What is the operator norm of this operator?












0












$begingroup$



Let $mathbb{X} = C^1([0,1], mathbb{R})$ and $|,f|_{mathbb{X}} = |,f|_{infty} + |,f'|_{infty}$. Let $A: mathbb{X} rightarrow mathbb{R}$ with $Af = f'left(frac{1}{2}right)$.

Find the operator norm of $A$




I am not too sure how to begin here. I have several definitions of the operator norm, the one I have been trying is that
$$|A|_{op} = infbig{c geq 0: |Av| leq c |v|big}$$



But I'm not even sure what the expression for $|Av|$ will be...



Any help is appreciated






operator-theory norm







share|cite|improve this question











$endgroup$

















    0












    $begingroup$



    Let $mathbb{X} = C^1([0,1], mathbb{R})$ and $|,f|_{mathbb{X}} = |,f|_{infty} + |,f'|_{infty}$. Let $A: mathbb{X} rightarrow mathbb{R}$ with $Af = f'left(frac{1}{2}right)$.

    Find the operator norm of $A$




    I am not too sure how to begin here. I have several definitions of the operator norm, the one I have been trying is that
    $$|A|_{op} = infbig{c geq 0: |Av| leq c |v|big}$$



    But I'm not even sure what the expression for $|Av|$ will be...



    Any help is appreciated






    operator-theory norm







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      Let $mathbb{X} = C^1([0,1], mathbb{R})$ and $|,f|_{mathbb{X}} = |,f|_{infty} + |,f'|_{infty}$. Let $A: mathbb{X} rightarrow mathbb{R}$ with $Af = f'left(frac{1}{2}right)$.

      Find the operator norm of $A$




      I am not too sure how to begin here. I have several definitions of the operator norm, the one I have been trying is that
      $$|A|_{op} = infbig{c geq 0: |Av| leq c |v|big}$$



      But I'm not even sure what the expression for $|Av|$ will be...



      Any help is appreciated






      operator-theory norm







      share|cite|improve this question











      $endgroup$





      Let $mathbb{X} = C^1([0,1], mathbb{R})$ and $|,f|_{mathbb{X}} = |,f|_{infty} + |,f'|_{infty}$. Let $A: mathbb{X} rightarrow mathbb{R}$ with $Af = f'left(frac{1}{2}right)$.

      Find the operator norm of $A$




      I am not too sure how to begin here. I have several definitions of the operator norm, the one I have been trying is that
      $$|A|_{op} = infbig{c geq 0: |Av| leq c |v|big}$$



      But I'm not even sure what the expression for $|Av|$ will be...



      Any help is appreciated






      operator-theory norm




      operator-theory norm






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 5:43









      Hanno

      2,091425




      2,091425










      asked Jan 27 '18 at 17:50









      PhysicsMathsLovePhysicsMathsLove

      1,189314




      1,189314






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Hints: Another way to look at the operator norm is to see what the operator does to a unit vector:
          $$
          |A|_{op}=sup{|Af|: |f|_{Bbb X}=1}
          $$
          Think about how large $|Af|$ could possibly be if we require that $|f|_{Bbb X}=1$.



          Some more pointers: $A$ takes in a function and spits out a real number. Which real number it spits out depends on the function, of course. For instance, $f(x)=x^2$ gives $Af=1$, and $g(x)=sin(x)$ gives $Ag=cos(0.5)$.



          But if $f$ is a function such that $|Af|=1$, what can you say about $|f|_{Bbb X}$? How small could it be? (This specific hint is more directed at the operator norm definition given in the question.)



