Taylor expansion for Gâteaux derivative












0












$begingroup$


Let $mathbb{X}$ be a normed Space and
$
f: mathbb{X}mapstomathbb{R}
$
is twice Gâteaux differentiable (not necessary Fréchet differentiable). Is it possible to build a Taylorexpansion for $f$ in the following sense



$
f(u)= f(bar{u}) + f'(bar{u})(u-bar{u})+frac{1}{2}f
''(bar{u}+theta(u-bar{u}))(u-bar{u})^2,
$



with $thetain(0,1)$?






functional-analysis taylor-expansion gateaux-derivative







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes. Just Taylor expand the function of one real variable $g(r)=f(overline{u}+r u), $ where $rinmathbb R$.
    $endgroup$
    – Giuseppe Negro
    Jan 10 at 13:56








  • 1




    $begingroup$
    I am not really sure why this should be the same, since the second Gâteaux dirivative f'' could be discontinuosly.
    $endgroup$
    – Bara
    Jan 10 at 14:07










  • $begingroup$
    That could be a problem even if X is $mathbb R$. My point is that this general case is actually exactly the same as the one-variable case.
    $endgroup$
    – Giuseppe Negro
    Jan 11 at 14:37
















0












$begingroup$


Let $mathbb{X}$ be a normed Space and
$
f: mathbb{X}mapstomathbb{R}
$
is twice Gâteaux differentiable (not necessary Fréchet differentiable). Is it possible to build a Taylorexpansion for $f$ in the following sense



$
f(u)= f(bar{u}) + f'(bar{u})(u-bar{u})+frac{1}{2}f
''(bar{u}+theta(u-bar{u}))(u-bar{u})^2,
$



with $thetain(0,1)$?






functional-analysis taylor-expansion gateaux-derivative







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes. Just Taylor expand the function of one real variable $g(r)=f(overline{u}+r u), $ where $rinmathbb R$.
    $endgroup$
    – Giuseppe Negro
    Jan 10 at 13:56








  • 1




    $begingroup$
    I am not really sure why this should be the same, since the second Gâteaux dirivative f'' could be discontinuosly.
    $endgroup$
    – Bara
    Jan 10 at 14:07










  • $begingroup$
    That could be a problem even if X is $mathbb R$. My point is that this general case is actually exactly the same as the one-variable case.
    $endgroup$
    – Giuseppe Negro
    Jan 11 at 14:37














0












0








0





$begingroup$


Let $mathbb{X}$ be a normed Space and
$
f: mathbb{X}mapstomathbb{R}
$
is twice Gâteaux differentiable (not necessary Fréchet differentiable). Is it possible to build a Taylorexpansion for $f$ in the following sense



$
f(u)= f(bar{u}) + f'(bar{u})(u-bar{u})+frac{1}{2}f
''(bar{u}+theta(u-bar{u}))(u-bar{u})^2,
$



with $thetain(0,1)$?






functional-analysis taylor-expansion gateaux-derivative







share|cite|improve this question











$endgroup$




Let $mathbb{X}$ be a normed Space and
$
f: mathbb{X}mapstomathbb{R}
$
is twice Gâteaux differentiable (not necessary Fréchet differentiable). Is it possible to build a Taylorexpansion for $f$ in the following sense



$
f(u)= f(bar{u}) + f'(bar{u})(u-bar{u})+frac{1}{2}f
''(bar{u}+theta(u-bar{u}))(u-bar{u})^2,
$



with $thetain(0,1)$?






functional-analysis taylor-expansion gateaux-derivative




functional-analysis taylor-expansion gateaux-derivative






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 13:56







Bara

















asked Jan 10 at 13:45









BaraBara

378




378








  • 1




    $begingroup$
    Yes. Just Taylor expand the function of one real variable $g(r)=f(overline{u}+r u), $ where $rinmathbb R$.
    $endgroup$
    – Giuseppe Negro
    Jan 10 at 13:56








  • 1




    $begingroup$
    I am not really sure why this should be the same, since the second Gâteaux dirivative f'' could be discontinuosly.
    $endgroup$
    – Bara
    Jan 10 at 14:07










  • $begingroup$
    That could be a problem even if X is $mathbb R$. My point is that this general case is actually exactly the same as the one-variable case.
    $endgroup$
    – Giuseppe Negro
    Jan 11 at 14:37














  • 1




    $begingroup$
    Yes. Just Taylor expand the function of one real variable $g(r)=f(overline{u}+r u), $ where $rinmathbb R$.
    $endgroup$
    – Giuseppe Negro
    Jan 10 at 13:56








