Calculate the derivative of $f(x) = | x|^2.$












0














Suppose we have a normed space $E$ with the induced inner product and a map $f:Eto R$ such that $f(x) = |x|^2.$ I want to compute the derivative of this map.



We have that
$$f(x+h)=|x|^2+|h|^2+2langle x,hrangle=f(x)+o(|h|)+df_{x}(h).$$



Clearly the derivative $df_x(h) = 2langle x,hrangle$ is linear in $h.$ We also want to show that $df_x(h)$ is continuous in $h$. This can be done by simply observing that $$|df_x(h)|leq 2|x|cdot |h|$$ and therefore
$$|df_x|leq 2|x|$$ implying that $df_x$ is continuous in $h.$



I also want to show that the derivative $df_x$ is $C^{1}.$ For this we have to show that the map $xto df_x$ is continuous. Therefore observe that for each $hin E$ we have that,
$$|df_x(h)-df_y(h)|=2|langle x-y,hrangle|leq |x-y|cdot2|h|.$$
I am not sure how to proceed after this step. Any hints will be much appreciated.










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  • 1




    What is a inner product induced by a norm? Most norms don't come from inner products.
    – José Carlos Santos
    Nov 26 at 17:43






  • 1




    Oh I think I meant the other way around.
    – Hello_World
    Nov 26 at 18:01










  • If you fix $varepsilon > 0$ and choose $delta = varepsilon/2lvert h rvert$ you are done: $2lVert h rVert cdot lVert x-y rVert < varepsilon$. But it seems you got the idea in the other steps. Is there anything I am missing?
    – Gibbs
    Nov 26 at 18:15
















0














Suppose we have a normed space $E$ with the induced inner product and a map $f:Eto R$ such that $f(x) = |x|^2.$ I want to compute the derivative of this map.



We have that
$$f(x+h)=|x|^2+|h|^2+2langle x,hrangle=f(x)+o(|h|)+df_{x}(h).$$



Clearly the derivative $df_x(h) = 2langle x,hrangle$ is linear in $h.$ We also want to show that $df_x(h)$ is continuous in $h$. This can be done by simply observing that $$|df_x(h)|leq 2|x|cdot |h|$$ and therefore
$$|df_x|leq 2|x|$$ implying that $df_x$ is continuous in $h.$



I also want to show that the derivative $df_x$ is $C^{1}.$ For this we have to show that the map $xto df_x$ is continuous. Therefore observe that for each $hin E$ we have that,
$$|df_x(h)-df_y(h)|=2|langle x-y,hrangle|leq |x-y|cdot2|h|.$$
I am not sure how to proceed after this step. Any hints will be much appreciated.










share|cite|improve this question




















  • 1




    What is a inner product induced by a norm? Most norms don't come from inner products.
    – José Carlos Santos
    Nov 26 at 17:43






  • 1




    Oh I think I meant the other way around.
    – Hello_World
    Nov 26 at 18:01










  • If you fix $varepsilon > 0$ and choose $delta = varepsilon/2lvert h rvert$ you are done: $2lVert h rVert cdot lVert x-y rVert < varepsilon$. But it seems you got the idea in the other steps. Is there anything I am missing?
    – Gibbs
    Nov 26 at 18:15














0












0








0







Suppose we have a normed space $E$ with the induced inner product and a map $f:Eto R$ such that $f(x) = |x|^2.$ I want to compute the derivative of this map.



We have that
$$f(x+h)=|x|^2+|h|^2+2langle x,hrangle=f(x)+o(|h|)+df_{x}(h).$$



Clearly the derivative $df_x(h) = 2langle x,hrangle$ is linear in $h.$ We also want to show that $df_x(h)$ is continuous in $h$. This can be done by simply observing that $$|df_x(h)|leq 2|x|cdot |h|$$ and therefore
$$|df_x|leq 2|x|$$ implying that $df_x$ is continuous in $h.$



I also want to show that the derivative $df_x$ is $C^{1}.$ For this we have to show that the map $xto df_x$ is continuous. Therefore observe that for each $hin E$ we have that,
$$|df_x(h)-df_y(h)|=2|langle x-y,hrangle|leq |x-y|cdot2|h|.$$
I am not sure how to proceed after this step. Any hints will be much appreciated.










share|cite|improve this question















Suppose we have a normed space $E$ with the induced inner product and a map $f:Eto R$ such that $f(x) = |x|^2.$ I want to compute the derivative of this map.



We have that
$$f(x+h)=|x|^2+|h|^2+2langle x,hrangle=f(x)+o(|h|)+df_{x}(h).$$



Clearly the derivative $df_x(h) = 2langle x,hrangle$ is linear in $h.$ We also want to show that $df_x(h)$ is continuous in $h$. This can be done by simply observing that $$|df_x(h)|leq 2|x|cdot |h|$$ and therefore
$$|df_x|leq 2|x|$$ implying that $df_x$ is continuous in $h.$



I also want to show that the derivative $df_x$ is $C^{1}.$ For this we have to show that the map $xto df_x$ is continuous. Therefore observe that for each $hin E$ we have that,
$$|df_x(h)-df_y(h)|=2|langle x-y,hrangle|leq |x-y|cdot2|h|.$$
I am not sure how to proceed after this step. Any hints will be much appreciated.







real-analysis functional-analysis






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edited Nov 26 at 18:18









Martin Sleziak

44.6k7115270




44.6k7115270










asked Nov 26 at 17:41









Hello_World

3,84721630




3,84721630








  • 1




    What is a inner product induced by a norm? Most norms don't come from inner products.
    – José Carlos Santos
    Nov 26 at 17:43






