Proving that $Vertcdot Vert$ defined on $C^{1}[a,b]$ is a norm
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In proving that $Vertcdot Vert$ defined on $C^{1}[a,b]$ by $Vert{f Vert}=maxlimits_{aleq tleq b}left|f(t)right|+maxlimits_{aleq tleq b}left|dfrac{d}{dt}f(t)right|$ is a norm, I encountered a problem. It's getting the equation to satisfy the first condition.
MY WORK
Let $fin C^{1}[a,b],$ then
begin{align} Vert{f Vert}=0 &leftrightarrow maxlimits_{aleq tleq b}left|f(t)right|+maxlimits_{aleq tleq b}left|dfrac{d}{dt}f(t)right|=0 \& leftrightarrow left|f(t)right|+left|dfrac{d}{dt}f(t)right|=0,;;tin [a,b] \& leftrightarrow f(t)+dfrac{d}{dt}f(t)=0,;;tin [a,b]\& leftrightarrow f(t)=e^{-t},;;tin [a,b]end{align}
I'm not getting $f(t)=0,;;forall,tin [a,b].$ Please, where did I get it wrong? Can someone help me? As to the other two conditions, I have no problems with them.
real-analysis functional-analysis analysis
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up vote
2
down vote
favorite
In proving that $Vertcdot Vert$ defined on $C^{1}[a,b]$ by $Vert{f Vert}=maxlimits_{aleq tleq b}left|f(t)right|+maxlimits_{aleq tleq b}left|dfrac{d}{dt}f(t)right|$ is a norm, I encountered a problem. It's getting the equation to satisfy the first condition.
MY WORK
Let $fin C^{1}[a,b],$ then
begin{align} Vert{f Vert}=0 &leftrightarrow maxlimits_{aleq tleq b}left|f(t)right|+maxlimits_{aleq tleq b}left|dfrac{d}{dt}f(t)right|=0 \& leftrightarrow left|f(t)right|+left|dfrac{d}{dt}f(t)right|=0,;;tin [a,b] \& leftrightarrow f(t)+dfrac{d}{dt}f(t)=0,;;tin [a,b]\& leftrightarrow f(t)=e^{-t},;;tin [a,b]end{align}
I'm not getting $f(t)=0,;;forall,tin [a,b].$ Please, where did I get it wrong? Can someone help me? As to the other two conditions, I have no problems with them.
real-analysis functional-analysis analysis
3
The last 2nd row. $|x|+|y|=0 iff x=y =0$
– xbh
17 hours ago
@xbh: Oh, thanks! Didn't realize that!
– Mike
17 hours ago
For example $|2|+|-2| ne 2+(-2)$, so $|a|+|b|=0$ cannot be deduced from $a+b=0$.
– GEdgar
14 hours ago
@GEdgar: Thanks for that!
– Mike
12 hours ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In proving that $Vertcdot Vert$ defined on $C^{1}[a,b]$ by $Vert{f Vert}=maxlimits_{aleq tleq b}left|f(t)right|+maxlimits_{aleq tleq b}left|dfrac{d}{dt}f(t)right|$ is a norm, I encountered a problem. It's getting the equation to satisfy the first condition.
MY WORK
Let $fin C^{1}[a,b],$ then
begin{align} Vert{f Vert}=0 &leftrightarrow maxlimits_{aleq tleq b}left|f(t)right|+maxlimits_{aleq tleq b}left|dfrac{d}{dt}f(t)right|=0 \& leftrightarrow left|f(t)right|+left|dfrac{d}{dt}f(t)right|=0,;;tin [a,b] \& leftrightarrow f(t)+dfrac{d}{dt}f(t)=0,;;tin [a,b]\& leftrightarrow f(t)=e^{-t},;;tin [a,b]end{align}
I'm not getting $f(t)=0,;;forall,tin [a,b].$ Please, where did I get it wrong? Can someone help me? As to the other two conditions, I have no problems with them.
real-analysis functional-analysis analysis
In proving that $Vertcdot Vert$ defined on $C^{1}[a,b]$ by $Vert{f Vert}=maxlimits_{aleq tleq b}left|f(t)right|+maxlimits_{aleq tleq b}left|dfrac{d}{dt}f(t)right|$ is a norm, I encountered a problem. It's getting the equation to satisfy the first condition.
MY WORK
Let $fin C^{1}[a,b],$ then
begin{align} Vert{f Vert}=0 &leftrightarrow maxlimits_{aleq tleq b}left|f(t)right|+maxlimits_{aleq tleq b}left|dfrac{d}{dt}f(t)right|=0 \& leftrightarrow left|f(t)right|+left|dfrac{d}{dt}f(t)right|=0,;;tin [a,b] \& leftrightarrow f(t)+dfrac{d}{dt}f(t)=0,;;tin [a,b]\& leftrightarrow f(t)=e^{-t},;;tin [a,b]end{align}
I'm not getting $f(t)=0,;;forall,tin [a,b].$ Please, where did I get it wrong? Can someone help me? As to the other two conditions, I have no problems with them.
real-analysis functional-analysis analysis
real-analysis functional-analysis analysis
edited 17 hours ago
asked 17 hours ago
Mike
994116
994116
3
The last 2nd row. $|x|+|y|=0 iff x=y =0$
– xbh
17 hours ago
@xbh: Oh, thanks! Didn't realize that!
