Regarding equivalent conditions of Frechet differentiability
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Gateaux and Frechet differentiability in a Banach space are defined as below.
Can you tell below how (ii) implies (i). The rest is easy.
functional-analysis banach-spaces
asked Jan 21 at 12:45
user31459user31459
284
$endgroup$
add a comment |
$begingroup$
Gateaux and Frechet differentiability in a Banach space are defined as below.
Can you tell below how (ii) implies (i). The rest is easy.
functional-analysis banach-spaces
asked Jan 21 at 12:45
user31459user31459
284
$endgroup$
$begingroup$
The equivalence of (i) and (ii) is just your given definition of Fréchet differentiability, isn't it?
$endgroup$
– gerw
Jan 21 at 13:49
$begingroup$
In the above, what is $S_X$?
$endgroup$
– kimchi lover
Jan 21 at 19:41
$begingroup$
I would guess that $S_X$ is the unit sphere of $X$, i.e., $S_X = {x in X mid |x|=1}$.
$endgroup$
– gerw
Jan 22 at 6:42
add a comment |
$begingroup$
Gateaux and Frechet differentiability in a Banach space are defined as below.
Can you tell below how (ii) implies (i). The rest is easy.
functional-analysis banach-spaces
asked Jan 21 at 12:45
user31459user31459
284
$endgroup$
Gateaux and Frechet differentiability in a Banach space are defined as below.
Can you tell below how (ii) implies (i). The rest is easy.
functional-analysis banach-spaces
functional-analysis banach-spaces
asked Jan 21 at 12:45
user31459user31459
284
asked Jan 21 at 12:45
user31459user31459
284
asked Jan 21 at 12:45
user31459user31459
284
asked Jan 21 at 12:45
user31459user31459
284
asked Jan 21 at 12:45
user31459user31459
284
284
$begingroup$
The equivalence of (i) and (ii) is just your given definition of Fréchet differentiability, isn't it?
$endgroup$
– gerw
Jan 21 at 13:49
$begingroup$
In the above, what is $S_X$?
$endgroup$
– kimchi lover
Jan 21 at 19:41
$begingroup$
I would guess that $S_X$ is the unit sphere of $X$, i.e., $S_X = {x in X mid |x|=1}$.
$endgroup$
– gerw
Jan 22 at 6:42
add a comment |
$begingroup$
The equivalence of (i) and (ii) is just your given definition of Fréchet differentiability, isn't it?
$endgroup$
– gerw
Jan 21 at 13:49
$begingroup$
In the above, what is $S_X$?
$endgroup$
– kimchi lover
Jan 21 at 19:41
$begingroup$
I would guess that $S_X$ is the unit sphere of $X$, i.e., $S_X = {x in X mid |x|=1}$.
$endgroup$
– gerw
Jan 22 at 6:42
$begingroup$
The equivalence of (i) and (ii) is just your given definition of Fréchet differentiability, isn't it?
$endgroup$
– gerw
Jan 21 at 13:49
$begingroup$
The equivalence of (i) and (ii) is just your given definition of Fréchet differentiability, isn't it?
$endgroup$
– gerw
Jan 21 at 13:49
$begingroup$
In the above, what is $S_X$?
$endgroup$
– kimchi lover
Jan 21 at 19:41
$begingroup$
In the above, what is $S_X$?
$endgroup$
– kimchi lover
Jan 21 at 19:41
$begingroup$
I would guess that $S_X$ is the unit sphere of $X$, i.e., $S_X = {x in X mid |x|=1}$.
$endgroup$
– gerw
Jan 22 at 6:42
$begingroup$
I would guess that $S_X$ is the unit sphere of $X$, i.e., $S_X = {x in X mid |x|=1}$.
$endgroup$
– gerw
Jan 22 at 6:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let me denote $f(x):=|x|$ and
the limit
$$
f'(x;h) = lim_{tto0} frac1t{f(x+th)-f(x)} quad xin S_X,hin X.
$$
Due to (ii), this limit exists for all such $x,t$.
