2

The statement of the exercise is pretty simple to understand, but I'm having trouble to prove it. The exercise is the following:

(Exercise 3, from Introduction to functional analysis, by A. E. Taylor, page 190).: If $X$ is a complex normed linear space and $X_r$ is the associated real linear space, we write $x'(x)=x_1'(x)-ix_1'(ix)$, where $x_1'\in (X_r)'$ and $x'\in X'$. Show that $\|x_1'\|=\|x'\|$.

Here, $X'$ denotes the dual space of $X$, i.e. the space of the linear continuous transformations $x:X\to \mathbb{K}$, where $\mathbb{K}$ is the field of the scalars. And $X_r$ denotes the same vector space $X$, but considering the scalars to be only in $\Bbb{R}\subset \Bbb{C}$.

Okay, we have to show that $\|x_1'\|=\|x'\|$, i.e. $$\sup\{\|x'(v)\|\,:\, \|v\|\leq 1\}=\sup\{\|x_1'(v)\|\,:\, \|v\|\leq 1\}.$$

I've thought to show that $\|x_1'\|\leq \|x'\|$ and $\|x'\|\leq \|x_1'\|$. The first one is easy, because, since $x_1'(v)\in \Bbb{R},\forall v \in X_r\approx X$, we have $$\|x_1'(v)\|\leq \|x_1'(v)-ix_1'(v)\|=\|x'(v)\|.$$

I could not prove the second one. Then I've tried by contradiction: supposed that $\|x_1'\|<\|x'\|$, made some computations, but could not find a contradiction at all...

Mancala
  • 796

2 Answers2

2

For given $x$, there is $\alpha \in \mathbb C$, $|\alpha|=1$ with $x_1'(\alpha \, i \, x) = 0$ (why?).

Then, $$|x'(x)| = |x'(\alpha \,x)| = \ldots.$$

gerw
  • 31,359
0

Let $x\in X$ be given. We can suppose that $x\neq0$, otherwise it is easy to work with $x=0$.

Define $A_x=\{\alpha i x\,:\,\alpha \in \Bbb{C} \}$. We can consider both $A_x\subset X$ and $A_x\subset X_r$ and, respectively, $\dim_{\Bbb{C}}A_x=1$ but $\dim_{\Bbb{R}}A_x=2$. Then we look to $$f:=x_1'|_{A_x}:A_x\to \Bbb{R}$$ with $A_x\subset X_r$: then $f$ is then a real linear transformation. Since we are now in finite dimension, we can use the rank-nulity theorem: $$2=\dim_\Bbb{R}A_x=\dim_\Bbb{R}\ker(f)+\dim_\Bbb{R}\mathrm{im}(f).$$

Since $0\leq \dim_\Bbb{R}\mathrm{im}(f)\leq 1$, we must have $1\leq \dim_\Bbb{R}\ker(f)\leq 2$.

Therefore, $\ker(f)$ is not the trivial space $\{0\}$. This means that exists $v\in A_x$, with $v\neq 0$, such that $f(v)=0$. But then $v=\alpha ix$, for some $\alpha\in \Bbb{C}-\{0\}$ and $f(v)=x_1'(\alpha ix)=0$.

Now, to obtain such an $\alpha$ with $|\alpha|=1$, just take $\dfrac{\alpha}{|\alpha|}$.

Now, use @gerw suggestion!

Derso
  • 2,685
  • 1
    It is much easier to obtain such an $\alpha$. You can just use the intermediate value theorem applied to a suitable scalar function. – gerw Mar 17 '17 at 07:20