Proving that $d(x,ker f) = frac{|f(x)|}{|f|}$ if $f in X^*$












3














Exercise :




Let $(X,|cdot|)$ be a normed space and $Vsubseteq X$. The distance of one point $x in X$ to $V$ is defined by :
$$d(x,V)=inf{|x-v|: v in V}$$
Show that if $f in X^*=B(X,mathbb R)$ which is the space of the bounded linear functionals, thefe for all $x in X$ it is :
$$d(x,ker f)=frac{|f(x)|}{|f|}$$




Attempt - thoughts :



So, first of all, since $f$ is a bounded linear functional, then there exists some $M>0$ such that :



$$|f(x)| leq M|x|$$



Also, the operator norm is given by :



$$|f| = supbigg{frac{|f(x)|}{|x|} : x in X, x neq 0bigg}$$



Now, the kernel of the function $f$ is defined as :



$$ker f = {x in X : f(x) = 0}$$



Essentialy, we need to calculate :



$$d(x,ker f) = inf{|x-v|:v in ker f}$$



Now, of course, the kernel of $f$ is a subspace of $X$ and also, we know that :
$$text{co}dim{ker f} = 1$$



I can't see how to combine these facts yielded by the hypothesis of the exercise, though, to continue to an attempted solution.



Any hints, tips or thorough elaborations will be greatly appreciated !










share|cite|improve this question





























    3














    Exercise :




    Let $(X,|cdot|)$ be a normed space and $Vsubseteq X$. The distance of one point $x in X$ to $V$ is defined by :
    $$d(x,V)=inf{|x-v|: v in V}$$
    Show that if $f in X^*=B(X,mathbb R)$ which is the space of the bounded linear functionals, thefe for all $x in X$ it is :
    $$d(x,ker f)=frac{|f(x)|}{|f|}$$




    Attempt - thoughts :



    So, first of all, since $f$ is a bounded linear functional, then there exists some $M>0$ such that :



    $$|f(x)| leq M|x|$$



    Also, the operator norm is given by :



    $$|f| = supbigg{frac{|f(x)|}{|x|} : x in X, x neq 0bigg}$$



    Now, the kernel of the function $f$ is defined as :



    $$ker f = {x in X : f(x) = 0}$$



    Essentialy, we need to calculate :



    $$d(x,ker f) = inf{|x-v|:v in ker f}$$



    Now, of course, the kernel of $f$ is a subspace of $X$ and also, we know that :
    $$text{co}dim{ker f} = 1$$



    I can't see how to combine these facts yielded by the hypothesis of the exercise, though, to continue to an attempted solution.



    Any hints, tips or thorough elaborations will be greatly appreciated !










    share|cite|improve this question



























      3












      3








      3







      Exercise :




      Let $(X,|cdot|)$ be a normed space and $Vsubseteq X$. The distance of one point $x in X$ to $V$ is defined by :
      $$d(x,V)=inf{|x-v|: v in V}$$
      Show that if $f in X^*=B(X,mathbb R)$ which is the space of the bounded linear functionals, thefe for all $x in X$ it is :
      $$d(x,ker f)=frac{|f(x)|}{|f|}$$




      Attempt - thoughts :



      So, first of all, since $f$ is a bounded linear functional, then there exists some $M>0$ such that :



      $$|f(x)| leq M|x|$$



      Also, the operator norm is given by :



      $$|f| = supbigg{frac{|f(x)|}{|x|} : x in X, x neq 0bigg}$$



      Now, the kernel of the function $f$ is defined as :



      $$ker f = {x in X : f(x) = 0}$$



      Essentialy, we need to calculate :



      $$d(x,ker f) = inf{|x-v|:v in ker f}$$



      Now, of course, the kernel of $f$ is a subspace of $X$ and also, we know that :
      $$text{co}dim{ker f} = 1$$



      I can't see how to combine these facts yielded by the hypothesis of the exercise, though, to continue to an attempted solution.



