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Suppose $X$ is a Banach space over the field $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$, and let $X'$ denote its dual. Fix $x_0 \in X$ and $\varphi_0 \in X'$. I want to prove that norm of linear map $T:B(X)\rightarrow\mathbb{F}$ given by $$T: A \mapsto \varphi_0(A(x_0))$$ equals $\|\varphi_0\| \|x_0\|$.

My work so far

My strategy is the following: I want to define $F : X' \ni A \rightarrow\varphi_0(A(x_0))$ and show that it's bounded and its norm equals to $\|\varphi_0\| \|x_0\|$. After that I'm going to use Hahn- Banach theorem to extend domain of $F$ from $X'$ to $B(X)$.

So:

$$A \in X' \Rightarrow \|A(x_0)\| \le \|A\|\cdot\|x_0\|$$

$$\varphi_0 \in X' \Rightarrow \|\varphi_0(A(x_0))\| \le \|\varphi_0\|\|A(x_0)\| \le \|\varphi_0\|\|A\|\|x_0\|$$

Let's observe that by this fact we have that:

$$\|F(A)\| = \|\varphi_0(A(x_0))\| \le \|\varphi_0\|\|A\|\|x_0\|$$

So we see that $F$ is bounded. Now we want to calculate norm $\|F\|$. To do this we want consider only $\|A\| \le 1$.

$$\|F(A)\| \le \|\varphi_0\|\|x_0\|$$

And it's very similar what I want to prove. I just need to come up with example of $A$ for which $||F(A)\| = \|\varphi_0 \|\|x_0\|$. I was trying to find such, starting from identity up to some more complex examples, but I didn't manage to figure out this particular map. Could you please give me a hint how to pick it?

Mittens
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Louis
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  • "extend" from $X'$ to $B(X)$??? this makes no sense, because $X'$ is not a subspace of $B(X)$. – David C. Ullrich Jun 23 '21 at 13:58
  • Take $\phi_1\in X'$, $x_1\in X$, then $A(x):= x_1 \cdot \phi_1(x)$ is a linear operator. Now choose $\phi_1,x_1$ to obtain the claim. – daw Jun 23 '21 at 14:06

1 Answers1

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It's evident that $\|T\|\leq\|\phi_0\|\cdot\|x_0\|$.

For the converse inequality, fix $\phi\in X'$ and $y\in X$ and consider the operator $A:X\to X$ defined by $A(x)=\phi(x)y$. Note that $A$ is linear obviously and $A\in B(X)$: $\|A(x)\|=\|\phi(x)y\|=|\phi(x)|\cdot\|y\|\leq\|\phi\|\cdot\|x\|\cdot\|y\|$, so $A$ is bounded and $\|A\|\leq\|\phi\|\|y\|$. Actually, $$\|A\|=\sup_{x\in X,\|x\|=1}\|A(x)\|=\sup_{\|x\|=1}|\phi(x)|\cdot\|y\|=\|\phi\|\cdot\|y\|$$

Now $$\|T\|=\sup_{B\in B(X), B\neq0}\frac{|T(B)|}{\|B\|}\geq\frac{|T(A)|}{\|A\|}=\frac{|\phi_0(A(x_0))|}{\|\phi_0\|\|x_0\|}=\frac{|\phi_0(\phi(x_0)y)|}{\|\phi\|\|y\|}=\frac{|\phi(x_0)|\cdot|\phi_0(y)|}{\|\phi\|\|y\|}$$

note that this inequality is true for all non-zero $y\in X$ and all non-zero $\phi\in X'$.

Use Hahn-Banach to obtain a functional $\phi\in X'$ such that $\phi(x_0)=\|x_0\|$ and $\|\phi\|=1$. Can you do that? After this, the above inequality gives $$\|T\|\geq\frac{\|x_0\|\cdot|\phi_0(y)|}{\|y\|}$$ and this is true for all non-zero $y\in X$. Thus $$\|T\|\geq\|x_0\|\cdot\sup_{y\in X,y\neq0}\frac{|\phi_0(y)|}{\|y\|}=\|x_0\|\|\phi_0\| $$ as we wanted.

Just dropped in
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