Proving that $Vertcdot Vert$ defined on $C^{1}[a,b]$ is a norm












2












$begingroup$


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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$endgroup$








  • 3




    $begingroup$
    The last 2nd row. $|x|+|y|=0 iff x=y =0$
    $endgroup$
    – xbh
    Nov 30 '18 at 9:44










  • $begingroup$
    @xbh: Oh, thanks! Didn't realize that!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 9:47










  • $begingroup$
    For example $|2|+|-2| ne 2+(-2)$, so $|a|+|b|=0$ cannot be deduced from $a+b=0$.
    $endgroup$
    – GEdgar
    Nov 30 '18 at 12:33










  • $begingroup$
    @GEdgar: Thanks for that!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 14:34
















2












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    The last 2nd row. $|x|+|y|=0 iff x=y =0$
    $endgroup$
    – xbh
    Nov 30 '18 at 9:44










  • $begingroup$
    @xbh: Oh, thanks! Didn't realize that!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 9:47










  • $begingroup$
    For example $|2|+|-2| ne 2+(-2)$, so $|a|+|b|=0$ cannot be deduced from $a+b=0$.
    $endgroup$
    – GEdgar
    Nov 30 '18 at 12:33










  • $begingroup$
    @GEdgar: Thanks for that!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 14:34














2












2








2


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 9:44







Omojola Micheal

















asked Nov 30 '18 at 9:43









Omojola MichealOmojola Micheal

1,753324




1,753324








  • 3




    $begingroup$
    The last 2nd row. $|x|+|y|=0 iff x=y =0$
    $endgroup$
    – xbh
    Nov 30 '18 at 9:44










  • $begingroup$
    @xbh: Oh, thanks! Didn't realize that!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 9:47










  • $begingroup$
    For example $|2|+|-2| ne 2+(-2)$, so $|a|+|b|=0$ cannot be deduced from $a+b=0$.
    $endgroup$
    – GEdgar
    Nov 30 '18 at 12:33










  • $begingroup$
    @GEdgar: Thanks for that!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 14:34














  • 3




    $begingroup$
    The last 2nd row. $|x|+|y|=0 iff x=y =0$
    $endgroup$
    – xbh
    Nov 30 '18 at 9:44










  • $begingroup$
    @xbh: Oh, thanks! Didn't realize that!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 9:47










  • $begingroup$
    For example $|2|+|-2| ne 2+(-2)$, so $|a|+|b|=0$ cannot be deduced from $a+b=0$.
    $endgroup$
    – GEdgar
    Nov 30 '18 at 12:33










  • $begingroup$
    @GEdgar: Thanks for that!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 14:34








3




3




$begingroup$
The last 2nd row. $|x|+|y|=0 iff x=y =0$
$endgroup$
– xbh
Nov 30 '18 at 9:44




$begingroup$
The last 2nd row. $|x|+|y|=0 iff x=y =0$
$endgroup$
– xbh
Nov 30 '18 at 9:44












$begingroup$
@xbh: Oh, thanks! Didn't realize that!
$endgroup$
– Omojola Micheal
Nov 30 '18 at 9:47




$begingroup$
@xbh: Oh, thanks! Didn't realize that!
$endgroup$
– Omojola Micheal
Nov 30 '18 at 9:47












$begingroup$
For example $|2|+|-2| ne 2+(-2)$, so $|a|+|b|=0$ cannot be deduced from $a+b=0$.
$endgroup$
– GEdgar
Nov 30 '18 at 12:33




$begingroup$
For example $|2|+|-2| ne 2+(-2)$, so $|a|+|b|=0$ cannot be deduced from $a+b=0$.
$endgroup$
– GEdgar
Nov 30 '18 at 12:33












$begingroup$
@GEdgar: Thanks for that!
$endgroup$
– Omojola Micheal
Nov 30 '18 at 14:34




$begingroup$
@GEdgar: Thanks for that!
$endgroup$
– Omojola Micheal
Nov 30 '18 at 14:34










1 Answer
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$begingroup$

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









$endgroup$









  • 1




    $begingroup$
    Thanks, platty!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 9:48











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1 Answer
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1 Answer
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active

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active

oldest

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active

oldest

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5












$begingroup$

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









$endgroup$









  • 1




    $begingroup$
    Thanks, platty!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 9:48
















5












$begingroup$

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









$endgroup$









  • 1




    $begingroup$
    Thanks, platty!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 9:48














5












5








5





$begingroup$

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









$endgroup$



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 Nov 30 '18 at 9:46









plattyplatty

3,370320




3,370320








  • 1




    $begingroup$
    Thanks, platty!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 9:48














  • 1




    $begingroup$
    Thanks, platty!
    $endgroup$
    – Omojola Micheal
    Nov 30 '18 at 9:48








1




1




$begingroup$
Thanks, platty!
$endgroup$
– Omojola Micheal
Nov 30 '18 at 9:48




$begingroup$
Thanks, platty!
$endgroup$
– Omojola Micheal
Nov 30 '18 at 9:48


















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