      Any hints, tips or thorough elaborations will be greatly appreciated !










      share|cite|improve this question















      Exercise :




      Let $(X,|cdot|)$ be a normed space and $Vsubseteq X$. The distance of one point $x in X$ to $V$ is defined by :
      $$d(x,V)=inf{|x-v|: v in V}$$
      Show that if $f in X^*=B(X,mathbb R)$ which is the space of the bounded linear functionals, thefe for all $x in X$ it is :
      $$d(x,ker f)=frac{|f(x)|}{|f|}$$




      Attempt - thoughts :



      So, first of all, since $f$ is a bounded linear functional, then there exists some $M>0$ such that :



      $$|f(x)| leq M|x|$$



      Also, the operator norm is given by :



      $$|f| = supbigg{frac{|f(x)|}{|x|} : x in X, x neq 0bigg}$$



      Now, the kernel of the function $f$ is defined as :



      $$ker f = {x in X : f(x) = 0}$$



      Essentialy, we need to calculate :



      $$d(x,ker f) = inf{|x-v|:v in ker f}$$



      Now, of course, the kernel of $f$ is a subspace of $X$ and also, we know that :
      $$text{co}dim{ker f} = 1$$



      I can't see how to combine these facts yielded by the hypothesis of the exercise, though, to continue to an attempted solution.



      Any hints, tips or thorough elaborations will be greatly appreciated !







      real-analysis functional-analysis operator-theory






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      edited Nov 18 at 16:30









      Xander Henderson

      14.1k103554




      14.1k103554










      asked Nov 18 at 14:08









      Rebellos

      14.4k31245




      14.4k31245






















          2 Answers
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          3














          hint: without loss of generality, assume $|f|=1$. It is clear that $|f(x)|le |v-x|$ for every $vin $ker$f$. For the reverse inequality, note that there is a vector $zin X$ such that $|f(z)|>1-epsilon$ for any $epsilon>0.$ For $xneq 0, $ set $v = x-frac{f(x)}{f(z)} z$ and observe that $vin $ker$f$.






          share|cite|improve this answer





















          • Why is it $|f(x) leq |v-x|$ ? Also why is there such a vector $z$ as you point out ? It may lead to the wanted result but it seems not that rigorous. I would really appriciate some clarification or elaboration. Also, $text{co}dim{ker f}=1$ wasn't used that is one of the basic facts (I think) of the information of the excercise and our current lesson section.
            – Rebellos
            Nov 18 at 22:23










          • The proof is correct and rigorous I think. To answer your questions: $frac{ |f(x-v)|}{|x-v|}=frac{ |f(x)|}{|x-v|}le |f|= 1$. There exists such a $z$ for the same reason: because $|f|=1$.
            – Matematleta
            Nov 18 at 22:51





















          1














          First of all, we can assume for simplicity $|f|=1$, by considering $f_1:=frac1{|f|}cdot f$ instead.

          Then $ker f=ker f_1$, $ |f_1|=1$, and we have to prove $d(x,ker f)=|f_1(x)|$ for all $xin V$.



          Assumed $|f|=1$, we have $|f(x)|le |x|$, apply it now to vectors $x-v$ with $vinker f$:
          $$|f(x)|= |f(x-v)|le|x-v|$$
          which shows $|f(x)|le d(x,ker f)$.



          For the other direction, by the definition of norm, there are unit vectors $a_1,a_2,dots$ such that $f(a_n) to 1$.

          Now, $y_n:= x - frac{f(x)}{f(a_n)}a_n inker f$, and hence
          $$d(x,ker f)le|x-y_n| = frac{|f(x)|}{|f(a_n)|}$$
          which tends to $|f(x)|$ as $ntoinfty$.






          share|cite|improve this answer





















          • Hi, thanks for your input. How do we yield that $d(x,ker f) = frac{|f(x)|}{|f|}$ from all this ? All I can see is an inequality at the end and not a strict equality. Also the fact about $text{co}dim {ker f} = 1$ wasn't used at all, which is one of the basic facts I assume.
            – Rebellos
            Nov 18 at 22:20











