Continuous dependence on initial data

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Let $L : mathbbR^d rightarrow mathbbR^d$ be a linear operator and $_infty $ be the maximum norm. Are the following conditions equivalent?



  1. For $x in mathbbR^d$ with $Lx = f$ it holds $_infty leq C _infty$.

  2. For $x, y in mathbbR^d$, with $Lx = f_1, Ly = f_2$ it holds $_infty leq C f_1 - f_2 _infty$.

Clearly taking $y = 0$ in the second statement yields $f_2 = 0$, which leads to the first statement. Is the converse true? If not, how can the second condition be modified so that the two are equivalent? I have considered the following:



beginalign
| x - y | = | x - y + Lx - Lx +Ly - Ly| &leq | Lx - x| + | Ly -y| + | Lx - Ly | \ &= | Lx - x| + | Ly -y| + | f_1 - f_2 |
endalign




differential-equations analysis numerical-methods




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  • 2




    Its just linearity, if we start with the first and the assumptions of $2$, then $L(x-y) = f_1 - f_2$, so say $X = x-y$ and $F = f_1 - f_2$ then 1 implies $$|X|_infty leq C|F|_infty$$ which is the conclusion of 2
    – Calvin Khor
    2 days ago







  • 2




    Indeed, so simple!
    – Holden
    2 days ago














up vote
1
down vote

favorite
2












Let $L : mathbbR^d rightarrow mathbbR^d$ be a linear operator and $_infty $ be the maximum norm. Are the following conditions equivalent?



  1. For $x in mathbbR^d$ with $Lx = f$ it holds $_infty leq C _infty$.

  2. For $x, y in mathbbR^d$, with $Lx = f_1, Ly = f_2$ it holds $_infty leq C f_1 - f_2 _infty$.

Clearly taking $y = 0$ in the second statement yields $f_2 = 0$, which leads to the first statement. Is the converse true? If not, how can the second condition be modified so that the two are equivalent? I have considered the following:



beginalign
| x - y | = | x - y + Lx - Lx +Ly - Ly| &leq | Lx - x| + | Ly -y| + | Lx - Ly | \ &= | Lx - x| + | Ly -y| + | f_1 - f_2 |
endalign




differential-equations analysis numerical-methods




share|cite|improve this question

















  • 2




    Its just linearity, if we start with the first and the assumptions of $2$, then $L(x-y) = f_1 - f_2$, so say $X = x-y$ and $F = f_1 - f_2$ then 1 implies $$|X|_infty leq C|F|_infty$$ which is the conclusion of 2
    – Calvin Khor
    2 days ago







  • 2




    Indeed, so simple!
    – Holden
    2 days ago












up vote
1
down vote

favorite
2









up vote
1
down vote

favorite
2






2





Let $L : mathbbR^d rightarrow mathbbR^d$ be a linear operator and $_infty $ be the maximum norm. Are the following conditions equivalent?



  1. For $x in mathbbR^d$ with $Lx = f$ it holds $_infty leq C _infty$.

  2. For $x, y in mathbbR^d$, with $Lx = f_1, Ly = f_2$ it holds $_infty leq C f_1 - f_2 _infty$.

Clearly taking $y = 0$ in the second statement yields $f_2 = 0$, which leads to the first statement. Is the converse true? If not, how can the second condition be modified so that the two are equivalent? I have considered the following:



beginalign
| x - y | = | x - y + Lx - Lx +Ly - Ly| &leq | Lx - x| + | Ly -y| + | Lx - Ly | \ &= | Lx - x| + | Ly -y| + | f_1 - f_2 |
endalign




differential-equations analysis numerical-methods




share|cite|improve this question













Let $L : mathbbR^d rightarrow mathbbR^d$ be a linear operator and $_infty $ be the maximum norm. Are the following conditions equivalent?



