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Differentiability of $bf x$ uniform in $bf y$
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Consider the real-valued function $f(bf x,bf y)$, with $bf x$ and $bf y $ vectors of length $n$ and $m$, respectively.
Suppose (1) $f$ is differentiable in $bf x$ at $(bf x_0, bf y)$ with $bf y$ on a neighborhood of $bf y_0$.
Question: Under which additional condition(s) is there a neighborhood $mathcalB$ of $bf y$ for which the differentiability in $x$ is uniform, that is,
$$ sup_bf y in mathcalB fracf(bf x, bf y) - f(bf x_0, bf y) - (bf x- bf x_0)bf J(bf x_0,bf y)rightbf x - bf x_0 = sup_bf y in mathcalB|R(bf x,bf y)| = o(1) $$
as $bf x to bf x_0$.
I know sufficient additional conditions are that (2) $f$ be differentiable in a neighborhood of $(bf x_0, bf y_0)$, and (3) $bf J(bf x,bf y)$ is continuous in $bf x$ at $bf x_0$ and with $bf y$ on a neighborhood of $bf y_0$
In this case, we have, through the mean value theorem,
$$sup_bf y in mathcalB|R(bf x,bf y)| = sup_bf y in mathcalBfrac leftbf x - bf x_0 = O( sup_bf y in mathcalBleft|bf Jleft(tildebf x,bf y) -bf J(bf x_0,bf y)right|right) =o(1)$$
where $tildebf x$ is between $bf x$ and $bf x_0$.
However I wonder if assumption (2) is necessarily. It was only used to apply the MVT, and a proof not requiring it could probably rely on assumption (1) and (3) only.
derivatives partial-derivative uniform-continuity
asked Jul 16 at 6:31
Guillaume F.
351211
add a comment |Â
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Consider the real-valued function $f(bf x,bf y)$, with $bf x$ and $bf y $ vectors of length $n$ and $m$, respectively.
Suppose (1) $f$ is differentiable in $bf x$ at $(bf x_0, bf y)$ with $bf y$ on a neighborhood of $bf y_0$.
Question: Under which additional condition(s) is there a neighborhood $mathcalB$ of $bf y$ for which the differentiability in $x$ is uniform, that is,
$$ sup_bf y in mathcalB fracf(bf x, bf y) - f(bf x_0, bf y) - (bf x- bf x_0)bf J(bf x_0,bf y)rightbf x - bf x_0 = sup_bf y in mathcalB|R(bf x,bf y)| = o(1) $$
as $bf x to bf x_0$.
I know sufficient additional conditions are that (2) $f$ be differentiable in a neighborhood of $(bf x_0, bf y_0)$, and (3) $bf J(bf x,bf y)$ is continuous in $bf x$ at $bf x_0$ and with $bf y$ on a neighborhood of $bf y_0$
In this case, we have, through the mean value theorem,
$$sup_bf y in mathcalB|R(bf x,bf y)| = sup_bf y in mathcalBfrac leftbf x - bf x_0 = O( sup_bf y in mathcalBleft|bf Jleft(tildebf x,bf y) -bf J(bf x_0,bf y)right|right) =o(1)$$
where $tildebf x$ is between $bf x$ and $bf x_0$.
However I wonder if assumption (2) is necessarily. It was only used to apply the MVT, and a proof not requiring it could probably rely on assumption (1) and (3) only.
derivatives partial-derivative uniform-continuity
asked Jul 16 at 6:31
Guillaume F.
351211
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider the real-valued function $f(bf x,bf y)$, with $bf x$ and $bf y $ vectors of length $n$ and $m$, respectively.
Suppose (1) $f$ is differentiable in $bf x$ at $(bf x_0, bf y)$ with $bf y$ on a neighborhood of $bf y_0$.
Question: Under which additional condition(s) is there a neighborhood $mathcalB$ of $bf y$ for which the differentiability in $x$ is uniform, that is,
$$ sup_bf y in mathcalB fracf(bf x, bf y) - f(bf x_0, bf y) - (bf x- bf x_0)bf J(bf x_0,bf y)rightbf x - bf x_0 = sup_bf y in mathcalB|R(bf x,bf y)| = o(1) $$
as $bf x to bf x_0$.
I know sufficient additional conditions are that (2) $f$ be differentiable in a neighborhood of $(bf x_0, bf y_0)$, and (3) $bf J(bf x,bf y)$ is continuous in $bf x$ at $bf x_0$ and with $bf y$ on a neighborhood of $bf y_0$
In this case, we have, through the mean value theorem,
$$sup_bf y in mathcalB|R(bf x,bf y)| = sup_bf y in mathcalBfrac leftbf x - bf x_0 = O( sup_bf y in mathcalBleft|bf Jleft(tildebf x,bf y) -bf J(bf x_0,bf y)right|right) =o(1)$$
where $tildebf x$ is between $bf x$ and $bf x_0$.
However I wonder if assumption (2) is necessarily. It was only used to apply the MVT, and a proof not requiring it could probably rely on assumption (1) and (3) only.
derivatives partial-derivative uniform-continuity
asked Jul 16 at 6:31
Guillaume F.
351211
Consider the real-valued function $f(bf x,bf y)$, with $bf x$ and $bf y $ vectors of length $n$ and $m$, respectively.
Suppose (1) $f$ is differentiable in $bf x$ at $(bf x_0, bf y)$ with $bf y$ on a neighborhood of $bf y_0$.
Question: Under which additional condition(s) is there a neighborhood $mathcalB$ of $bf y$ for which the differentiability in $x$ is uniform, that is,
$$ sup_bf y in mathcalB fracf(bf x, bf y) - f(bf x_0, bf y) - (bf x- bf x_0)bf J(bf x_0,bf y)rightbf x - bf x_0 = sup_bf y in mathcalB|R(bf x,bf y)| = o(1) $$
as $bf x to bf x_0$.
I know sufficient additional conditions are that (2) $f$ be differentiable in a neighborhood of $(bf x_0, bf y_0)$, and (3) $bf J(bf x,bf y)$ is continuous in $bf x$ at $bf x_0$ and with $bf y$ on a neighborhood of $bf y_0$
In this case, we have, through the mean value theorem,
$$sup_bf y in mathcalB|R(bf x,bf y)| = sup_bf y in mathcalBfrac leftbf x - bf x_0 = O( sup_bf y in mathcalBleft|bf Jleft(tildebf x,bf y) -bf J(bf x_0,bf y)right|right) =o(1)$$
where $tildebf x$ is between $bf x$ and $bf x_0$.
However I wonder if assumption (2) is necessarily. It was only used to apply the MVT, and a proof not requiring it could probably rely on assumption (1) and (3) only.
derivatives partial-derivative uniform-continuity
asked Jul 16 at 6:31
Guillaume F.
351211
edited Jul 16 at 6:40
asked Jul 16 at 6:31
Guillaume F.
351211
asked Jul 16 at 6:31
Guillaume F.
351211
asked Jul 16 at 6:31
Guillaume F.
351211
351211
add a comment |Â
add a comment |Â
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