$g_t = frac{e^{f_t}}{int e^{f_t}}$ is differentiable












0












$begingroup$


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







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$endgroup$












  • $begingroup$
    $e^{f_t}$ 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
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    $e^{f_t}$ 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














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    $e^{f_t}$ 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










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