Mixing
The derivative of a real valued function whose domain in a function space
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I am struggling to follow an example of the derivative of a function whose domain is a function space. My question concerns in particular Example 1.1.6 taken from Pierre Schapira's online lecture notes for differential calculus on Banach spaces, which can be found here
https://webusers.imj-prg.fr/~pierre.schapira/lectnotes/
The passage I am struggling with is at the top of page 10; in particular, I cannot obtain the constant C with the property that $|epsilon(phi(t), psi(t))| leq C|psi(t)|$. Applying the mean value theorem he quotes above seems to yield
$|ln(phi(t)+psi(t)) - ln(phi(t)) - frac{psi(t)}{phi(t)}| leq |psi(t)|cdot M$
where $M = sup_{xi in [-1, 1]}|ln'(phi(t)+xipsi(t)) - ln'(phi(t))|$.
Now from (1.3) we have that $|ln(phi(t)+psi(t)) - ln(phi(t)) - frac{psi(t)}{phi(t)}| = |psi(t)||epsilon(phi(t), psi(t))|$
and hence
$|psi(t)||epsilon(phi(t), psi(t))| leq |psi(t)|cdot M$.
Now since $psi > 0$ this gives us $|epsilon(phi(t), psi(t))| leq M$ and I cannot seem to convert this into an expression of the form $|epsilon(phi(t), psi(t))| leq C|psi(t)|$ where C does not depend on $t$.
In my frustration, I attempted a more direct approach to the obtaining the result
$ln(phi + psi) - ln(phi) - frac{psi}{phi} = ||psi||alpha(phi, psi)$
where $alpha(phi, psi) to 0$ as $psi to 0$.
Let, as suggested, $alpha$ be the function $alpha(phi, psi)(t) = frac{|psi(t)|}{||psi||}epsilon(phi(t), psi(t))$. We have, since $frac{|psi(t)|}{||psi||}$ is bounded by 1, that
$|alpha(phi, psi)(t)| leq |epsilon(phi(t), psi(t))|$.
Now since $|psi(t)| leq ||psi||$ it seems clear that as $||psi|| to 0$ we must have $|psi(t)| to 0$ independently of $t$ and thus, by the definition of $epsilon(phi(t), psi(t))$ we have that
$|alpha(phi, psi)(t)| leq |epsilon(phi(t), psi(t))| to 0$ as $||psi|| to 0$
as required.
I've been playing around with this for a while now and I'm sure my error will be suitably embarrassing, but any help with either my original problem or comments regarding my alternative would be greatly appreciated.
Thanks in advance.
calculus analysis banach-spaces
asked Jan 18 at 16:00
SamSam
515
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add a comment |
$begingroup$
I am struggling to follow an example of the derivative of a function whose domain is a function space. My question concerns in particular Example 1.1.6 taken from Pierre Schapira's online lecture notes for differential calculus on Banach spaces, which can be found here
https://webusers.imj-prg.fr/~pierre.schapira/lectnotes/
The passage I am struggling with is at the top of page 10; in particular, I cannot obtain the constant C with the property that $|epsilon(phi(t), psi(t))| leq C|psi(t)|$. Applying the mean value theorem he quotes above seems to yield
$|ln(phi(t)+psi(t)) - ln(phi(t)) - frac{psi(t)}{phi(t)}| leq |psi(t)|cdot M$
where $M = sup_{xi in [-1, 1]}|ln'(phi(t)+xipsi(t)) - ln'(phi(t))|$.
Now from (1.3) we have that $|ln(phi(t)+psi(t)) - ln(phi(t)) - frac{psi(t)}{phi(t)}| = |psi(t)||epsilon(phi(t), psi(t))|$
and hence
$|psi(t)||epsilon(phi(t), psi(t))| leq |psi(t)|cdot M$.
Now since $psi > 0$ this gives us $|epsilon(phi(t), psi(t))| leq M$ and I cannot seem to convert this into an expression of the form $|epsilon(phi(t), psi(t))| leq C|psi(t)|$ where C does not depend on $t$.
