How to conclude from a homogeneous function $f$ of degree $1$ differentiable at $0$ that all norms are never...












1












$begingroup$


Assume the fact that a homogeneous function $f$ of degree 1 (i.e. a map from $E$ to $F$ such that $f(tx) = tf(x)$ for all $t > 0$) that is differentiable at $0$ is necessarily linear.



How do we conclude from this that all norms are never differentiable at the origin?






functional-analysis differential-geometry







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  • 2




    $begingroup$
    Is the norm a homogeneous function of degree $1$? Is it linear?
    $endgroup$
    – Robert Israel
    Jan 8 at 18:00










  • $begingroup$
    @Robert Oh I see, yes the norm is a homogeneous function of deg 1 and no it is not linear.
    $endgroup$
    – metalder9
    Jan 8 at 18:32
















1












$begingroup$


Assume the fact that a homogeneous function $f$ of degree 1 (i.e. a map from $E$ to $F$ such that $f(tx) = tf(x)$ for all $t > 0$) that is differentiable at $0$ is necessarily linear.



How do we conclude from this that all norms are never differentiable at the origin?






functional-analysis differential-geometry







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Is the norm a homogeneous function of degree $1$? Is it linear?
    $endgroup$
    – Robert Israel
    Jan 8 at 18:00










  • $begingroup$
    @Robert Oh I see, yes the norm is a homogeneous function of deg 1 and no it is not linear.
    $endgroup$
    – metalder9
    Jan 8 at 18:32














1












1








1





$begingroup$


Assume the fact that a homogeneous function $f$ of degree 1 (i.e. a map from $E$ to $F$ such that $f(tx) = tf(x)$ for all $t > 0$) that is differentiable at $0$ is necessarily linear.



How do we conclude from this that all norms are never differentiable at the origin?






functional-analysis differential-geometry







share|cite|improve this question











$endgroup$




Assume the fact that a homogeneous function $f$ of degree 1 (i.e. a map from $E$ to $F$ such that $f(tx) = tf(x)$ for all $t > 0$) that is differentiable at $0$ is necessarily linear.



How do we conclude from this that all norms are never differentiable at the origin?






functional-analysis differential-geometry




functional-analysis differential-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 19:45









mechanodroid

28.9k62648




28.9k62648










asked Jan 8 at 17:54









metalder9metalder9

697




697








  • 2




    $begingroup$
    Is the norm a homogeneous function of degree $1$? Is it linear?
    $endgroup$
    – Robert Israel
    Jan 8 at 18:00










  • $begingroup$
    @Robert Oh I see, yes the norm is a homogeneous function of deg 1 and no it is not linear.
    $endgroup$
    – metalder9
    Jan 8 at 18:32














  • 2




    $begingroup$
    Is the norm a homogeneous function of degree $1$? Is it linear?
    $endgroup$
    – Robert Israel
    Jan 8 at 18:00










  • $begingroup$
    @Robert Oh I see, yes the norm is a homogeneous function of deg 1 and no it is not linear.
    $endgroup$
    – metalder9
    Jan 8 at 18:32








2




2




$begingroup$
Is the norm a homogeneous function of degree $1$? Is it linear?
$endgroup$
– Robert Israel
Jan 8 at 18:00




$begingroup$
Is the norm a homogeneous function of degree $1$? Is it linear?
$endgroup$
– Robert Israel
Jan 8 at 18:00












$begingroup$
@Robert Oh I see, yes the norm is a homogeneous function of deg 1 and no it is not linear.
$endgroup$
– metalder9
Jan 8 at 18:32




$begingroup$
@Robert Oh I see, yes the norm is a homogeneous function of deg 1 and no it is not linear.
$endgroup$
– metalder9
Jan 8 at 18:32










1 Answer
1






active

oldest

votes


















1












$begingroup$

To illustrate the problem, let's prove the claim. Let $f : E to F$ be homogeneous of degree $1$ and differentiable at the origin. Hence there exists a bounded linear map $A : E to F$ such that
$$0 = lim_{hto 0} frac{f(0+h) - f(0) - Ah}{|h|} = lim_{hto 0} frac{f(h) - Ah}{|h|}$$



Note that $f(0) = 0$, this follows from $f(0) = f(t0) = tf(0), forall t > 0$.



