
1,192515
1,192515
1
The first "norm" is not a norm, please put the modulus inside the integral.
– dan_fulea
Nov 23 at 10:54
The idea is, I think, that if $max|f(t)|$ is much larger than $int |f(t)|dt$, then $max|f'(t)|$ must be large.
– Arthur
Nov 23 at 10:55
Looks like Mean Value Thm
– Richard Martin
Nov 23 at 10:55
what have you tried?
– supinf
Nov 23 at 10:57
add a comment |
1
The first "norm" is not a norm, please put the modulus inside the integral.
– dan_fulea
Nov 23 at 10:54
The idea is, I think, that if $max|f(t)|$ is much larger than $int |f(t)|dt$, then $max|f'(t)|$ must be large.
– Arthur
Nov 23 at 10:55
Looks like Mean Value Thm
– Richard Martin
Nov 23 at 10:55
what have you tried?
– supinf
Nov 23 at 10:57
1
1
The first "norm" is not a norm, please put the modulus inside the integral.
– dan_fulea
Nov 23 at 10:54
The first "norm" is not a norm, please put the modulus inside the integral.
– dan_fulea
Nov 23 at 10:54
The idea is, I think, that if $max|f(t)|$ is much larger than $int |f(t)|dt$, then $max|f'(t)|$ must be large.
– Arthur
Nov 23 at 10:55
The idea is, I think, that if $max|f(t)|$ is much larger than $int |f(t)|dt$, then $max|f'(t)|$ must be large.
– Arthur
Nov 23 at 10:55
Looks like Mean Value Thm
– Richard Martin
Nov 23 at 10:55
Looks like Mean Value Thm
– Richard Martin
Nov 23 at 10:55
what have you tried?
– supinf
Nov 23 at 10:57
what have you tried?
– supinf
Nov 23 at 10:57
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Suppose that$$lvert frvertleqslant1.tag1$$Then$$displaystylemax_{tin[0,1]}bigllvert f'(t)bigrrvertleqslant1.tag2$$Suppose that $displaystylemax_{tin[0,1]}bigllvert f(t)bigrrvert>2$. Then, by $(2)$ and by the Mean Value Theorem, $displaystylemin_{tin[0,1]}bigllvert f(t)bigrrvert>1$ and so $displaystyleint_0^1bigllvert f(t)bigrrvert,mathrm dt>1$. This is impossible, since we are assuming $(1)$. This proves that $lvert frvertleqslant1implieslVert frVertleqslant2$. Can you take it from here?
answered Nov 23 at 11:04
José Carlos Santos
147k22117218
What is the value of $K$?
– Mittal G
Nov 23 at 11:28
1
@MittalG $K=2$ will do.
– José Carlos Santos
Nov 23 at 11:29
@ José Carlos Santos Also what about other cases i.e. when $|f| > 1$?
– Mittal G
Nov 23 at 11:30
1
If $lvert frvert=Mneq0$, then $leftlvertfrac fMrightrvert=1$. So, $leftlVertfrac fMrightrVertleqslant2$, and therefore $lVert frVertleqslant2M$.
– José Carlos Santos
Nov 23 at 11:32
Great. Thanks a lot.
– Mittal G
Nov 23 at 11:38
add a comment |
up vote
0
down vote
Let $f$ be a function in the given space. It reaches the minimal and the maximal values, $m$ and $M$ say, at two points $a,b$ say inside $I=[0,1]$. Then the (only) term we have to control using $int_I |f(x)|; dx$ and $|f'|_infty$ is $|f|_infty=max(|m|,|M|)$.
In the case $m,M$ have the same sign, we may and do assume (after possibly passing from $f$ to $-f$) $0le mle M$. Then
$$ M
=m+(M-m)le int_I|f(x)|; dx+|f'(xi)|,|b-a|le |f| .
$$
In the remained case $mle 0le M$ we find an intermediate $c$ between $a$, $b$ with $f(c)=0$. So
$$
begin{aligned}
|M| &= (M-0) = |f'(xi_1)|,|b-c|le |f'|_infty le |f| , \
|m| &= (0-m) = |f'(xi_2)|,|c-a|le |f'|_infty le |f| .
