$g_t = frac{e^{f_t}}{int e^{f_t}}$ is differentiable
$begingroup$
Let $Omega$ be a metric space and $f_0, f_1: Omega rightarrow mathbb{R}$ be any two differentiable functions. For each $t in [0,1]$, define the function $f_t = tf_1 + (1-t)f_0$. I need to show that that $g_t = frac{e^{f_t}}{int e^{f_t}}$ is differentiable.
It's clear that $f_t$ is differentiable and as a composition of differentiable functions is also differentiable, we get that $e^{f_t}$ is differentiable. Thus, this problem is easily reduced to show that $int e^{f_t}$ is differentiable.
I think I have to use the Mean Value theorem and some convergence theorem (as monotone convergence or dominated one), but I am having some trouble doing that.
I would appreciate very much if someone could help me :)
integration functions derivatives exponential-function
asked Dec 11 '18 at 10:39
Luísa BorsatoLuísa Borsato
1,501315
$endgroup$
add a comment |
$begingroup$
Let $Omega$ be a metric space and $f_0, f_1: Omega rightarrow mathbb{R}$ be any two differentiable functions. For each $t in [0,1]$, define the function $f_t = tf_1 + (1-t)f_0$. I need to show that that $g_t = frac{e^{f_t}}{int e^{f_t}}$ is differentiable.
It's clear that $f_t$ is differentiable and as a composition of differentiable functions is also differentiable, we get that $e^{f_t}$ is differentiable. Thus, this problem is easily reduced to show that $int e^{f_t}$ is differentiable.
I think I have to use the Mean Value theorem and some convergence theorem (as monotone convergence or dominated one), but I am having some trouble doing that.
I would appreciate very much if someone could help me :)
integration functions derivatives exponential-function
asked Dec 11 '18 at 10:39
Luísa BorsatoLuísa Borsato
1,501315
$endgroup$
$begingroup$
$e^{f_t}$ is differentiable, so it has best linear approximation, whereas integral is a linear operator on functions.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 11 '18 at 10:56
add a comment |
$begingroup$
Let $Omega$ be a metric space and $f_0, f_1: Omega rightarrow mathbb{R}$ be any two differentiable functions. For each $t in [0,1]$, define the function $f_t = tf_1 + (1-t)f_0$. I need to show that that $g_t = frac{e^{f_t}}{int e^{f_t}}$ is differentiable.
It's clear that $f_t$ is differentiable and as a composition of differentiable functions is also differentiable, we get that $e^{f_t}$ is differentiable. Thus, this problem is easily reduced to show that $int e^{f_t}$ is differentiable.
I think I have to use the Mean Value theorem and some convergence theorem (as monotone convergence or dominated one), but I am having some trouble doing that.
I would appreciate very much if someone could help me :)
integration functions derivatives exponential-function
asked Dec 11 '18 at 10:39
Luísa BorsatoLuísa Borsato
1,501315
$endgroup$
Let $Omega$ be a metric space and $f_0, f_1: Omega rightarrow mathbb{R}$ be any two differentiable functions. For each $t in [0,1]$, define the function $f_t = tf_1 + (1-t)f_0$. I need to show that that $g_t = frac{e^{f_t}}{int e^{f_t}}$ is differentiable.
It's clear that $f_t$ is differentiable and as a composition of differentiable functions is also differentiable, we get that $e^{f_t}$ is differentiable. Thus, this problem is easily reduced to show that $int e^{f_t}$ is differentiable.
I think I have to use the Mean Value theorem and some convergence theorem (as monotone convergence or dominated one), but I am having some trouble doing that.
I would appreciate very much if someone could help me :)
integration functions derivatives exponential-function
integration functions derivatives exponential-function
asked Dec 11 '18 at 10:39
Luísa BorsatoLuísa Borsato
1,501315
asked Dec 11 '18 at 10:39
Luísa BorsatoLuísa Borsato
1,501315
asked Dec 11 '18 at 10:39
Luísa BorsatoLuísa Borsato
1,501315
asked Dec 11 '18 at 10:39
Luísa BorsatoLuísa Borsato
1,501315
asked Dec 11 '18 at 10:39
Luísa BorsatoLuísa Borsato
1,501315
1,501315
$begingroup$
$e^{f_t}$ is differentiable, so it has best linear approximation, whereas integral is a linear operator on functions.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 11 '18 at 10:56
add a comment |
$begingroup$
$e^{f_t}$ is differentiable, so it has best linear approximation, whereas integral is a linear operator on functions.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 11 '18 at 10:56
$begingroup$
$e^{f_t}$ is differentiable, so it has best linear approximation, whereas integral is a linear operator on functions.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 11 '18 at 10:56
$begingroup$
$e^{f_t}$ is differentiable, so it has best linear approximation, whereas integral is a linear operator on functions.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 11 '18 at 10:56
add a comment |
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$begingroup$
$e^{f_t}$ is differentiable, so it has best linear approximation, whereas integral is a linear operator on functions.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 11 '18 at 10:56