In my frustration, I attempted a more direct approach to the obtaining the result
$ln(phi + psi) - ln(phi) - frac{psi}{phi} = ||psi||alpha(phi, psi)$
where $alpha(phi, psi) to 0$ as $psi to 0$.
Let, as suggested, $alpha$ be the function $alpha(phi, psi)(t) = frac{|psi(t)|}{||psi||}epsilon(phi(t), psi(t))$. We have, since $frac{|psi(t)|}{||psi||}$ is bounded by 1, that
$|alpha(phi, psi)(t)| leq |epsilon(phi(t), psi(t))|$.
Now since $|psi(t)| leq ||psi||$ it seems clear that as $||psi|| to 0$ we must have $|psi(t)| to 0$ independently of $t$ and thus, by the definition of $epsilon(phi(t), psi(t))$ we have that
$|alpha(phi, psi)(t)| leq |epsilon(phi(t), psi(t))| to 0$ as $||psi|| to 0$
as required.
I've been playing around with this for a while now and I'm sure my error will be suitably embarrassing, but any help with either my original problem or comments regarding my alternative would be greatly appreciated.
Thanks in advance.
calculus analysis banach-spaces
asked Jan 18 at 16:00
SamSam
515
$endgroup$
$begingroup$
I don't understand any of this sort of mathematics, but +1 for a question with lots of faithful work shown.
$endgroup$
– Randall
Jan 18 at 16:02
add a comment |
$begingroup$
I am struggling to follow an example of the derivative of a function whose domain is a function space. My question concerns in particular Example 1.1.6 taken from Pierre Schapira's online lecture notes for differential calculus on Banach spaces, which can be found here
https://webusers.imj-prg.fr/~pierre.schapira/lectnotes/
The passage I am struggling with is at the top of page 10; in particular, I cannot obtain the constant C with the property that $|epsilon(phi(t), psi(t))| leq C|psi(t)|$. Applying the mean value theorem he quotes above seems to yield
$|ln(phi(t)+psi(t)) - ln(phi(t)) - frac{psi(t)}{phi(t)}| leq |psi(t)|cdot M$
where $M = sup_{xi in [-1, 1]}|ln'(phi(t)+xipsi(t)) - ln'(phi(t))|$.
Now from (1.3) we have that $|ln(phi(t)+psi(t)) - ln(phi(t)) - frac{psi(t)}{phi(t)}| = |psi(t)||epsilon(phi(t), psi(t))|$
and hence
$|psi(t)||epsilon(phi(t), psi(t))| leq |psi(t)|cdot M$.
Now since $psi > 0$ this gives us $|epsilon(phi(t), psi(t))| leq M$ and I cannot seem to convert this into an expression of the form $|epsilon(phi(t), psi(t))| leq C|psi(t)|$ where C does not depend on $t$.
In my frustration, I attempted a more direct approach to the obtaining the result
$ln(phi + psi) - ln(phi) - frac{psi}{phi} = ||psi||alpha(phi, psi)$
where $alpha(phi, psi) to 0$ as $psi to 0$.
Let, as suggested, $alpha$ be the function $alpha(phi, psi)(t) = frac{|psi(t)|}{||psi||}epsilon(phi(t), psi(t))$. We have, since $frac{|psi(t)|}{||psi||}$ is bounded by 1, that
$|alpha(phi, psi)(t)| leq |epsilon(phi(t), psi(t))|$.
Now since $|psi(t)| leq ||psi||$ it seems clear that as $||psi|| to 0$ we must have $|psi(t)| to 0$ independently of $t$ and thus, by the definition of $epsilon(phi(t), psi(t))$ we have that
$|alpha(phi, psi)(t)| leq |epsilon(phi(t), psi(t))| to 0$ as $||psi|| to 0$
as required.