end{aligned}
$$
Summing up:
$$
|f| =|f|_infty + |f'|_infty le |f| + |f'|_infty le |f|+|f|= 2|f| .
$$
answered Nov 23 at 17:40
dan_fulea
6,2401312
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
up vote
2
down vote
accepted
Suppose that$$lvert frvertleqslant1.tag1$$Then$$displaystylemax_{tin[0,1]}bigllvert f'(t)bigrrvertleqslant1.tag2$$Suppose that $displaystylemax_{tin[0,1]}bigllvert f(t)bigrrvert>2$. Then, by $(2)$ and by the Mean Value Theorem, $displaystylemin_{tin[0,1]}bigllvert f(t)bigrrvert>1$ and so $displaystyleint_0^1bigllvert f(t)bigrrvert,mathrm dt>1$. This is impossible, since we are assuming $(1)$. This proves that $lvert frvertleqslant1implieslVert frVertleqslant2$. Can you take it from here?
answered Nov 23 at 11:04
José Carlos Santos
147k22117218
What is the value of $K$?
– Mittal G
Nov 23 at 11:28
1
@MittalG $K=2$ will do.
– José Carlos Santos
Nov 23 at 11:29
@ José Carlos Santos Also what about other cases i.e. when $|f| > 1$?
– Mittal G
Nov 23 at 11:30
1
If $lvert frvert=Mneq0$, then $leftlvertfrac fMrightrvert=1$. So, $leftlVertfrac fMrightrVertleqslant2$, and therefore $lVert frVertleqslant2M$.
– José Carlos Santos
Nov 23 at 11:32
Great. Thanks a lot.
– Mittal G
Nov 23 at 11:38
add a comment |
up vote
2
down vote
accepted
Suppose that$$lvert frvertleqslant1.tag1$$Then$$displaystylemax_{tin[0,1]}bigllvert f'(t)bigrrvertleqslant1.tag2$$Suppose that $displaystylemax_{tin[0,1]}bigllvert f(t)bigrrvert>2$. Then, by $(2)$ and by the Mean Value Theorem, $displaystylemin_{tin[0,1]}bigllvert f(t)bigrrvert>1$ and so $displaystyleint_0^1bigllvert f(t)bigrrvert,mathrm dt>1$. This is impossible, since we are assuming $(1)$. This proves that $lvert frvertleqslant1implieslVert frVertleqslant2$. Can you take it from here?
answered Nov 23 at 11:04
José Carlos Santos
147k22117218
What is the value of $K$?
– Mittal G
Nov 23 at 11:28
1
@MittalG $K=2$ will do.
– José Carlos Santos
Nov 23 at 11:29
@ José Carlos Santos Also what about other cases i.e. when $|f| > 1$?
– Mittal G
Nov 23 at 11:30
1
If $lvert frvert=Mneq0$, then $leftlvertfrac fMrightrvert=1$. So, $leftlVertfrac fMrightrVertleqslant2$, and therefore $lVert frVertleqslant2M$.
– José Carlos Santos
Nov 23 at 11:32
Great. Thanks a lot.
– Mittal G
Nov 23 at 11:38
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Suppose that$$lvert frvertleqslant1.tag1$$Then$$displaystylemax_{tin[0,1]}bigllvert f'(t)bigrrvertleqslant1.tag2$$Suppose that $displaystylemax_{tin[0,1]}bigllvert f(t)bigrrvert>2$. Then, by $(2)$ and by the Mean Value Theorem, $displaystylemin_{tin[0,1]}bigllvert f(t)bigrrvert>1$ and so $displaystyleint_0^1bigllvert f(t)bigrrvert,mathrm dt>1$. This is impossible, since we are assuming $(1)$. This proves that $lvert frvertleqslant1implieslVert frVertleqslant2$. Can you take it from here?
answered Nov 23 at 11:04
José Carlos Santos
147k22117218
Suppose that$$lvert frvertleqslant1.tag1$$Then$$displaystylemax_{tin[0,1]}bigllvert f'(t)bigrrvertleqslant1.tag2$$Suppose that $displaystylemax_{tin[0,1]}bigllvert f(t)bigrrvert>2$. Then, by $(2)$ and by the Mean Value Theorem, $displaystylemin_{tin[0,1]}bigllvert f(t)bigrrvert>1$ and so $displaystyleint_0^1bigllvert f(t)bigrrvert,mathrm dt>1$. This is impossible, since we are assuming $(1)$. This proves that $lvert frvertleqslant1implieslVert frVertleqslant2$. Can you take it from here?
answered Nov 23 at 11:04
José Carlos Santos
147k22117218
answered Nov 23 at 11:04
José Carlos Santos
147k22117218
answered Nov 23 at 11:04
José Carlos Santos
147k22117218
answered Nov 23 at 11:04
José Carlos Santos
147k22117218
147k22117218
What is the value of $K$?
