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Theorem: Let $X$ be a normed space and $0 \not= x_0 \in X$ be a arbitrary. Then there exists a bounded linear functional $\bar f$ on $X$ such that $$\|\bar f \|=1, \quad \bar f(x_0) =\|x_0\|.$$

Problem: Find $\bar f$ when $X$ is the Euclidean plane $\mathbb{R}^2$.

My interpretation of the task is that I have to explicitly specify a functional that satisfies the conditions and works for arbitrary non-zero points in the plane. I've been told however that such a functional need not be unique and may depend on each $x_0$. I'm confused now about how to go about solving this problem.

I'd also appreciate if someone can comment on if the following solution to finding some $f \in X'$ such that $\|f\|=\|x_0\|^{-1}$ and $f(x)=1$ under the assumptions is correct?

I've defined $f(x)=\frac{1}{\|x_0\|} \bar f(x).$ Then $f$ is a bounded linear functional on $X$ because of $\bar f$ of the theorem.

$$f(x_0)=\frac{1}{\|x_0\|}\bar f(x_0)=\frac{1}{\|x_0\|} \|x_0\|=1.$$

and since $f(x)=\frac{1}{\|x_0\|} \bar f(x)$,

$\|f\|=\frac{1}{\|x_0\|} \|\bar f\|$ (Does this equality require any further justification? Intuitively I can see it has to hold, but I'm unable to find a good reason for why and would like someone to explain this to me.)

$=\frac{1}{\|x_0\|}$.

Mark
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When $X$ is $\mathbb{R}^2$, it is a Hilbert space. (Assuming the Euclidean Norm is used)

Every functional on a Hilbert space can be written as $\bar f(x)=\langle x, y \rangle = x\cdot y$ for some $y \in X$. This is the Riesz Representation Theorem.

So our goal is to then find a $y$ that gives $\| \bar f \| = 1$ and $\bar f(x_0) = \|x_0\|$. Somehow our $y$ should incorporate $x_0$. So we can try $y=x_0$.

Here we find $\bar f(x_0) = x_0 \cdot x_0 = \|x_0\|^2$, and $|\bar f(x)| \le \|x\|\cdot \|x_0\|$ (Cauchy Schwarz) so $\| \bar f \| = \|x_0\|$. This is a little too big.

Using $y=x_0/\|x_0\|$ is then sufficient. You can check this.

As for your second question, $\| \lambda f\| = | \lambda | \|f\|$ for all $\lambda$ in our field (either $\mathbb{C}$ or $\mathbb{R}$). We can prove this quickly.

Note that $\| f\| = \sup_{\|x\| =1} |f(x)|$ by definition.

Thus $$\| \lambda f\| = \sup_{\|x\|=1} |\lambda f(x)| = |\lambda| \sup_{\|x\|=1} |f(x)| = |\lambda| \| f\|.$$

You reasoning is correct.

Joel
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  • Sorry to bother you with this again. I have a question concerning Riesz's representation theorem: The statement to this theorem in our text (Functional Analysis - Kreeyszig) is: "Every bounded linear functional on a Hilbert space X has a representation $f(x)=\langle x, y \rangle$, where $y$ depends on $f$ and is uniquely determined by $f$." If $f$ determines $y$, then how are we able to freely choose $y$ without changing the functional? – Mark Dec 02 '15 at 15:13
  • What I'm trying to say is how do we know $f \in X'$ when $f(x)= \langle x, x_0 / |x_0| \rangle $? – Mark Dec 02 '15 at 15:16
  • The most straight forward way to see that this is a linear functional: if we write $x=(x_1,x_2)$ and $y=x_0/|x_0|=(a,b)$ then $f(x) = a x_1 + b x_2$. This is a linear function, is continuous, and $f: X \to \mathbb{R}$. – Joel Dec 02 '15 at 15:59
  • Alternatively, if you know that every bounded functional is continuous, then $|f(x)| = \langle x, x_0/|x_0| \rangle \le | x | \cdot ( 1 )$ by Cauchy Schwarz, and thus $f$ is a bounded functional. – Joel Dec 02 '15 at 19:26