  • 1




    Oh I think I meant the other way around.
    – Hello_World
    Nov 26 at 18:01










  • If you fix $varepsilon > 0$ and choose $delta = varepsilon/2lvert h rvert$ you are done: $2lVert h rVert cdot lVert x-y rVert < varepsilon$. But it seems you got the idea in the other steps. Is there anything I am missing?
    – Gibbs
    Nov 26 at 18:15














  • 1




    What is a inner product induced by a norm? Most norms don't come from inner products.
    – José Carlos Santos
    Nov 26 at 17:43






  • 1




    Oh I think I meant the other way around.
    – Hello_World
    Nov 26 at 18:01










  • If you fix $varepsilon > 0$ and choose $delta = varepsilon/2lvert h rvert$ you are done: $2lVert h rVert cdot lVert x-y rVert < varepsilon$. But it seems you got the idea in the other steps. Is there anything I am missing?
    – Gibbs
    Nov 26 at 18:15








1




1




What is a inner product induced by a norm? Most norms don't come from inner products.
– José Carlos Santos
Nov 26 at 17:43




What is a inner product induced by a norm? Most norms don't come from inner products.
– José Carlos Santos
Nov 26 at 17:43




1




1




Oh I think I meant the other way around.
– Hello_World
Nov 26 at 18:01




Oh I think I meant the other way around.
– Hello_World
Nov 26 at 18:01












If you fix $varepsilon > 0$ and choose $delta = varepsilon/2lvert h rvert$ you are done: $2lVert h rVert cdot lVert x-y rVert < varepsilon$. But it seems you got the idea in the other steps. Is there anything I am missing?
– Gibbs
Nov 26 at 18:15




If you fix $varepsilon > 0$ and choose $delta = varepsilon/2lvert h rvert$ you are done: $2lVert h rVert cdot lVert x-y rVert < varepsilon$. But it seems you got the idea in the other steps. Is there anything I am missing?
– Gibbs
Nov 26 at 18:15










2 Answers
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Alternatively, you can use that the inner product is bilinear and continuous, so $C^1$. Now, write $f$ as a composition:
$$xlongmapsto(x,x)longmapstolangle x,xrangle = f(x)$$
and apply the chain rule.






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    0














    I assume you use the norm $|f|=sup |f(x)|$ where the supremum runs over all $|x| = 1.$ (If such set of $x$ is empty, you are dealing with a trivial case; the details are left to you.) So, set $|h| = 1$ and get that the linear transformation $d_xf-d_yf$ has norm $leq 2|x - y|.$ So, when $y to x,$ $d_yf to d_xf.$ Q.E.D.






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      1














      Alternatively, you can use that the inner product is bilinear and continuous, so $C^1$. Now, write $f$ as a composition:
      $$xlongmapsto(x,x)longmapstolangle x,xrangle = f(x)$$
      and apply the chain rule.






      share|cite|improve this answer


























        1














        Alternatively, you can use that the inner product is bilinear and continuous, so $C^1$. Now, write $f$ as a composition:
        $$xlongmapsto(x,x)longmapstolangle x,xrangle = f(x)$$
        and apply the chain rule.






        share|cite|improve this answer
























          1












          1








          1






          Alternatively, you can use that the inner product is bilinear and continuous, so $C^1$. Now, write $f$ as a composition:
          $$xlongmapsto(x,x)longmapstolangle x,xrangle = f(x)$$
          and apply the chain rule.






          share|cite|improve this answer












          Alternatively, you can use that the inner product is bilinear and continuous, so $C^1$. Now, write $f$ as a composition:
          $$xlongmapsto(x,x)longmapstolangle x,xrangle = f(x)$$
          and apply the chain rule.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 at 18:50









          Martín-Blas Pérez Pinilla

          34.1k42771




          34.1k42771























              0














              I assume you use the norm $|f|=sup |f(x)|$ where the supremum runs over all $|x| = 1.$ (If such set of $x$ is empty, you are dealing with a trivial case; the details are left to you.) So, set $|h| = 1$ and get that the linear transformation $d_xf-d_yf$ has norm $leq 2|x - y|.$ So, when $y to x,$ $d_yf to d_xf.$ Q.E.D.






              share|cite|improve this answer


























                0














                I assume you use the norm $|f|=sup |f(x)|$ where the supremum runs over all $|x| = 1.$ (If such set of $x$ is empty, you are dealing with a trivial case; the details are left to you.) So, set $|h| = 1$ and get that the linear transformation $d_xf-d_yf$ has norm $leq 2|x - y|.$ So, when $y to x,$ $d_yf to d_xf.$ Q.E.D.






                share|cite|improve this answer
























                  0












                  0








                  0






                  I assume you use the norm $|f|=sup |f(x)|$ where the supremum runs over all $|x| = 1.$ (If such set of $x$ is empty, you are dealing with a trivial case; the details are left to you.) So, set $|h| = 1$ and get that the linear transformation $d_xf-d_yf$ has norm $leq 2|x - y|.$ So, when $y to x,$ $d_yf to d_xf.$ Q.E.D.






                  share|cite|improve this answer












                  I assume you use the norm $|f|=sup |f(x)|$ where the supremum runs over all $|x| = 1.$ (If such set of $x$ is empty, you are dealing with a trivial case; the details are left to you.) So, set $|h| = 1$ and get that the linear transformation $d_xf-d_yf$ has norm $leq 2|x - y|.$ So, when $y to x,$ $d_yf to d_xf.$ Q.E.D.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 26 at 18:21









                  Will M.

                  2,377314




                  2,377314






























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