First, let me show that $hmapsto f'(x;h)$ is a linear map. Let $hin X$, $sin mathbb Rsetminus{0}$ be given. Then
$$
f'(x; sh) = lim_{tto0} frac1t{f(x+sth)-f(x)} = s lim_{tto0} frac1{st}{f(x+sth)-f(x)} = s f'(x;h).
$$
Let now $h_1,h_2$ be given. Then by convexity (or triangle inequality)
$$
|x + frac12t(h_1+h_2)| le frac12 |x+th_1| + frac12 |x+th_2|.
$$
Subtracting $|x|$ on both sides, dividing by $t>0$, passing to the limit $tsearrow0$ yields
$$
f'(x,frac12(h_1+h_2))le frac12 f'(x,h_1+h_2)+frac12 f'(x,h_1+h_2),
$$
which implies $f'(x;h_1+h_2) le f'(x;h_1)+f'(x;h_2)$. Replacing $h_i$ by $-h_i$ implies the reverse inequality, hence $f'(x;h_1+h_2) = f'(x;h_1)+f'(x;h_2)$.
This proves Gateaux differentiability of the norm.
The uniformity of the limit then gives the required $o$-property to obtain Frechet differentiability.
answered Jan 23 at 7:38
dawdaw
24.6k1645
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1 Answer
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1 Answer
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$begingroup$
Let me denote $f(x):=|x|$ and
the limit
$$
f'(x;h) = lim_{tto0} frac1t{f(x+th)-f(x)} quad xin S_X,hin X.
$$
Due to (ii), this limit exists for all such $x,t$.
First, let me show that $hmapsto f'(x;h)$ is a linear map. Let $hin X$, $sin mathbb Rsetminus{0}$ be given. Then
$$
f'(x; sh) = lim_{tto0} frac1t{f(x+sth)-f(x)} = s lim_{tto0} frac1{st}{f(x+sth)-f(x)} = s f'(x;h).
$$
Let now $h_1,h_2$ be given. Then by convexity (or triangle inequality)
$$
|x + frac12t(h_1+h_2)| le frac12 |x+th_1| + frac12 |x+th_2|.
$$
Subtracting $|x|$ on both sides, dividing by $t>0$, passing to the limit $tsearrow0$ yields
$$
f'(x,frac12(h_1+h_2))le frac12 f'(x,h_1+h_2)+frac12 f'(x,h_1+h_2),
$$
which implies $f'(x;h_1+h_2) le f'(x;h_1)+f'(x;h_2)$. Replacing $h_i$ by $-h_i$ implies the reverse inequality, hence $f'(x;h_1+h_2) = f'(x;h_1)+f'(x;h_2)$.
This proves Gateaux differentiability of the norm.
The uniformity of the limit then gives the required $o$-property to obtain Frechet differentiability.
answered Jan 23 at 7:38
dawdaw
24.6k1645
$endgroup$
add a comment |
$begingroup$
Let me denote $f(x):=|x|$ and
the limit
$$
f'(x;h) = lim_{tto0} frac1t{f(x+th)-f(x)} quad xin S_X,hin X.
$$
Due to (ii), this limit exists for all such $x,t$.
First, let me show that $hmapsto f'(x;h)$ is a linear map. Let $hin X$, $sin mathbb Rsetminus{0}$ be given. Then
$$
f'(x; sh) = lim_{tto0} frac1t{f(x+sth)-f(x)} = s lim_{tto0} frac1{st}{f(x+sth)-f(x)} = s f'(x;h).
$$
Let now $h_1,h_2$ be given. Then by convexity (or triangle inequality)
$$
|x + frac12t(h_1+h_2)| le frac12 |x+th_1| + frac12 |x+th_2|.
$$
Subtracting $|x|$ on both sides, dividing by $t>0$, passing to the limit $tsearrow0$ yields
$$
f'(x,frac12(h_1+h_2))le frac12 f'(x,h_1+h_2)+frac12 f'(x,h_1+h_2),
$$
which implies $f'(x;h_1+h_2) le f'(x;h_1)+f'(x;h_2)$. Replacing $h_i$ by $-h_i$ implies the reverse inequality, hence $f'(x;h_1+h_2) = f'(x;h_1)+f'(x;h_2)$.