  • 1




    $begingroup$
    I am not really sure why this should be the same, since the second Gâteaux dirivative f'' could be discontinuosly.
    $endgroup$
    – Bara
    Jan 10 at 14:07










  • $begingroup$
    That could be a problem even if X is $mathbb R$. My point is that this general case is actually exactly the same as the one-variable case.
    $endgroup$
    – Giuseppe Negro
    Jan 11 at 14:37








1




1




$begingroup$
Yes. Just Taylor expand the function of one real variable $g(r)=f(overline{u}+r u), $ where $rinmathbb R$.
$endgroup$
– Giuseppe Negro
Jan 10 at 13:56






$begingroup$
Yes. Just Taylor expand the function of one real variable $g(r)=f(overline{u}+r u), $ where $rinmathbb R$.
$endgroup$
– Giuseppe Negro
Jan 10 at 13:56






1




1




$begingroup$
I am not really sure why this should be the same, since the second Gâteaux dirivative f'' could be discontinuosly.
$endgroup$
– Bara
Jan 10 at 14:07




$begingroup$
I am not really sure why this should be the same, since the second Gâteaux dirivative f'' could be discontinuosly.
$endgroup$
– Bara
Jan 10 at 14:07












$begingroup$
That could be a problem even if X is $mathbb R$. My point is that this general case is actually exactly the same as the one-variable case.
$endgroup$
– Giuseppe Negro
Jan 11 at 14:37




$begingroup$
That could be a problem even if X is $mathbb R$. My point is that this general case is actually exactly the same as the one-variable case.
$endgroup$
– Giuseppe Negro
Jan 11 at 14:37










1 Answer
1






active

oldest

votes


















1












$begingroup$

Yes, this holds. It is basically a consequence of Darboux's theorem and the mean value theorem.



As indicated by Giuseppe Negro, it is sufficient to discuss the case of real arguments. Further, by simple transformation, it is sufficient to consider the situation
$$f(0) = f'(0) = 0, quad f(1) = 1$$
and we will show the existence of $t in (0,1)$ with $f''(t) = 2$.



First, we claim that there exists $t_1 in (0,1)$ with $f''(t_1) ge 2$.
We argue by contradiction and assume $f''(t) < 2$ for all $t in (0,1)$.
The function $f'$ is differentiable, hence it has the mean value property. For every $t in (0,1)$ we have $hat t in (0,t)$ with
$$ frac{f'(t)-f'(0)}{t} = f''(hat t) < 2,$$
i.e., $f'(t) < 2 , t$. Now, the fundamental theorem of calculus yields
$$f(1) - f(0) = int_0^1 f'(t) , mathrm{d}t < int_0^1 2 , t ,mathrm{d}t < 1$$
which is a contradiction. This shows the existence of $t_1$.



Similarly, we can show the existence of $t_2 in (0,1)$ with $f''(t_2) le 2$.



If $f''(t_1) = 2$ or $f''(t_2) = 2$, we are finished. Otherwise,
$f''(t_1) > 2$ and $f''(t_2) < 2$. Hence, Darboux's theorem implies the existence of $t$ between $t_1$ and $t_2$ such that $f''(t) = 2$.






share|cite|improve this answer









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    active

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    1












    $begingroup$

    Yes, this holds. It is basically a consequence of Darboux's theorem and the mean value theorem.



    As indicated by Giuseppe Negro, it is sufficient to discuss the case of real arguments. Further, by simple transformation, it is sufficient to consider the situation
    $$f(0) = f'(0) = 0, quad f(1) = 1$$
    and we will show the existence of $t in (0,1)$ with $f''(t) = 2$.



    First, we claim that there exists $t_1 in (0,1)$ with $f''(t_1) ge 2$.
    We argue by contradiction and assume $f''(t) < 2$ for all $t in (0,1)$.
    The function $f'$ is differentiable, hence it has the mean value property. For every $t in (0,1)$ we have $hat t in (0,t)$ with
    $$ frac{f'(t)-f'(0)}{t} = f''(hat t) < 2,$$
    i.e., $f'(t) < 2 , t$. Now, the fundamental theorem of calculus yields
    $$f(1) - f(0) = int_0^1 f'(t) , mathrm{d}t < int_0^1 2 , t ,mathrm{d}t < 1$$
    which is a contradiction. This shows the existence of $t_1$.



    Similarly, we can show the existence of $t_2 in (0,1)$ with $f''(t_2) le 2$.



    If $f''(t_1) = 2$ or $f''(t_2) = 2$, we are finished. Otherwise,
    $f''(t_1) > 2$ and $f''(t_2) < 2$. Hence, Darboux's theorem implies the existence of $t$ between $t_1$ and $t_2$ such that $f''(t) = 2$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Yes, this holds. It is basically a consequence of Darboux's theorem and the mean value theorem.