– Mittal G
Nov 23 at 11:28
1
@MittalG $K=2$ will do.
– José Carlos Santos
Nov 23 at 11:29
@ José Carlos Santos Also what about other cases i.e. when $|f| > 1$?
– Mittal G
Nov 23 at 11:30
1
If $lvert frvert=Mneq0$, then $leftlvertfrac fMrightrvert=1$. So, $leftlVertfrac fMrightrVertleqslant2$, and therefore $lVert frVertleqslant2M$.
– José Carlos Santos
Nov 23 at 11:32
Great. Thanks a lot.
– Mittal G
Nov 23 at 11:38
add a comment |
What is the value of $K$?
– Mittal G
Nov 23 at 11:28
1
@MittalG $K=2$ will do.
– José Carlos Santos
Nov 23 at 11:29
@ José Carlos Santos Also what about other cases i.e. when $|f| > 1$?
– Mittal G
Nov 23 at 11:30
1
If $lvert frvert=Mneq0$, then $leftlvertfrac fMrightrvert=1$. So, $leftlVertfrac fMrightrVertleqslant2$, and therefore $lVert frVertleqslant2M$.
– José Carlos Santos
Nov 23 at 11:32
Great. Thanks a lot.
– Mittal G
Nov 23 at 11:38
What is the value of $K$?
– Mittal G
Nov 23 at 11:28
What is the value of $K$?
– Mittal G
Nov 23 at 11:28
1
1
@MittalG $K=2$ will do.
– José Carlos Santos
Nov 23 at 11:29
@MittalG $K=2$ will do.
– José Carlos Santos
Nov 23 at 11:29
@ José Carlos Santos Also what about other cases i.e. when $|f| > 1$?
– Mittal G
Nov 23 at 11:30
@ José Carlos Santos Also what about other cases i.e. when $|f| > 1$?
– Mittal G
Nov 23 at 11:30
1
1
If $lvert frvert=Mneq0$, then $leftlvertfrac fMrightrvert=1$. So, $leftlVertfrac fMrightrVertleqslant2$, and therefore $lVert frVertleqslant2M$.
– José Carlos Santos
Nov 23 at 11:32
If $lvert frvert=Mneq0$, then $leftlvertfrac fMrightrvert=1$. So, $leftlVertfrac fMrightrVertleqslant2$, and therefore $lVert frVertleqslant2M$.
– José Carlos Santos
Nov 23 at 11:32
Great. Thanks a lot.
– Mittal G
Nov 23 at 11:38
Great. Thanks a lot.
– Mittal G
Nov 23 at 11:38
add a comment |
up vote
0
down vote
Let $f$ be a function in the given space. It reaches the minimal and the maximal values, $m$ and $M$ say, at two points $a,b$ say inside $I=[0,1]$. Then the (only) term we have to control using $int_I |f(x)|; dx$ and $|f'|_infty$ is $|f|_infty=max(|m|,|M|)$.
In the case $m,M$ have the same sign, we may and do assume (after possibly passing from $f$ to $-f$) $0le mle M$. Then
$$ M
=m+(M-m)le int_I|f(x)|; dx+|f'(xi)|,|b-a|le |f| .
$$
In the remained case $mle 0le M$ we find an intermediate $c$ between $a$, $b$ with $f(c)=0$. So
$$
begin{aligned}
|M| &= (M-0) = |f'(xi_1)|,|b-c|le |f'|_infty le |f| , \
|m| &= (0-m) = |f'(xi_2)|,|c-a|le |f'|_infty le |f| .
end{aligned}
$$
Summing up:
$$
|f| =|f|_infty + |f'|_infty le |f| + |f'|_infty le |f|+|f|= 2|f| .
$$
answered Nov 23 at 17:40
dan_fulea
6,2401312
add a comment |
up vote
0
down vote
Let $f$ be a function in the given space. It reaches the minimal and the maximal values, $m$ and $M$ say, at two points $a,b$ say inside $I=[0,1]$. Then the (only) term we have to control using $int_I |f(x)|; dx$ and $|f'|_infty$ is $|f|_infty=max(|m|,|M|)$.