Fix $x ne 0$. Taking $h = tx$ for $t > 0$ we get
$$0 = lim_{tto 0^+} frac{f(tx) - A(tx)}{|tx|} = lim_{tto 0^+} frac{tf(x) - tA(x)}{t|x|} = frac{f(x)-Ax}{|x|}$$



so $f(x) = Ax$. Therefore $f equiv A$ so it is linear.



The norm $|cdot|$ is homogeneous of degree $1$ but not linear so it cannot be differentiable at the origin.






share|cite|improve this answer









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    $begingroup$

    To illustrate the problem, let's prove the claim. Let $f : E to F$ be homogeneous of degree $1$ and differentiable at the origin. Hence there exists a bounded linear map $A : E to F$ such that
    $$0 = lim_{hto 0} frac{f(0+h) - f(0) - Ah}{|h|} = lim_{hto 0} frac{f(h) - Ah}{|h|}$$



    Note that $f(0) = 0$, this follows from $f(0) = f(t0) = tf(0), forall t > 0$.



    Fix $x ne 0$. Taking $h = tx$ for $t > 0$ we get
    $$0 = lim_{tto 0^+} frac{f(tx) - A(tx)}{|tx|} = lim_{tto 0^+} frac{tf(x) - tA(x)}{t|x|} = frac{f(x)-Ax}{|x|}$$



    so $f(x) = Ax$. Therefore $f equiv A$ so it is linear.



    The norm $|cdot|$ is homogeneous of degree $1$ but not linear so it cannot be differentiable at the origin.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      To illustrate the problem, let's prove the claim. Let $f : E to F$ be homogeneous of degree $1$ and differentiable at the origin. Hence there exists a bounded linear map $A : E to F$ such that
      $$0 = lim_{hto 0} frac{f(0+h) - f(0) - Ah}{|h|} = lim_{hto 0} frac{f(h) - Ah}{|h|}$$



      Note that $f(0) = 0$, this follows from $f(0) = f(t0) = tf(0), forall t > 0$.



      Fix $x ne 0$. Taking $h = tx$ for $t > 0$ we get
      $$0 = lim_{tto 0^+} frac{f(tx) - A(tx)}{|tx|} = lim_{tto 0^+} frac{tf(x) - tA(x)}{t|x|} = frac{f(x)-Ax}{|x|}$$



      so $f(x) = Ax$. Therefore $f equiv A$ so it is linear.



      The norm $|cdot|$ is homogeneous of degree $1$ but not linear so it cannot be differentiable at the origin.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        To illustrate the problem, let's prove the claim. Let $f : E to F$ be homogeneous of degree $1$ and differentiable at the origin. Hence there exists a bounded linear map $A : E to F$ such that
        $$0 = lim_{hto 0} frac{f(0+h) - f(0) - Ah}{|h|} = lim_{hto 0} frac{f(h) - Ah}{|h|}$$



        Note that $f(0) = 0$, this follows from $f(0) = f(t0) = tf(0), forall t > 0$.



        Fix $x ne 0$. Taking $h = tx$ for $t > 0$ we get
        $$0 = lim_{tto 0^+} frac{f(tx) - A(tx)}{|tx|} = lim_{tto 0^+} frac{tf(x) - tA(x)}{t|x|} = frac{f(x)-Ax}{|x|}$$



        so $f(x) = Ax$. Therefore $f equiv A$ so it is linear.



        The norm $|cdot|$ is homogeneous of degree $1$ but not linear so it cannot be differentiable at the origin.






        share|cite|improve this answer









        $endgroup$



        To illustrate the problem, let's prove the claim. Let $f : E to F$ be homogeneous of degree $1$ and differentiable at the origin. Hence there exists a bounded linear map $A : E to F$ such that
        $$0 = lim_{hto 0} frac{f(0+h) - f(0) - Ah}{|h|} = lim_{hto 0} frac{f(h) - Ah}{|h|}$$



        Note that $f(0) = 0$, this follows from $f(0) = f(t0) = tf(0), forall t > 0$.



        Fix $x ne 0$. Taking $h = tx$ for $t > 0$ we get
        $$0 = lim_{tto 0^+} frac{f(tx) - A(tx)}{|tx|} = lim_{tto 0^+} frac{tf(x) - tA(x)}{t|x|} = frac{f(x)-Ax}{|x|}$$



        so $f(x) = Ax$. Therefore $f equiv A$ so it is linear.



        The norm $|cdot|$ is homogeneous of degree $1$ but not linear so it cannot be differentiable at the origin.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 19:43









        mechanodroidmechanodroid

        28.9k62648




        28.9k62648






























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