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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 $f:M\rightarrow \mathbb{K}$ 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?

user289143
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1 Answers1

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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 $(M\oplus N)^* = M^* \oplus N^*$ 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 $g = 0$.

Robert Israel
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  • 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. – SmileyCraft Dec 14 '18 at 16:36
  • Then you are saying that the extension if $f$ itself? And that $f(0,0,0,x_4,x_5,..)=0$? – user289143 Dec 14 '18 at 16:42
  • The extension is $f + 0$, i.e. the functional taking $x+y$ to $f(x)$ where $x \in M$, $y \in N$ – Robert Israel Dec 14 '18 at 18:02
  • So $0+y$ is sent to $0$ for every $y \in N$ – user289143 Dec 14 '18 at 18:07