      As indicated by Giuseppe Negro, it is sufficient to discuss the case of real arguments. Further, by simple transformation, it is sufficient to consider the situation
      $$f(0) = f'(0) = 0, quad f(1) = 1$$
      and we will show the existence of $t in (0,1)$ with $f''(t) = 2$.



      First, we claim that there exists $t_1 in (0,1)$ with $f''(t_1) ge 2$.
      We argue by contradiction and assume $f''(t) < 2$ for all $t in (0,1)$.
      The function $f'$ is differentiable, hence it has the mean value property. For every $t in (0,1)$ we have $hat t in (0,t)$ with
      $$ frac{f'(t)-f'(0)}{t} = f''(hat t) < 2,$$
      i.e., $f'(t) < 2 , t$. Now, the fundamental theorem of calculus yields
      $$f(1) - f(0) = int_0^1 f'(t) , mathrm{d}t < int_0^1 2 , t ,mathrm{d}t < 1$$
      which is a contradiction. This shows the existence of $t_1$.



      Similarly, we can show the existence of $t_2 in (0,1)$ with $f''(t_2) le 2$.



      If $f''(t_1) = 2$ or $f''(t_2) = 2$, we are finished. Otherwise,
      $f''(t_1) > 2$ and $f''(t_2) < 2$. Hence, Darboux's theorem implies the existence of $t$ between $t_1$ and $t_2$ such that $f''(t) = 2$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Yes, this holds. It is basically a consequence of Darboux's theorem and the mean value theorem.



        As indicated by Giuseppe Negro, it is sufficient to discuss the case of real arguments. Further, by simple transformation, it is sufficient to consider the situation
        $$f(0) = f'(0) = 0, quad f(1) = 1$$
        and we will show the existence of $t in (0,1)$ with $f''(t) = 2$.



        First, we claim that there exists $t_1 in (0,1)$ with $f''(t_1) ge 2$.
        We argue by contradiction and assume $f''(t) < 2$ for all $t in (0,1)$.
        The function $f'$ is differentiable, hence it has the mean value property. For every $t in (0,1)$ we have $hat t in (0,t)$ with
        $$ frac{f'(t)-f'(0)}{t} = f''(hat t) < 2,$$
        i.e., $f'(t) < 2 , t$. Now, the fundamental theorem of calculus yields
        $$f(1) - f(0) = int_0^1 f'(t) , mathrm{d}t < int_0^1 2 , t ,mathrm{d}t < 1$$
        which is a contradiction. This shows the existence of $t_1$.



        Similarly, we can show the existence of $t_2 in (0,1)$ with $f''(t_2) le 2$.



        If $f''(t_1) = 2$ or $f''(t_2) = 2$, we are finished. Otherwise,
        $f''(t_1) > 2$ and $f''(t_2) < 2$. Hence, Darboux's theorem implies the existence of $t$ between $t_1$ and $t_2$ such that $f''(t) = 2$.






        share|cite|improve this answer









        $endgroup$



        Yes, this holds. It is basically a consequence of Darboux's theorem and the mean value theorem.



        As indicated by Giuseppe Negro, it is sufficient to discuss the case of real arguments. Further, by simple transformation, it is sufficient to consider the situation
        $$f(0) = f'(0) = 0, quad f(1) = 1$$
        and we will show the existence of $t in (0,1)$ with $f''(t) = 2$.



        First, we claim that there exists $t_1 in (0,1)$ with $f''(t_1) ge 2$.
        We argue by contradiction and assume $f''(t) < 2$ for all $t in (0,1)$.
        The function $f'$ is differentiable, hence it has the mean value property. For every $t in (0,1)$ we have $hat t in (0,t)$ with
        $$ frac{f'(t)-f'(0)}{t} = f''(hat t) < 2,$$
        i.e., $f'(t) < 2 , t$. Now, the fundamental theorem of calculus yields
        $$f(1) - f(0) = int_0^1 f'(t) , mathrm{d}t < int_0^1 2 , t ,mathrm{d}t < 1$$
        which is a contradiction. This shows the existence of $t_1$.



        Similarly, we can show the existence of $t_2 in (0,1)$ with $f''(t_2) le 2$.



        If $f''(t_1) = 2$ or $f''(t_2) = 2$, we are finished. Otherwise,
        $f''(t_1) > 2$ and $f''(t_2) < 2$. Hence, Darboux's theorem implies the existence of $t$ between $t_1$ and $t_2$ such that $f''(t) = 2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 17 at 8:08









        gerwgerw

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