Uniqueness of Hahn-Banach extension












3












$begingroup$


Let $M={(x,y,z,0,0,...): x,y,z, in mathbb{K}}$ be a subspace of $(l_p(mathbb{N}),||cdot||_p)$ I already proved that $f:Mrightarrow mathbb{K}$ where $f(x,y,z,0,...)=x-y-z$ is bounded with $||f||=3^{frac{p-1}{p}}$ for $1 leq p < infty$. I already proved that for $p=1$ there are infinitely many extensions of $f$ and that for $p=2$ there is only one since $l_2(mathbb{N})$ is an Hilbert space. But I should prove that for $1 < p <infty$ there is a unique extension. My attempt:

Suppose there are two such extensions $F,G$ such that $||F||=||G||=3^frac{p-1}{p}$. I would like to find a contradiction by taking an $x$ outside $M$ such that $F neq G$ and I have $||(F-G)x|| leq 2||f||||x||_p$but then I'm stuck. Any suggestions?






functional-analysis







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    3












    $begingroup$


    Let $M={(x,y,z,0,0,...): x,y,z, in mathbb{K}}$ be a subspace of $(l_p(mathbb{N}),||cdot||_p)$ I already proved that $f:Mrightarrow mathbb{K}$ where $f(x,y,z,0,...)=x-y-z$ is bounded with $||f||=3^{frac{p-1}{p}}$ for $1 leq p < infty$. I already proved that for $p=1$ there are infinitely many extensions of $f$ and that for $p=2$ there is only one since $l_2(mathbb{N})$ is an Hilbert space. But I should prove that for $1 < p <infty$ there is a unique extension. My attempt:

    Suppose there are two such extensions $F,G$ such that $||F||=||G||=3^frac{p-1}{p}$. I would like to find a contradiction by taking an $x$ outside $M$ such that $F neq G$ and I have $||(F-G)x|| leq 2||f||||x||_p$but then I'm stuck. Any suggestions?






    functional-analysis







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      Let $M={(x,y,z,0,0,...): x,y,z, in mathbb{K}}$ be a subspace of $(l_p(mathbb{N}),||cdot||_p)$ I already proved that $f:Mrightarrow mathbb{K}$ where $f(x,y,z,0,...)=x-y-z$ is bounded with $||f||=3^{frac{p-1}{p}}$ for $1 leq p < infty$. I already proved that for $p=1$ there are infinitely many extensions of $f$ and that for $p=2$ there is only one since $l_2(mathbb{N})$ is an Hilbert space. But I should prove that for $1 < p <infty$ there is a unique extension. My attempt:

      Suppose there are two such extensions $F,G$ such that $||F||=||G||=3^frac{p-1}{p}$. I would like to find a contradiction by taking an $x$ outside $M$ such that $F neq G$ and I have $||(F-G)x|| leq 2||f||||x||_p$but then I'm stuck. Any suggestions?






      functional-analysis







      share|cite|improve this question









      $endgroup$




      Let $M={(x,y,z,0,0,...): x,y,z, in mathbb{K}}$ be a subspace of $(l_p(mathbb{N}),||cdot||_p)$ I already proved that $f:Mrightarrow mathbb{K}$ where $f(x,y,z,0,...)=x-y-z$ is bounded with $||f||=3^{frac{p-1}{p}}$ for $1 leq p < infty$. I already proved that for $p=1$ there are infinitely many extensions of $f$ and that for $p=2$ there is only one since $l_2(mathbb{N})$ is an Hilbert space. But I should prove that for $1 < p <infty$ there is a unique extension. My attempt:

      Suppose there are two such extensions $F,G$ such that $||F||=||G||=3^frac{p-1}{p}$. I would like to find a contradiction by taking an $x$ outside $M$ such that $F neq G$ and I have $||(F-G)x|| leq 2||f||||x||_p$but then I'm stuck. Any suggestions?






      functional-analysis




      functional-analysis






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      asked Dec 14 '18 at 15:49









      user289143user289143

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          $begingroup$

          Write $l_p(mathbb N) = l_p({1,2,3}) oplus l_p({4,5,ldots}) = M oplus N$ where
          $$|x + y|^p = |x|^p + |y|^p, x in M,; y in N$$
          Correspondingly we can write the dual $(Moplus N)^* = M^* oplus N^*$
          with $$|f + g|^q = |f|^q + |g|^q, f in M^*,; y in N^* $$
          where $1/p + 1/q = 1$. Thus any extension $f + g$ of $f in M^*$ that has the same norm must have $g = 0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Very neat argument! I would like to add that this argument fails if $p=1$ because then $q=infty$ and thus $|f+g|^q=max{|x|^q,|y|^q}$. The argument fails if $p=infty$ because then the dual would not be $ell^1$, but something much larger.
            $endgroup$
            – SmileyCraft
            Dec 14 '18 at 16:36