  1. For $x in mathbbR^d$ with $Lx = f$ it holds $_infty leq C _infty$.

  2. For $x, y in mathbbR^d$, with $Lx = f_1, Ly = f_2$ it holds $_infty leq C f_1 - f_2 _infty$.

Clearly taking $y = 0$ in the second statement yields $f_2 = 0$, which leads to the first statement. Is the converse true? If not, how can the second condition be modified so that the two are equivalent? I have considered the following:



beginalign
| x - y | = | x - y + Lx - Lx +Ly - Ly| &leq | Lx - x| + | Ly -y| + | Lx - Ly | \ &= | Lx - x| + | Ly -y| + | f_1 - f_2 |
endalign





differential-equations analysis numerical-methods





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 2 days ago
























asked 2 days ago









Holden

343112




343112







  • 2




    Its just linearity, if we start with the first and the assumptions of $2$, then $L(x-y) = f_1 - f_2$, so say $X = x-y$ and $F = f_1 - f_2$ then 1 implies $$|X|_infty leq C|F|_infty$$ which is the conclusion of 2
    – Calvin Khor
    2 days ago







  • 2




    Indeed, so simple!
    – Holden
    2 days ago












  • 2




    Its just linearity, if we start with the first and the assumptions of $2$, then $L(x-y) = f_1 - f_2$, so say $X = x-y$ and $F = f_1 - f_2$ then 1 implies $$|X|_infty leq C|F|_infty$$ which is the conclusion of 2
    – Calvin Khor
    2 days ago







  • 2




    Indeed, so simple!
    – Holden
    2 days ago







2




2




Its just linearity, if we start with the first and the assumptions of $2$, then $L(x-y) = f_1 - f_2$, so say $X = x-y$ and $F = f_1 - f_2$ then 1 implies $$|X|_infty leq C|F|_infty$$ which is the conclusion of 2
– Calvin Khor
2 days ago





Its just linearity, if we start with the first and the assumptions of $2$, then $L(x-y) = f_1 - f_2$, so say $X = x-y$ and $F = f_1 - f_2$ then 1 implies $$|X|_infty leq C|F|_infty$$ which is the conclusion of 2
– Calvin Khor
2 days ago





2




2




Indeed, so simple!
– Holden
2 days ago




Indeed, so simple!
– Holden
2 days ago










1 Answer
1






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oldest

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0
down vote













(1) $Longrightarrow$ (2):



Set



$z = x - y; tag 1$



set



$g = f_1 - f_2; tag 2$



then



$Lz = L(x - y) = Lx - Ly = f_1 - f_2 = g; tag 3$



thus, via our OP Holden's stipulated item (1),



$Vert z Vert_infty le C Vert g Vert_infty; tag 4$



therefore,



$Vert x - y Vert_infty = Vert z Vert_infty le C Vert g Vert_infty = CVert f_1 - f_2 Vert_infty, tag 5$



which is Holden's item (2).



(2) $Longrightarrow$ (1):



As Holden him/herself points out, taking



$y = 0, ; f_1 = f; tag 6$



yields



$f_2 = Ly = L0 = 0, tag 7$



whence



$Vert x Vert_infty = Vert x - y Vert_infty le C Vert f_1 - f_2 Vert_infty = C Vert f - 0 Vert_infty = C Vert f Vert_infty, tag 8$



which is Holden's item (1).



Thus the two stipulated hypotheses are, in fact, equivalent.






share|cite|improve this answer























  • You can use $...$ for small mathematical formulas so that not too much vertical space is used. Also, you don't need to number every line. Indeed in this case it only adds confusion since there are now two things with the name "1" and the name "2".
    – Calvin Khor
    2 days ago










  • @Calvin Khor: which is why I consistently referred to 'Holden's item (1)" und so weiter, But thanks for the stylistic tips; I'll bear then in mind. Cheers!
    – Robert Lewis
    2 days ago










  • Well (1) appears in the first 3 lines with different meanings without mention of OP, right? I wasn't confused though, and its not a big deal. You're welcome, have a good day
    – Calvin Khor
    2 days ago










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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote













(1) $Longrightarrow$ (2):



Set



$z = x - y; tag 1$



set



$g = f_1 - f_2; tag 2$



then



$Lz = L(x - y) = Lx - Ly = f_1 - f_2 = g; tag 3$



thus, via our OP Holden's stipulated item (1),



$Vert z Vert_infty le C Vert g Vert_infty; tag 4$



therefore,



$Vert x - y Vert_infty = Vert z Vert_infty le C Vert g Vert_infty = CVert f_1 - f_2 Vert_infty, tag 5$



which is Holden's item (2).