          Finally, since you were wondering, the expression for $|Av|$ is $|v'(0.5)|$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Is the expression for $||Af||_{mathbb{X}} = ||Af||_{infty} + ||Af'||_{infty} = ||f'(frac{1}{2})||_{infty} + ||f''(frac{1}{2})||_{infty}$? If so, I have no idea where to go from here, even with the restriction $||f||_{mathbb{X}} = 1$, as we are taking infinity norms now...
            $endgroup$
            – PhysicsMathsLove
            Jan 27 '18 at 17:59












          • $begingroup$
            No. $Af$ is a real number and the corresponding norm is the usual absolute value.
            $endgroup$
            – Paul K
            Jan 27 '18 at 17:59










          • $begingroup$
            I am confused... Please can you add an answer?
            $endgroup$
            – PhysicsMathsLove
            Jan 27 '18 at 18:00










          • $begingroup$
            @PhyaicsMathsLove I added another paragraph for you to think about.
            $endgroup$
            – Arthur
            Jan 27 '18 at 18:07










          • $begingroup$
            It is apparent that $|Af|leq|f|_{X}$, so $|A|leq 1$. But I wonder if $|A|=1$. To do this, let $epsilon>0$ be small, I fail to find some $f_{epsilon}$ with $|f|_{epsilon}leq 1$ and such that $|Af_{epsilon}|>1-epsilon$.
            $endgroup$
            – user284331
            Jan 27 '18 at 22:55











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Hints: Another way to look at the operator norm is to see what the operator does to a unit vector:
          $$
          |A|_{op}=sup{|Af|: |f|_{Bbb X}=1}
          $$
          Think about how large $|Af|$ could possibly be if we require that $|f|_{Bbb X}=1$.



          Some more pointers: $A$ takes in a function and spits out a real number. Which real number it spits out depends on the function, of course. For instance, $f(x)=x^2$ gives $Af=1$, and $g(x)=sin(x)$ gives $Ag=cos(0.5)$.



          But if $f$ is a function such that $|Af|=1$, what can you say about $|f|_{Bbb X}$? How small could it be? (This specific hint is more directed at the operator norm definition given in the question.)



          Finally, since you were wondering, the expression for $|Av|$ is $|v'(0.5)|$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Is the expression for $||Af||_{mathbb{X}} = ||Af||_{infty} + ||Af'||_{infty} = ||f'(frac{1}{2})||_{infty} + ||f''(frac{1}{2})||_{infty}$? If so, I have no idea where to go from here, even with the restriction $||f||_{mathbb{X}} = 1$, as we are taking infinity norms now...
            $endgroup$
            – PhysicsMathsLove
            Jan 27 '18 at 17:59












          • $begingroup$
            No. $Af$ is a real number and the corresponding norm is the usual absolute value.
            $endgroup$
            – Paul K
            Jan 27 '18 at 17:59










          • $begingroup$
            I am confused... Please can you add an answer?
            $endgroup$
            – PhysicsMathsLove
            Jan 27 '18 at 18:00










          • $begingroup$
            @PhyaicsMathsLove I added another paragraph for you to think about.
            $endgroup$
            – Arthur
            Jan 27 '18 at 18:07










          • $begingroup$
            It is apparent that $|Af|leq|f|_{X}$, so $|A|leq 1$. But I wonder if $|A|=1$. To do this, let $epsilon>0$ be small, I fail to find some $f_{epsilon}$ with $|f|_{epsilon}leq 1$ and such that $|Af_{epsilon}|>1-epsilon$.
            $endgroup$
            – user284331
            Jan 27 '18 at 22:55
















          1












          $begingroup$

          Hints: Another way to look at the operator norm is to see what the operator does to a unit vector:
          $$
          |A|_{op}=sup{|Af|: |f|_{Bbb X}=1}
          $$
          Think about how large $|Af|$ could possibly be if we require that $|f|_{Bbb X}=1$.



          Some more pointers: $A$ takes in a function and spits out a real number. Which real number it spits out depends on the function, of course. For instance, $f(x)=x^2$ gives $Af=1$, and $g(x)=sin(x)$ gives $Ag=cos(0.5)$.



          But if $f$ is a function such that $|Af|=1$, what can you say about $|f|_{Bbb X}$? How small could it be? (This specific hint is more directed at the operator norm definition given in the question.)