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          2 Answers
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          active

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          2 Answers
          2






          active

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          active

          oldest

          votes






          active

          oldest

          votes









          3














          hint: without loss of generality, assume $|f|=1$. It is clear that $|f(x)|le |v-x|$ for every $vin $ker$f$. For the reverse inequality, note that there is a vector $zin X$ such that $|f(z)|>1-epsilon$ for any $epsilon>0.$ For $xneq 0, $ set $v = x-frac{f(x)}{f(z)} z$ and observe that $vin $ker$f$.






          share|cite|improve this answer





















          • Why is it $|f(x) leq |v-x|$ ? Also why is there such a vector $z$ as you point out ? It may lead to the wanted result but it seems not that rigorous. I would really appriciate some clarification or elaboration. Also, $text{co}dim{ker f}=1$ wasn't used that is one of the basic facts (I think) of the information of the excercise and our current lesson section.
            – Rebellos
            Nov 18 at 22:23










          • The proof is correct and rigorous I think. To answer your questions: $frac{ |f(x-v)|}{|x-v|}=frac{ |f(x)|}{|x-v|}le |f|= 1$. There exists such a $z$ for the same reason: because $|f|=1$.
            – Matematleta
            Nov 18 at 22:51


















          3














          hint: without loss of generality, assume $|f|=1$. It is clear that $|f(x)|le |v-x|$ for every $vin $ker$f$. For the reverse inequality, note that there is a vector $zin X$ such that $|f(z)|>1-epsilon$ for any $epsilon>0.$ For $xneq 0, $ set $v = x-frac{f(x)}{f(z)} z$ and observe that $vin $ker$f$.






          share|cite|improve this answer





















          • Why is it $|f(x) leq |v-x|$ ? Also why is there such a vector $z$ as you point out ? It may lead to the wanted result but it seems not that rigorous. I would really appriciate some clarification or elaboration. Also, $text{co}dim{ker f}=1$ wasn't used that is one of the basic facts (I think) of the information of the excercise and our current lesson section.
            – Rebellos
            Nov 18 at 22:23










          • The proof is correct and rigorous I think. To answer your questions: $frac{ |f(x-v)|}{|x-v|}=frac{ |f(x)|}{|x-v|}le |f|= 1$. There exists such a $z$ for the same reason: because $|f|=1$.
            – Matematleta
            Nov 18 at 22:51
















          3












          3








          3






          hint: without loss of generality, assume $|f|=1$. It is clear that $|f(x)|le |v-x|$ for every $vin $ker$f$. For the reverse inequality, note that there is a vector $zin X$ such that $|f(z)|>1-epsilon$ for any $epsilon>0.$ For $xneq 0, $ set $v = x-frac{f(x)}{f(z)} z$ and observe that $vin $ker$f$.






          share|cite|improve this answer












          hint: without loss of generality, assume $|f|=1$. It is clear that $|f(x)|le |v-x|$ for every $vin $ker$f$. For the reverse inequality, note that there is a vector $zin X$ such that $|f(z)|>1-epsilon$ for any $epsilon>0.$ For $xneq 0, $ set $v = x-frac{f(x)}{f(z)} z$ and observe that $vin $ker$f$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 18 at 15:22









          Matematleta

          9,9522918




          9,9522918












          • Why is it $|f(x) leq |v-x|$ ? Also why is there such a vector $z$ as you point out ? It may lead to the wanted result but it seems not that rigorous. I would really appriciate some clarification or elaboration. Also, $text{co}dim{ker f}=1$ wasn't used that is one of the basic facts (I think) of the information of the excercise and our current lesson section.
            – Rebellos
            Nov 18 at 22:23










          • The proof is correct and rigorous I think. To answer your questions: $frac{ |f(x-v)|}{|x-v|}=frac{ |f(x)|}{|x-v|}le |f|= 1$. There exists such a $z$ for the same reason: because $|f|=1$.
            – Matematleta
            Nov 18 at 22:51




