          • $begingroup$
            Then you are saying that the extension if $f$ itself? And that $f(0,0,0,x_4,x_5,..)=0$?
            $endgroup$
            – user289143
            Dec 14 '18 at 16:42












          • $begingroup$
            The extension is $f + 0$, i.e. the functional taking $x+y$ to $f(x)$ where $x in M$, $y in N$
            $endgroup$
            – Robert Israel
            Dec 14 '18 at 18:02










          • $begingroup$
            So $0+y$ is sent to $0$ for every $y in N$
            $endgroup$
            – user289143
            Dec 14 '18 at 18:07











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          $begingroup$

          Write $l_p(mathbb N) = l_p({1,2,3}) oplus l_p({4,5,ldots}) = M oplus N$ where
          $$|x + y|^p = |x|^p + |y|^p, x in M,; y in N$$
          Correspondingly we can write the dual $(Moplus N)^* = M^* oplus N^*$
          with $$|f + g|^q = |f|^q + |g|^q, f in M^*,; y in N^* $$
          where $1/p + 1/q = 1$. Thus any extension $f + g$ of $f in M^*$ that has the same norm must have $g = 0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Very neat argument! I would like to add that this argument fails if $p=1$ because then $q=infty$ and thus $|f+g|^q=max{|x|^q,|y|^q}$. The argument fails if $p=infty$ because then the dual would not be $ell^1$, but something much larger.
            $endgroup$
            – SmileyCraft
            Dec 14 '18 at 16:36












          • $begingroup$
            Then you are saying that the extension if $f$ itself? And that $f(0,0,0,x_4,x_5,..)=0$?
            $endgroup$
            – user289143
            Dec 14 '18 at 16:42












          • $begingroup$
            The extension is $f + 0$, i.e. the functional taking $x+y$ to $f(x)$ where $x in M$, $y in N$
            $endgroup$
            – Robert Israel
            Dec 14 '18 at 18:02










          • $begingroup$
            So $0+y$ is sent to $0$ for every $y in N$
            $endgroup$
            – user289143
            Dec 14 '18 at 18:07
















          2












          $begingroup$

          Write $l_p(mathbb N) = l_p({1,2,3}) oplus l_p({4,5,ldots}) = M oplus N$ where
          $$|x + y|^p = |x|^p + |y|^p, x in M,; y in N$$
          Correspondingly we can write the dual $(Moplus N)^* = M^* oplus N^*$
          with $$|f + g|^q = |f|^q + |g|^q, f in M^*,; y in N^* $$
          where $1/p + 1/q = 1$. Thus any extension $f + g$ of $f in M^*$ that has the same norm must have $g = 0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Very neat argument! I would like to add that this argument fails if $p=1$ because then $q=infty$ and thus $|f+g|^q=max{|x|^q,|y|^q}$. The argument fails if $p=infty$ because then the dual would not be $ell^1$, but something much larger.
            $endgroup$
            – SmileyCraft
            Dec 14 '18 at 16:36












          • $begingroup$
            Then you are saying that the extension if $f$ itself? And that $f(0,0,0,x_4,x_5,..)=0$?
            $endgroup$
            – user289143
            Dec 14 '18 at 16:42












          • $begingroup$
            The extension is $f + 0$, i.e. the functional taking $x+y$ to $f(x)$ where $x in M$, $y in N$
            $endgroup$
            – Robert Israel
            Dec 14 '18 at 18:02










          • $begingroup$
            So $0+y$ is sent to $0$ for every $y in N$
            $endgroup$
            – user289143
            Dec 14 '18 at 18:07














          2












          2








          2





          $begingroup$

          Write $l_p(mathbb N) = l_p({1,2,3}) oplus l_p({4,5,ldots}) = M oplus N$ where
          $$|x + y|^p = |x|^p + |y|^p, x in M,; y in N$$
          Correspondingly we can write the dual $(Moplus N)^* = M^* oplus N^*$
          with $$|f + g|^q = |f|^q + |g|^q, f in M^*,; y in N^* $$
          where $1/p + 1/q = 1$. Thus any extension $f + g$ of $f in M^*$ that has the same norm must have $g = 0$.






