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Let $(V, \|\cdot \|)$ be a normed vector space over $\mathbb{F}$, $W \subset V$ be a linear subspace. If $\psi_o \in W^* := \{\text{ linear, continouos functionals }: V \to \mathbb{F} \}$ then there is $\psi \in V^{*}$ s.t i) $\psi (x) = \psi_o (x) \forall x \in W$ ii) $\|\psi \| = \| \psi_0 \|$

The proof goes like this.

Let $x \in V$, $p(x) = \| \psi_o \| \|x \|$, then $p$ is a seminorm on $V$ and $| \psi_o | \leq p(x) \ \forall x \in W$ since $\psi_o$ is continuous. By Hahn-Banach theorem there is $\psi \in V' := \{\text{Linear functionals } : V \to \mathbb{F} \} $ s.t $\psi = \psi_0$ on $W$, and $|\psi (x)| \leq p(x) \ \forall x \in V$. Thus we have $\|\psi\|\leq \| \psi_o \|$ (why?, I Believe it follows from the def. of $p(x)$ but I just dont see it)

Then the proofs end with

$$ \| \psi_o \| = \sup_{x \in W, \|x \| = 1 }|\psi_o(x) | \leq \sup_{x \in W, \|x \| = 1 }|\psi(x) | = \|\psi\| \text{ *}$$

  • why do we have that $\|\psi_0 \| = \sup_{x \in W, \|x \| = 1 } |\psi_o(x) |$?
Olba12
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1 Answers1

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The fact that $\|\psi\|\leq \|\psi_0\|$ follows from

$$\|\psi\| = \sup_{\|x\|=1} |\psi(x)| \leq \sup_{\|x\|=1}|p(x)| = \sup_{\|x\|=1} ||\psi_0\|\|x\| = \|\psi_0\|.$$

I believe the last line of inequalities is incorrect. Perhaps you mean to write:

$$\|\psi_0\| = \sup_{x\in W,\|x\|=1} |\psi_0(x)| = \sup_{x\in W,\|x\|=1} |\psi(x)| \leq \sup_{x\in V, \|x\|=1} |\psi(x)|= \|\psi\|.$$

We use that $\psi=\psi_0$ on $W$ for the second equality. The inequality follows simply from the fact that we are expanding the set of $x$ from $W$ to $V$.

Matt
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  • First, I did not mean to write what you wrote, I just copied from my lecture notes but, I'm not saying it is correct. What you wrote makes more sence to me, which I understand. But then, are my notes wrong, or are they correct... – Olba12 Dec 09 '16 at 16:44
  • I believe it is incorrect as written. The norm of a functional cannot be determined from a subspace. The norm of $\psi_0$ is determined from the subspace $W$ since it is an element of $W^$ only. The norm of $\psi$ must be determined as an element of $V^$, so we must consider all $x\in V$. – Matt Dec 09 '16 at 16:49
  • If you look at http://math.stackexchange.com/questions/1206549/corollary-of-hahn-banach-theorem?rq=1 which is the same problem, the also use $|\psi|$, if you look at the comments of the answer. – Olba12 Dec 09 '16 at 16:50
  • Sorry, it's not exactly clear to me what you mean. Is you question about the inequality $|\psi|\leq |\psi_0|$ or $|\psi_0|\leq |\psi|$? Or is there a question about notation? Above, I abuse notation whereby $|f(x)|=|f(x)|$. – Matt Dec 09 '16 at 16:57
  • If we look at my second inequality. $|\psi_o | \leq | \psi |$ I read the part $| \psi_0 | = \sup_{x\in W, |x| = 1} |\psi_o (x)|$. As "the norm of $\psi_o$ is equal to the supremum of the absolute value $|\psi_0|$ where $x \in W$ and the norm of $x$ is equal to 1. I did NOT understand why the norm of $\psi_o$ was equal to the supremum of this particual absolute value. – Olba12 Dec 09 '16 at 17:03
  • So the question is why is $| \psi_0 | = \sup_{x \in W, | x | = 1 } |\psi_0(x) |$? – Olba12 Dec 09 '16 at 17:09
  • I see. Thanks for the clarification. This follows from the definition of operator norm (or at least one of the equivalent formations of the definition). Since $\psi_0\in W^*$, this means that $|\psi_0| = \sup_{x\in W} |\psi_0(x)|/|x| = \sup_{x\in W,|x|=1} |\psi_0(x)| = \sup_{x\in W,|x|\leq 1} |\psi_0(x)|.$ The equivalence of these definitions should be in any functional analysis textbook. Keep in mind that $\psi_0$ is not defined on $V$ but only on $W$, so it is important here that $x\in W$. – Matt Dec 09 '16 at 17:11
  • Does this also hold for $W'$? Or is it just $W^*$, my class does not follow a book, so Im stuck with the lecture notes, and he has not mentioned this. – Olba12 Dec 09 '16 at 17:14
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    Yes. sorry... again notation abuse: $W' = W^*$. They are the same. – Matt Dec 09 '16 at 17:15