          • Why is it $|f(x) leq |v-x|$ ? Also why is there such a vector $z$ as you point out ? It may lead to the wanted result but it seems not that rigorous. I would really appriciate some clarification or elaboration. Also, $text{co}dim{ker f}=1$ wasn't used that is one of the basic facts (I think) of the information of the excercise and our current lesson section.
            – Rebellos
            Nov 18 at 22:23










          • The proof is correct and rigorous I think. To answer your questions: $frac{ |f(x-v)|}{|x-v|}=frac{ |f(x)|}{|x-v|}le |f|= 1$. There exists such a $z$ for the same reason: because $|f|=1$.
            – Matematleta
            Nov 18 at 22:51


















          Why is it $|f(x) leq |v-x|$ ? Also why is there such a vector $z$ as you point out ? It may lead to the wanted result but it seems not that rigorous. I would really appriciate some clarification or elaboration. Also, $text{co}dim{ker f}=1$ wasn't used that is one of the basic facts (I think) of the information of the excercise and our current lesson section.
          – Rebellos
          Nov 18 at 22:23




          Why is it $|f(x) leq |v-x|$ ? Also why is there such a vector $z$ as you point out ? It may lead to the wanted result but it seems not that rigorous. I would really appriciate some clarification or elaboration. Also, $text{co}dim{ker f}=1$ wasn't used that is one of the basic facts (I think) of the information of the excercise and our current lesson section.
          – Rebellos
          Nov 18 at 22:23












          The proof is correct and rigorous I think. To answer your questions: $frac{ |f(x-v)|}{|x-v|}=frac{ |f(x)|}{|x-v|}le |f|= 1$. There exists such a $z$ for the same reason: because $|f|=1$.
          – Matematleta
          Nov 18 at 22:51






          The proof is correct and rigorous I think. To answer your questions: $frac{ |f(x-v)|}{|x-v|}=frac{ |f(x)|}{|x-v|}le |f|= 1$. There exists such a $z$ for the same reason: because $|f|=1$.
          – Matematleta
          Nov 18 at 22:51













          1














          First of all, we can assume for simplicity $|f|=1$, by considering $f_1:=frac1{|f|}cdot f$ instead.

          Then $ker f=ker f_1$, $ |f_1|=1$, and we have to prove $d(x,ker f)=|f_1(x)|$ for all $xin V$.



          Assumed $|f|=1$, we have $|f(x)|le |x|$, apply it now to vectors $x-v$ with $vinker f$:
          $$|f(x)|= |f(x-v)|le|x-v|$$
          which shows $|f(x)|le d(x,ker f)$.



          For the other direction, by the definition of norm, there are unit vectors $a_1,a_2,dots$ such that $f(a_n) to 1$.

          Now, $y_n:= x - frac{f(x)}{f(a_n)}a_n inker f$, and hence
          $$d(x,ker f)le|x-y_n| = frac{|f(x)|}{|f(a_n)|}$$
          which tends to $|f(x)|$ as $ntoinfty$.






          share|cite|improve this answer





















          • Hi, thanks for your input. How do we yield that $d(x,ker f) = frac{|f(x)|}{|f|}$ from all this ? All I can see is an inequality at the end and not a strict equality. Also the fact about $text{co}dim {ker f} = 1$ wasn't used at all, which is one of the basic facts I assume.
            – Rebellos
            Nov 18 at 22:20
















          1














          First of all, we can assume for simplicity $|f|=1$, by considering $f_1:=frac1{|f|}cdot f$ instead.

          Then $ker f=ker f_1$, $ |f_1|=1$, and we have to prove $d(x,ker f)=|f_1(x)|$ for all $xin V$.



          Assumed $|f|=1$, we have $|f(x)|le |x|$, apply it now to vectors $x-v$ with $vinker f$:
          $$|f(x)|= |f(x-v)|le|x-v|$$
          which shows $|f(x)|le d(x,ker f)$.



          For the other direction, by the definition of norm, there are unit vectors $a_1,a_2,dots$ such that $f(a_n) to 1$.