          share|cite|improve this answer









          $endgroup$



          Write $l_p(mathbb N) = l_p({1,2,3}) oplus l_p({4,5,ldots}) = M oplus N$ where
          $$|x + y|^p = |x|^p + |y|^p, x in M,; y in N$$
          Correspondingly we can write the dual $(Moplus N)^* = M^* oplus N^*$
          with $$|f + g|^q = |f|^q + |g|^q, f in M^*,; y in N^* $$
          where $1/p + 1/q = 1$. Thus any extension $f + g$ of $f in M^*$ that has the same norm must have $g = 0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 16:27









          Robert IsraelRobert Israel

          325k23215469




          325k23215469












          • $begingroup$
            Very neat argument! I would like to add that this argument fails if $p=1$ because then $q=infty$ and thus $|f+g|^q=max{|x|^q,|y|^q}$. The argument fails if $p=infty$ because then the dual would not be $ell^1$, but something much larger.
            $endgroup$
            – SmileyCraft
            Dec 14 '18 at 16:36












          • $begingroup$
            Then you are saying that the extension if $f$ itself? And that $f(0,0,0,x_4,x_5,..)=0$?
            $endgroup$
            – user289143
            Dec 14 '18 at 16:42












          • $begingroup$
            The extension is $f + 0$, i.e. the functional taking $x+y$ to $f(x)$ where $x in M$, $y in N$
            $endgroup$
            – Robert Israel
            Dec 14 '18 at 18:02










          • $begingroup$
            So $0+y$ is sent to $0$ for every $y in N$
            $endgroup$
            – user289143
            Dec 14 '18 at 18:07


















          • $begingroup$
            Very neat argument! I would like to add that this argument fails if $p=1$ because then $q=infty$ and thus $|f+g|^q=max{|x|^q,|y|^q}$. The argument fails if $p=infty$ because then the dual would not be $ell^1$, but something much larger.
            $endgroup$
            – SmileyCraft
            Dec 14 '18 at 16:36












          • $begingroup$
            Then you are saying that the extension if $f$ itself? And that $f(0,0,0,x_4,x_5,..)=0$?
            $endgroup$
            – user289143
            Dec 14 '18 at 16:42












          • $begingroup$
            The extension is $f + 0$, i.e. the functional taking $x+y$ to $f(x)$ where $x in M$, $y in N$
            $endgroup$
            – Robert Israel
            Dec 14 '18 at 18:02










          • $begingroup$
            So $0+y$ is sent to $0$ for every $y in N$
            $endgroup$
            – user289143
            Dec 14 '18 at 18:07
















          $begingroup$
          Very neat argument! I would like to add that this argument fails if $p=1$ because then $q=infty$ and thus $|f+g|^q=max{|x|^q,|y|^q}$. The argument fails if $p=infty$ because then the dual would not be $ell^1$, but something much larger.
          $endgroup$
          – SmileyCraft
          Dec 14 '18 at 16:36






          $begingroup$
          Very neat argument! I would like to add that this argument fails if $p=1$ because then $q=infty$ and thus $|f+g|^q=max{|x|^q,|y|^q}$. The argument fails if $p=infty$ because then the dual would not be $ell^1$, but something much larger.
          $endgroup$
          – SmileyCraft
          Dec 14 '18 at 16:36














          $begingroup$
          Then you are saying that the extension if $f$ itself? And that $f(0,0,0,x_4,x_5,..)=0$?
          $endgroup$
          – user289143
          Dec 14 '18 at 16:42






          $begingroup$
          Then you are saying that the extension if $f$ itself? And that $f(0,0,0,x_4,x_5,..)=0$?
          $endgroup$
          – user289143
          Dec 14 '18 at 16:42














          $begingroup$
          The extension is $f + 0$, i.e. the functional taking $x+y$ to $f(x)$ where $x in M$, $y in N$
          $endgroup$
          – Robert Israel
          Dec 14 '18 at 18:02




          $begingroup$
          The extension is $f + 0$, i.e. the functional taking $x+y$ to $f(x)$ where $x in M$, $y in N$
          $endgroup$
          – Robert Israel
          Dec 14 '18 at 18:02












          $begingroup$
          So $0+y$ is sent to $0$ for every $y in N$
          $endgroup$
          – user289143
          Dec 14 '18 at 18:07




          $begingroup$
          So $0+y$ is sent to $0$ for every $y in N$
          $endgroup$
          – user289143
          Dec 14 '18 at 18:07


















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