In the case $m,M$ have the same sign, we may and do assume (after possibly passing from $f$ to $-f$) $0le mle M$. Then
$$ M
=m+(M-m)le int_I|f(x)|; dx+|f'(xi)|,|b-a|le |f| .
$$
In the remained case $mle 0le M$ we find an intermediate $c$ between $a$, $b$ with $f(c)=0$. So
$$
begin{aligned}
|M| &= (M-0) = |f'(xi_1)|,|b-c|le |f'|_infty le |f| , \
|m| &= (0-m) = |f'(xi_2)|,|c-a|le |f'|_infty le |f| .
end{aligned}
$$
Summing up:
$$
|f| =|f|_infty + |f'|_infty le |f| + |f'|_infty le |f|+|f|= 2|f| .
$$
answered Nov 23 at 17:40
dan_fulea
6,2401312
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $f$ be a function in the given space. It reaches the minimal and the maximal values, $m$ and $M$ say, at two points $a,b$ say inside $I=[0,1]$. Then the (only) term we have to control using $int_I |f(x)|; dx$ and $|f'|_infty$ is $|f|_infty=max(|m|,|M|)$.
In the case $m,M$ have the same sign, we may and do assume (after possibly passing from $f$ to $-f$) $0le mle M$. Then
$$ M
=m+(M-m)le int_I|f(x)|; dx+|f'(xi)|,|b-a|le |f| .
$$
In the remained case $mle 0le M$ we find an intermediate $c$ between $a$, $b$ with $f(c)=0$. So
$$
begin{aligned}
|M| &= (M-0) = |f'(xi_1)|,|b-c|le |f'|_infty le |f| , \
|m| &= (0-m) = |f'(xi_2)|,|c-a|le |f'|_infty le |f| .
end{aligned}
$$
Summing up:
$$
|f| =|f|_infty + |f'|_infty le |f| + |f'|_infty le |f|+|f|= 2|f| .
$$
answered Nov 23 at 17:40
dan_fulea
6,2401312
Let $f$ be a function in the given space. It reaches the minimal and the maximal values, $m$ and $M$ say, at two points $a,b$ say inside $I=[0,1]$. Then the (only) term we have to control using $int_I |f(x)|; dx$ and $|f'|_infty$ is $|f|_infty=max(|m|,|M|)$.
In the case $m,M$ have the same sign, we may and do assume (after possibly passing from $f$ to $-f$) $0le mle M$. Then
$$ M
=m+(M-m)le int_I|f(x)|; dx+|f'(xi)|,|b-a|le |f| .
$$
In the remained case $mle 0le M$ we find an intermediate $c$ between $a$, $b$ with $f(c)=0$. So
$$
begin{aligned}
|M| &= (M-0) = |f'(xi_1)|,|b-c|le |f'|_infty le |f| , \
|m| &= (0-m) = |f'(xi_2)|,|c-a|le |f'|_infty le |f| .
end{aligned}
$$
Summing up:
$$
|f| =|f|_infty + |f'|_infty le |f| + |f'|_infty le |f|+|f|= 2|f| .
$$
answered Nov 23 at 17:40
dan_fulea
6,2401312
answered Nov 23 at 17:40
dan_fulea
6,2401312
answered Nov 23 at 17:40
dan_fulea
6,2401312
answered Nov 23 at 17:40
dan_fulea
6,2401312
6,2401312
add a comment |
add a comment |
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1
The first "norm" is not a norm, please put the modulus inside the integral.
– dan_fulea
Nov 23 at 10:54
The idea is, I think, that if $max|f(t)|$ is much larger than $int |f(t)|dt$, then $max|f'(t)|$ must be large.
– Arthur
Nov 23 at 10:55
Looks like Mean Value Thm
– Richard Martin
Nov 23 at 10:55
what have you tried?
– supinf
Nov 23 at 10:57