– Mike
17 hours ago
For example $|2|+|-2| ne 2+(-2)$, so $|a|+|b|=0$ cannot be deduced from $a+b=0$.
– GEdgar
14 hours ago
@GEdgar: Thanks for that!
– Mike
12 hours ago
add a comment |
3
The last 2nd row. $|x|+|y|=0 iff x=y =0$
– xbh
17 hours ago
@xbh: Oh, thanks! Didn't realize that!
– Mike
17 hours ago
For example $|2|+|-2| ne 2+(-2)$, so $|a|+|b|=0$ cannot be deduced from $a+b=0$.
– GEdgar
14 hours ago
@GEdgar: Thanks for that!
– Mike
12 hours ago
3
3
The last 2nd row. $|x|+|y|=0 iff x=y =0$
– xbh
17 hours ago
The last 2nd row. $|x|+|y|=0 iff x=y =0$
– xbh
17 hours ago
@xbh: Oh, thanks! Didn't realize that!
– Mike
17 hours ago
@xbh: Oh, thanks! Didn't realize that!
– Mike
17 hours ago
For example $|2|+|-2| ne 2+(-2)$, so $|a|+|b|=0$ cannot be deduced from $a+b=0$.
– GEdgar
14 hours ago
For example $|2|+|-2| ne 2+(-2)$, so $|a|+|b|=0$ cannot be deduced from $a+b=0$.
– GEdgar
14 hours ago
@GEdgar: Thanks for that!
– Mike
12 hours ago
@GEdgar: Thanks for that!
– Mike
12 hours ago
add a comment |
1 Answer
1
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up vote
5
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accepted
You dropped the absolute value bars between the second and third lines. If you have $|f(t)| + left| frac{d}{dt} f(t) right| = 0$ for all $t in [a,b]$, then it must be the case that both $f(t) = 0$ and $frac{d}{dt} f(t) = 0$. Really, the first suffices, as this gives $f(t) = 0$ for all $t in [a,b]$.
1
Thanks, platty!
– Mike
17 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
You dropped the absolute value bars between the second and third lines. If you have $|f(t)| + left| frac{d}{dt} f(t) right| = 0$ for all $t in [a,b]$, then it must be the case that both $f(t) = 0$ and $frac{d}{dt} f(t) = 0$. Really, the first suffices, as this gives $f(t) = 0$ for all $t in [a,b]$.
1
Thanks, platty!
– Mike
17 hours ago
add a comment |
up vote
5
down vote
accepted
You dropped the absolute value bars between the second and third lines. If you have $|f(t)| + left| frac{d}{dt} f(t) right| = 0$ for all $t in [a,b]$, then it must be the case that both $f(t) = 0$ and $frac{d}{dt} f(t) = 0$. Really, the first suffices, as this gives $f(t) = 0$ for all $t in [a,b]$.
1
Thanks, platty!
– Mike
17 hours ago
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
You dropped the absolute value bars between the second and third lines. If you have $|f(t)| + left| frac{d}{dt} f(t) right| = 0$ for all $t in [a,b]$, then it must be the case that both $f(t) = 0$ and $frac{d}{dt} f(t) = 0$. Really, the first suffices, as this gives $f(t) = 0$ for all $t in [a,b]$.
You dropped the absolute value bars between the second and third lines. If you have $|f(t)| + left| frac{d}{dt} f(t) right| = 0$ for all $t in [a,b]$, then it must be the case that both $f(t) = 0$ and $frac{d}{dt} f(t) = 0$. Really, the first suffices, as this gives $f(t) = 0$ for all $t in [a,b]$.
answered 17 hours ago
platty
2,323214
2,323214
1
Thanks, platty!
– Mike
17 hours ago
add a comment |
1
Thanks, platty!
– Mike
17 hours ago
1
1
Thanks, platty!
– Mike
17 hours ago
Thanks, platty!
– Mike
17 hours ago
add a comment |
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3
The last 2nd row. $|x|+|y|=0 iff x=y =0$
– xbh
17 hours ago
@xbh: Oh, thanks! Didn't realize that!
– Mike
17 hours ago
For example $|2|+|-2| ne 2+(-2)$, so $|a|+|b|=0$ cannot be deduced from $a+b=0$.
– GEdgar
14 hours ago
@GEdgar: Thanks for that!
– Mike
12 hours ago