          Now, $y_n:= x - frac{f(x)}{f(a_n)}a_n inker f$, and hence
          $$d(x,ker f)le|x-y_n| = frac{|f(x)|}{|f(a_n)|}$$
          which tends to $|f(x)|$ as $ntoinfty$.






          share|cite|improve this answer





















          • Hi, thanks for your input. How do we yield that $d(x,ker f) = frac{|f(x)|}{|f|}$ from all this ? All I can see is an inequality at the end and not a strict equality. Also the fact about $text{co}dim {ker f} = 1$ wasn't used at all, which is one of the basic facts I assume.
            – Rebellos
            Nov 18 at 22:20














          1












          1








          1






          First of all, we can assume for simplicity $|f|=1$, by considering $f_1:=frac1{|f|}cdot f$ instead.

          Then $ker f=ker f_1$, $ |f_1|=1$, and we have to prove $d(x,ker f)=|f_1(x)|$ for all $xin V$.



          Assumed $|f|=1$, we have $|f(x)|le |x|$, apply it now to vectors $x-v$ with $vinker f$:
          $$|f(x)|= |f(x-v)|le|x-v|$$
          which shows $|f(x)|le d(x,ker f)$.



          For the other direction, by the definition of norm, there are unit vectors $a_1,a_2,dots$ such that $f(a_n) to 1$.

          Now, $y_n:= x - frac{f(x)}{f(a_n)}a_n inker f$, and hence
          $$d(x,ker f)le|x-y_n| = frac{|f(x)|}{|f(a_n)|}$$
          which tends to $|f(x)|$ as $ntoinfty$.






          share|cite|improve this answer












          First of all, we can assume for simplicity $|f|=1$, by considering $f_1:=frac1{|f|}cdot f$ instead.

          Then $ker f=ker f_1$, $ |f_1|=1$, and we have to prove $d(x,ker f)=|f_1(x)|$ for all $xin V$.



          Assumed $|f|=1$, we have $|f(x)|le |x|$, apply it now to vectors $x-v$ with $vinker f$:
          $$|f(x)|= |f(x-v)|le|x-v|$$
          which shows $|f(x)|le d(x,ker f)$.



          For the other direction, by the definition of norm, there are unit vectors $a_1,a_2,dots$ such that $f(a_n) to 1$.

          Now, $y_n:= x - frac{f(x)}{f(a_n)}a_n inker f$, and hence
          $$d(x,ker f)le|x-y_n| = frac{|f(x)|}{|f(a_n)|}$$
          which tends to $|f(x)|$ as $ntoinfty$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 18 at 15:29









          Berci

          59.6k23672




          59.6k23672












          • Hi, thanks for your input. How do we yield that $d(x,ker f) = frac{|f(x)|}{|f|}$ from all this ? All I can see is an inequality at the end and not a strict equality. Also the fact about $text{co}dim {ker f} = 1$ wasn't used at all, which is one of the basic facts I assume.
            – Rebellos
            Nov 18 at 22:20


















          • Hi, thanks for your input. How do we yield that $d(x,ker f) = frac{|f(x)|}{|f|}$ from all this ? All I can see is an inequality at the end and not a strict equality. Also the fact about $text{co}dim {ker f} = 1$ wasn't used at all, which is one of the basic facts I assume.
            – Rebellos
            Nov 18 at 22:20
















          Hi, thanks for your input. How do we yield that $d(x,ker f) = frac{|f(x)|}{|f|}$ from all this ? All I can see is an inequality at the end and not a strict equality. Also the fact about $text{co}dim {ker f} = 1$ wasn't used at all, which is one of the basic facts I assume.
          – Rebellos
          Nov 18 at 22:20




          Hi, thanks for your input. How do we yield that $d(x,ker f) = frac{|f(x)|}{|f|}$ from all this ? All I can see is an inequality at the end and not a strict equality. Also the fact about $text{co}dim {ker f} = 1$ wasn't used at all, which is one of the basic facts I assume.
          – Rebellos
          Nov 18 at 22:20


















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