(2) $Longrightarrow$ (1):



As Holden him/herself points out, taking



$y = 0, ; f_1 = f; tag 6$



yields



$f_2 = Ly = L0 = 0, tag 7$



whence



$Vert x Vert_infty = Vert x - y Vert_infty le C Vert f_1 - f_2 Vert_infty = C Vert f - 0 Vert_infty = C Vert f Vert_infty, tag 8$



which is Holden's item (1).



Thus the two stipulated hypotheses are, in fact, equivalent.






share|cite|improve this answer























  • You can use $...$ for small mathematical formulas so that not too much vertical space is used. Also, you don't need to number every line. Indeed in this case it only adds confusion since there are now two things with the name "1" and the name "2".
    – Calvin Khor
    2 days ago










  • @Calvin Khor: which is why I consistently referred to 'Holden's item (1)" und so weiter, But thanks for the stylistic tips; I'll bear then in mind. Cheers!
    – Robert Lewis
    2 days ago










  • Well (1) appears in the first 3 lines with different meanings without mention of OP, right? I wasn't confused though, and its not a big deal. You're welcome, have a good day
    – Calvin Khor
    2 days ago














up vote
0
down vote













(1) $Longrightarrow$ (2):



Set



$z = x - y; tag 1$



set



$g = f_1 - f_2; tag 2$



then



$Lz = L(x - y) = Lx - Ly = f_1 - f_2 = g; tag 3$



thus, via our OP Holden's stipulated item (1),



$Vert z Vert_infty le C Vert g Vert_infty; tag 4$



therefore,



$Vert x - y Vert_infty = Vert z Vert_infty le C Vert g Vert_infty = CVert f_1 - f_2 Vert_infty, tag 5$



which is Holden's item (2).



(2) $Longrightarrow$ (1):



As Holden him/herself points out, taking



$y = 0, ; f_1 = f; tag 6$



yields



$f_2 = Ly = L0 = 0, tag 7$



whence



$Vert x Vert_infty = Vert x - y Vert_infty le C Vert f_1 - f_2 Vert_infty = C Vert f - 0 Vert_infty = C Vert f Vert_infty, tag 8$



which is Holden's item (1).



Thus the two stipulated hypotheses are, in fact, equivalent.






share|cite|improve this answer























  • You can use $...$ for small mathematical formulas so that not too much vertical space is used. Also, you don't need to number every line. Indeed in this case it only adds confusion since there are now two things with the name "1" and the name "2".
    – Calvin Khor
    2 days ago










  • @Calvin Khor: which is why I consistently referred to 'Holden's item (1)" und so weiter, But thanks for the stylistic tips; I'll bear then in mind. Cheers!
    – Robert Lewis
    2 days ago










  • Well (1) appears in the first 3 lines with different meanings without mention of OP, right? I wasn't confused though, and its not a big deal. You're welcome, have a good day
    – Calvin Khor
    2 days ago












up vote
0
down vote










up vote
0
down vote









(1) $Longrightarrow$ (2):



Set



$z = x - y; tag 1$



set



$g = f_1 - f_2; tag 2$



then



$Lz = L(x - y) = Lx - Ly = f_1 - f_2 = g; tag 3$



thus, via our OP Holden's stipulated item (1),



$Vert z Vert_infty le C Vert g Vert_infty; tag 4$



therefore,



$Vert x - y Vert_infty = Vert z Vert_infty le C Vert g Vert_infty = CVert f_1 - f_2 Vert_infty, tag 5$



which is Holden's item (2).