This proves Gateaux differentiability of the norm.
The uniformity of the limit then gives the required $o$-property to obtain Frechet differentiability.
answered Jan 23 at 7:38
dawdaw
24.6k1645
$endgroup$
add a comment |
$begingroup$
Let me denote $f(x):=|x|$ and
the limit
$$
f'(x;h) = lim_{tto0} frac1t{f(x+th)-f(x)} quad xin S_X,hin X.
$$
Due to (ii), this limit exists for all such $x,t$.
First, let me show that $hmapsto f'(x;h)$ is a linear map. Let $hin X$, $sin mathbb Rsetminus{0}$ be given. Then
$$
f'(x; sh) = lim_{tto0} frac1t{f(x+sth)-f(x)} = s lim_{tto0} frac1{st}{f(x+sth)-f(x)} = s f'(x;h).
$$
Let now $h_1,h_2$ be given. Then by convexity (or triangle inequality)
$$
|x + frac12t(h_1+h_2)| le frac12 |x+th_1| + frac12 |x+th_2|.
$$
Subtracting $|x|$ on both sides, dividing by $t>0$, passing to the limit $tsearrow0$ yields
$$
f'(x,frac12(h_1+h_2))le frac12 f'(x,h_1+h_2)+frac12 f'(x,h_1+h_2),
$$
which implies $f'(x;h_1+h_2) le f'(x;h_1)+f'(x;h_2)$. Replacing $h_i$ by $-h_i$ implies the reverse inequality, hence $f'(x;h_1+h_2) = f'(x;h_1)+f'(x;h_2)$.
This proves Gateaux differentiability of the norm.
The uniformity of the limit then gives the required $o$-property to obtain Frechet differentiability.
answered Jan 23 at 7:38
dawdaw
24.6k1645
$endgroup$
Let me denote $f(x):=|x|$ and
the limit
$$
f'(x;h) = lim_{tto0} frac1t{f(x+th)-f(x)} quad xin S_X,hin X.
$$
Due to (ii), this limit exists for all such $x,t$.
First, let me show that $hmapsto f'(x;h)$ is a linear map. Let $hin X$, $sin mathbb Rsetminus{0}$ be given. Then
$$
f'(x; sh) = lim_{tto0} frac1t{f(x+sth)-f(x)} = s lim_{tto0} frac1{st}{f(x+sth)-f(x)} = s f'(x;h).
$$
Let now $h_1,h_2$ be given. Then by convexity (or triangle inequality)
$$
|x + frac12t(h_1+h_2)| le frac12 |x+th_1| + frac12 |x+th_2|.
$$
Subtracting $|x|$ on both sides, dividing by $t>0$, passing to the limit $tsearrow0$ yields
$$
f'(x,frac12(h_1+h_2))le frac12 f'(x,h_1+h_2)+frac12 f'(x,h_1+h_2),
$$
which implies $f'(x;h_1+h_2) le f'(x;h_1)+f'(x;h_2)$. Replacing $h_i$ by $-h_i$ implies the reverse inequality, hence $f'(x;h_1+h_2) = f'(x;h_1)+f'(x;h_2)$.
This proves Gateaux differentiability of the norm.
The uniformity of the limit then gives the required $o$-property to obtain Frechet differentiability.
answered Jan 23 at 7:38
dawdaw
24.6k1645
answered Jan 23 at 7:38
dawdaw
24.6k1645
answered Jan 23 at 7:38
dawdaw
24.6k1645
answered Jan 23 at 7:38
dawdaw
24.6k1645
24.6k1645
add a comment |
add a comment |
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$begingroup$
The equivalence of (i) and (ii) is just your given definition of Fréchet differentiability, isn't it?
$endgroup$
– gerw
Jan 21 at 13:49
$begingroup$
In the above, what is $S_X$?
$endgroup$
– kimchi lover
Jan 21 at 19:41
$begingroup$
I would guess that $S_X$ is the unit sphere of $X$, i.e., $S_X = {x in X mid |x|=1}$.
$endgroup$
– gerw
Jan 22 at 6:42