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.










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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















up vote
2
down vote

favorite
1












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.










share|cite|improve this question




















  • 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













up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





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.










share|cite|improve this question















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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edited 17 hours ago

























asked 17 hours ago









Mike

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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














  • 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










1 Answer
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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 $t in [a,b]$.






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    Thanks, platty!
    – Mike
    17 hours ago











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1 Answer
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active

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1 Answer
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active

oldest

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oldest

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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 $t in [a,b]$.






share|cite|improve this answer

















  • 1




    Thanks, platty!
    – Mike
    17 hours ago















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 $t in [a,b]$.






share|cite|improve this answer

















  • 1




    Thanks, platty!
    – Mike
    17 hours ago













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 $t in [a,b]$.






share|cite|improve this answer












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 $t in [a,b]$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 17 hours ago









platty

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  • 1




    Thanks, platty!
    – Mike
    17 hours ago














  • 1




    Thanks, platty!
    – Mike
    17 hours ago








1




1




Thanks, platty!
– Mike
17 hours ago




Thanks, platty!
– Mike
17 hours ago


















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