I've been playing around with this for a while now and I'm sure my error will be suitably embarrassing, but any help with either my original problem or comments regarding my alternative would be greatly appreciated.
Thanks in advance.
calculus analysis banach-spaces
asked Jan 18 at 16:00
SamSam
515
$endgroup$
I am struggling to follow an example of the derivative of a function whose domain is a function space. My question concerns in particular Example 1.1.6 taken from Pierre Schapira's online lecture notes for differential calculus on Banach spaces, which can be found here
https://webusers.imj-prg.fr/~pierre.schapira/lectnotes/
The passage I am struggling with is at the top of page 10; in particular, I cannot obtain the constant C with the property that $|epsilon(phi(t), psi(t))| leq C|psi(t)|$. Applying the mean value theorem he quotes above seems to yield
$|ln(phi(t)+psi(t)) - ln(phi(t)) - frac{psi(t)}{phi(t)}| leq |psi(t)|cdot M$
where $M = sup_{xi in [-1, 1]}|ln'(phi(t)+xipsi(t)) - ln'(phi(t))|$.
Now from (1.3) we have that $|ln(phi(t)+psi(t)) - ln(phi(t)) - frac{psi(t)}{phi(t)}| = |psi(t)||epsilon(phi(t), psi(t))|$
and hence
$|psi(t)||epsilon(phi(t), psi(t))| leq |psi(t)|cdot M$.
Now since $psi > 0$ this gives us $|epsilon(phi(t), psi(t))| leq M$ and I cannot seem to convert this into an expression of the form $|epsilon(phi(t), psi(t))| leq C|psi(t)|$ where C does not depend on $t$.
In my frustration, I attempted a more direct approach to the obtaining the result
$ln(phi + psi) - ln(phi) - frac{psi}{phi} = ||psi||alpha(phi, psi)$
where $alpha(phi, psi) to 0$ as $psi to 0$.
Let, as suggested, $alpha$ be the function $alpha(phi, psi)(t) = frac{|psi(t)|}{||psi||}epsilon(phi(t), psi(t))$. We have, since $frac{|psi(t)|}{||psi||}$ is bounded by 1, that
$|alpha(phi, psi)(t)| leq |epsilon(phi(t), psi(t))|$.
Now since $|psi(t)| leq ||psi||$ it seems clear that as $||psi|| to 0$ we must have $|psi(t)| to 0$ independently of $t$ and thus, by the definition of $epsilon(phi(t), psi(t))$ we have that
$|alpha(phi, psi)(t)| leq |epsilon(phi(t), psi(t))| to 0$ as $||psi|| to 0$
as required.
I've been playing around with this for a while now and I'm sure my error will be suitably embarrassing, but any help with either my original problem or comments regarding my alternative would be greatly appreciated.
Thanks in advance.
calculus analysis banach-spaces
calculus analysis banach-spaces
asked Jan 18 at 16:00
SamSam
515
asked Jan 18 at 16:00
SamSam
515
asked Jan 18 at 16:00
SamSam
515
asked Jan 18 at 16:00
SamSam
515
asked Jan 18 at 16:00
SamSam
515
515
$begingroup$
I don't understand any of this sort of mathematics, but +1 for a question with lots of faithful work shown.
$endgroup$
– Randall
Jan 18 at 16:02
add a comment |
$begingroup$
I don't understand any of this sort of mathematics, but +1 for a question with lots of faithful work shown.
$endgroup$
– Randall
Jan 18 at 16:02
$begingroup$
I don't understand any of this sort of mathematics, but +1 for a question with lots of faithful work shown.
$endgroup$
– Randall
Jan 18 at 16:02
$begingroup$
I don't understand any of this sort of mathematics, but +1 for a question with lots of faithful work shown.
$endgroup$
– Randall
Jan 18 at 16:02
add a comment |
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$begingroup$
I don't understand any of this sort of mathematics, but +1 for a question with lots of faithful work shown.
$endgroup$
– Randall
Jan 18 at 16:02