          Finally, since you were wondering, the expression for $|Av|$ is $|v'(0.5)|$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Is the expression for $||Af||_{mathbb{X}} = ||Af||_{infty} + ||Af'||_{infty} = ||f'(frac{1}{2})||_{infty} + ||f''(frac{1}{2})||_{infty}$? If so, I have no idea where to go from here, even with the restriction $||f||_{mathbb{X}} = 1$, as we are taking infinity norms now...
            $endgroup$
            – PhysicsMathsLove
            Jan 27 '18 at 17:59












          • $begingroup$
            No. $Af$ is a real number and the corresponding norm is the usual absolute value.
            $endgroup$
            – Paul K
            Jan 27 '18 at 17:59










          • $begingroup$
            I am confused... Please can you add an answer?
            $endgroup$
            – PhysicsMathsLove
            Jan 27 '18 at 18:00










          • $begingroup$
            @PhyaicsMathsLove I added another paragraph for you to think about.
            $endgroup$
            – Arthur
            Jan 27 '18 at 18:07










          • $begingroup$
            It is apparent that $|Af|leq|f|_{X}$, so $|A|leq 1$. But I wonder if $|A|=1$. To do this, let $epsilon>0$ be small, I fail to find some $f_{epsilon}$ with $|f|_{epsilon}leq 1$ and such that $|Af_{epsilon}|>1-epsilon$.
            $endgroup$
            – user284331
            Jan 27 '18 at 22:55














          1












          1








          1





          $begingroup$

          Hints: Another way to look at the operator norm is to see what the operator does to a unit vector:
          $$
          |A|_{op}=sup{|Af|: |f|_{Bbb X}=1}
          $$
          Think about how large $|Af|$ could possibly be if we require that $|f|_{Bbb X}=1$.



          Some more pointers: $A$ takes in a function and spits out a real number. Which real number it spits out depends on the function, of course. For instance, $f(x)=x^2$ gives $Af=1$, and $g(x)=sin(x)$ gives $Ag=cos(0.5)$.



          But if $f$ is a function such that $|Af|=1$, what can you say about $|f|_{Bbb X}$? How small could it be? (This specific hint is more directed at the operator norm definition given in the question.)



          Finally, since you were wondering, the expression for $|Av|$ is $|v'(0.5)|$.






          share|cite|improve this answer











          $endgroup$



          Hints: Another way to look at the operator norm is to see what the operator does to a unit vector:
          $$
          |A|_{op}=sup{|Af|: |f|_{Bbb X}=1}
          $$
          Think about how large $|Af|$ could possibly be if we require that $|f|_{Bbb X}=1$.



          Some more pointers: $A$ takes in a function and spits out a real number. Which real number it spits out depends on the function, of course. For instance, $f(x)=x^2$ gives $Af=1$, and $g(x)=sin(x)$ gives $Ag=cos(0.5)$.



          But if $f$ is a function such that $|Af|=1$, what can you say about $|f|_{Bbb X}$? How small could it be? (This specific hint is more directed at the operator norm definition given in the question.)



          Finally, since you were wondering, the expression for $|Av|$ is $|v'(0.5)|$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 27 '18 at 18:07

























          answered Jan 27 '18 at 17:54









          ArthurArthur

          112k7107191




          112k7107191












          • $begingroup$
            Is the expression for $||Af||_{mathbb{X}} = ||Af||_{infty} + ||Af'||_{infty} = ||f'(frac{1}{2})||_{infty} + ||f''(frac{1}{2})||_{infty}$? If so, I have no idea where to go from here, even with the restriction $||f||_{mathbb{X}} = 1$, as we are taking infinity norms now...
            $endgroup$
            – PhysicsMathsLove
            Jan 27 '18 at 17:59












          • $begingroup$
            No. $Af$ is a real number and the corresponding norm is the usual absolute value.
            $endgroup$
            – Paul K
            Jan 27 '18 at 17:59










          • $begingroup$
            I am confused... Please can you add an answer?
            $endgroup$
            – PhysicsMathsLove
            Jan 27 '18 at 18:00










          • $begingroup$
            @PhyaicsMathsLove I added another paragraph for you to think about.
            $endgroup$
            – Arthur
            Jan 27 '18 at 18:07










          • $begingroup$
            It is apparent that $|Af|leq|f|_{X}$, so $|A|leq 1$. But I wonder if $|A|=1$. To do this, let $epsilon>0$ be small, I fail to find some $f_{epsilon}$ with $|f|_{epsilon}leq 1$ and such that $|Af_{epsilon}|>1-epsilon$.
            $endgroup$
            – user284331
            Jan 27 '18 at 22:55


