(2) $Longrightarrow$ (1):



As Holden him/herself points out, taking



$y = 0, ; f_1 = f; tag 6$



yields



$f_2 = Ly = L0 = 0, tag 7$



whence



$Vert x Vert_infty = Vert x - y Vert_infty le C Vert f_1 - f_2 Vert_infty = C Vert f - 0 Vert_infty = C Vert f Vert_infty, tag 8$



which is Holden's item (1).



Thus the two stipulated hypotheses are, in fact, equivalent.






share|cite|improve this answer















(1) $Longrightarrow$ (2):



Set



$z = x - y; tag 1$



set



$g = f_1 - f_2; tag 2$



then



$Lz = L(x - y) = Lx - Ly = f_1 - f_2 = g; tag 3$



thus, via our OP Holden's stipulated item (1),



$Vert z Vert_infty le C Vert g Vert_infty; tag 4$



therefore,



$Vert x - y Vert_infty = Vert z Vert_infty le C Vert g Vert_infty = CVert f_1 - f_2 Vert_infty, tag 5$



which is Holden's item (2).



(2) $Longrightarrow$ (1):



As Holden him/herself points out, taking



$y = 0, ; f_1 = f; tag 6$



yields



$f_2 = Ly = L0 = 0, tag 7$



whence



$Vert x Vert_infty = Vert x - y Vert_infty le C Vert f_1 - f_2 Vert_infty = C Vert f - 0 Vert_infty = C Vert f Vert_infty, tag 8$



which is Holden's item (1).



Thus the two stipulated hypotheses are, in fact, equivalent.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago


























answered 2 days ago









Robert Lewis

36.7k22155




36.7k22155











  • You can use $...$ for small mathematical formulas so that not too much vertical space is used. Also, you don't need to number every line. Indeed in this case it only adds confusion since there are now two things with the name "1" and the name "2".
    – Calvin Khor
    2 days ago










  • @Calvin Khor: which is why I consistently referred to 'Holden's item (1)" und so weiter, But thanks for the stylistic tips; I'll bear then in mind. Cheers!
    – Robert Lewis
    2 days ago










  • Well (1) appears in the first 3 lines with different meanings without mention of OP, right? I wasn't confused though, and its not a big deal. You're welcome, have a good day
    – Calvin Khor
    2 days ago
















  • You can use $...$ for small mathematical formulas so that not too much vertical space is used. Also, you don't need to number every line. Indeed in this case it only adds confusion since there are now two things with the name "1" and the name "2".
    – Calvin Khor
    2 days ago










  • @Calvin Khor: which is why I consistently referred to 'Holden's item (1)" und so weiter, But thanks for the stylistic tips; I'll bear then in mind. Cheers!
    – Robert Lewis
    2 days ago










  • Well (1) appears in the first 3 lines with different meanings without mention of OP, right? I wasn't confused though, and its not a big deal. You're welcome, have a good day
    – Calvin Khor
    2 days ago















You can use $...$ for small mathematical formulas so that not too much vertical space is used. Also, you don't need to number every line. Indeed in this case it only adds confusion since there are now two things with the name "1" and the name "2".
– Calvin Khor
2 days ago




You can use $...$ for small mathematical formulas so that not too much vertical space is used. Also, you don't need to number every line. Indeed in this case it only adds confusion since there are now two things with the name "1" and the name "2".
– Calvin Khor
2 days ago












@Calvin Khor: which is why I consistently referred to 'Holden's item (1)" und so weiter, But thanks for the stylistic tips; I'll bear then in mind. Cheers!
– Robert Lewis
2 days ago




@Calvin Khor: which is why I consistently referred to 'Holden's item (1)" und so weiter, But thanks for the stylistic tips; I'll bear then in mind. Cheers!
– Robert Lewis
2 days ago












Well (1) appears in the first 3 lines with different meanings without mention of OP, right? I wasn't confused though, and its not a big deal. You're welcome, have a good day
– Calvin Khor
2 days ago




Well (1) appears in the first 3 lines with different meanings without mention of OP, right? I wasn't confused though, and its not a big deal. You're welcome, have a good day
– Calvin Khor
2 days ago












 

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