          • $begingroup$
            Is the expression for $||Af||_{mathbb{X}} = ||Af||_{infty} + ||Af'||_{infty} = ||f'(frac{1}{2})||_{infty} + ||f''(frac{1}{2})||_{infty}$? If so, I have no idea where to go from here, even with the restriction $||f||_{mathbb{X}} = 1$, as we are taking infinity norms now...
            $endgroup$
            – PhysicsMathsLove
            Jan 27 '18 at 17:59












          • $begingroup$
            No. $Af$ is a real number and the corresponding norm is the usual absolute value.
            $endgroup$
            – Paul K
            Jan 27 '18 at 17:59










          • $begingroup$
            I am confused... Please can you add an answer?
            $endgroup$
            – PhysicsMathsLove
            Jan 27 '18 at 18:00










          • $begingroup$
            @PhyaicsMathsLove I added another paragraph for you to think about.
            $endgroup$
            – Arthur
            Jan 27 '18 at 18:07










          • $begingroup$
            It is apparent that $|Af|leq|f|_{X}$, so $|A|leq 1$. But I wonder if $|A|=1$. To do this, let $epsilon>0$ be small, I fail to find some $f_{epsilon}$ with $|f|_{epsilon}leq 1$ and such that $|Af_{epsilon}|>1-epsilon$.
            $endgroup$
            – user284331
            Jan 27 '18 at 22:55
















          $begingroup$
          Is the expression for $||Af||_{mathbb{X}} = ||Af||_{infty} + ||Af'||_{infty} = ||f'(frac{1}{2})||_{infty} + ||f''(frac{1}{2})||_{infty}$? If so, I have no idea where to go from here, even with the restriction $||f||_{mathbb{X}} = 1$, as we are taking infinity norms now...
          $endgroup$
          – PhysicsMathsLove
          Jan 27 '18 at 17:59






          $begingroup$
          Is the expression for $||Af||_{mathbb{X}} = ||Af||_{infty} + ||Af'||_{infty} = ||f'(frac{1}{2})||_{infty} + ||f''(frac{1}{2})||_{infty}$? If so, I have no idea where to go from here, even with the restriction $||f||_{mathbb{X}} = 1$, as we are taking infinity norms now...
          $endgroup$
          – PhysicsMathsLove
          Jan 27 '18 at 17:59














          $begingroup$
          No. $Af$ is a real number and the corresponding norm is the usual absolute value.
          $endgroup$
          – Paul K
          Jan 27 '18 at 17:59




          $begingroup$
          No. $Af$ is a real number and the corresponding norm is the usual absolute value.
          $endgroup$
          – Paul K
          Jan 27 '18 at 17:59












          $begingroup$
          I am confused... Please can you add an answer?
          $endgroup$
          – PhysicsMathsLove
          Jan 27 '18 at 18:00




          $begingroup$
          I am confused... Please can you add an answer?
          $endgroup$
          – PhysicsMathsLove
          Jan 27 '18 at 18:00












          $begingroup$
          @PhyaicsMathsLove I added another paragraph for you to think about.
          $endgroup$
          – Arthur
          Jan 27 '18 at 18:07




          $begingroup$
          @PhyaicsMathsLove I added another paragraph for you to think about.
          $endgroup$
          – Arthur
          Jan 27 '18 at 18:07












          $begingroup$
          It is apparent that $|Af|leq|f|_{X}$, so $|A|leq 1$. But I wonder if $|A|=1$. To do this, let $epsilon>0$ be small, I fail to find some $f_{epsilon}$ with $|f|_{epsilon}leq 1$ and such that $|Af_{epsilon}|>1-epsilon$.
          $endgroup$
          – user284331
          Jan 27 '18 at 22:55




          $begingroup$
          It is apparent that $|Af|leq|f|_{X}$, so $|A|leq 1$. But I wonder if $|A|=1$. To do this, let $epsilon>0$ be small, I fail to find some $f_{epsilon}$ with $|f|_{epsilon}leq 1$ and such that $|Af_{epsilon}|>1-epsilon$.
          $endgroup$
          – user284331
          Jan 27 '18 at 22:55


















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