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Let $(\mathbb R^2, \|\cdot\|_1)$ be normed space and $E$ be a 1-dimensional subspace. If $f:E \to \mathbb R$ is linear functional with induced 1-norm one, i.e $\|f\|_{1,E}=1$ on $E$ extend it to linear functional $f:\mathbb R^2 \to \mathbb R$, with the same norm, $\|f\|_1 = 1$.

Some definitions:

  • For a vector $x \in \mathbb R^2$ the norm is defined as $\|x\|_1 := |x_1| + |x_2|$.
  • For a linear operator the norm $\displaystyle\|f\|_1 := \sup_{\|x\|_1 = 1} | f(x) |$
  • The induced norm of $f$ is defined as $\displaystyle\|f\|_{1,E} := \sup_{x \in E,\ \|x\|_1 = 1} | f(x) |$

Attempt 1: A vector $x \in E$ would have components related as $x_1 a_1 + x_2 a_2 = 0$ for some non-vanishig $a = (a_1,a_2)$. If $\|f\|_{1,E} = 1$ then \begin{align} 1 = \sup_{x \in E,\ \|x\|_1 = 1} | f(x) | &= \sup_{x_1 a_1 + x_2 a_2 = 0,\ \|x\|_1 = 1}\left| x_1 f_1 + x_2 f_2 \right| \\ &= \sup_{\|x\|_1 = 1}\left| x_1 f_1 -\frac{a_1}{a_2} x_1 f_2 \right| \\ &= \left| f_1 -\frac{a_1}{a_2} f_2 \right|\sup_{\|x\|_1 = 1} |x_1| \\ &= \left| f_1 -\frac{a_1}{a_2} f_2 \right| \frac{|a_2|}{|a_1| + |a_2|} = \frac{ |a_2 f_1 - a_1 f_2| }{|a_1| + |a_2|} = \frac{|\det(f,a)|}{|a_1| + |a_2|} \end{align} or more concisely $$ |\det(f,a)| = \| a \|_1 $$ But that doesn't determine $f$.

Attempt 2:

I wanted to find a linear functional such that $f(v) = v_1 f_1 + v_2 f_2 = 1$, for $v\in E$ and $f(u) = u_1 f_1 + u_2 f_2 = 0$ for linearly independent vector, where $\|v\|_1 = \|u\|_1 = 1$. Then we get the solutions \begin{align} f_1 &= \frac{u_2}{\det(v,u)} & f_2 &= -\frac{u_1}{\det(v,u)} \end{align} But that stll doesn't determine $u$ nor $f$.


By the way, I found the solutions geometrically, i.e. without computations as above, but I don't know how to justify them with computation. Any hints would be helpful. It will be great to learn something for the infinite dimensional case from this exercise.

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1 Answer 1

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There are no many options for a subspace, so lets assume that $E=\{(x_1, x_2)\in \mathbb{R}^2: x_2=ax_1\}$ for some $a\in \mathbb{R}$, then we also may assume that the linear functional $f: E\to \mathbb{R}$ must be of the form \begin{align*} f(x_1, x_2)=\eta x_1, \end{align*} for some $\eta\in \mathbb{R}$. Now we observe that \begin{align*} |f(x_1, x_2)|=|\eta||x_1|=\dfrac{|\eta|}{1+|a|}\|(x_1, x_2)\|_1 \end{align*} hence by assumption the relation between the data $\eta$ and $a$ must be \begin{align*} \dfrac{|\eta|}{1+|a|}=1. \end{align*} Now, by Hahn-Banach, $f$ admits an extension to the whole $\mathbb{R}^2$, denoted by $F$. By letting $\beta_1=F(1, 0)$ and $\beta_2=F(0, 1)$, we can describe $F$ by \begin{align*} F(x_1, x_2)=\beta_1x_1+\beta_2x_2,\ \forall x_1, x_2\in\mathbb{R}. \end{align*} Since $F$ extends $f$, over $E$ we must have \begin{align*} \eta x_1=f(x_1, ax_1)=F(x_1, ax_1)=(\beta_1+a\beta_2)x_1, \end{align*} which results on the first condition for the extension: $\beta_1+a\beta_2=\eta$. The induced $1$-norm of $F$ is given by \begin{align*} \|F\|_1=\max\{|\beta_1|, |\beta_2|\}, \end{align*} therefore the second conditions for the extension is \begin{align*} \max\{|\beta_1|, |\beta_2|\}=1. \end{align*} since $F$ has norm $1$. Hence by solving the system \begin{align*} \begin{cases}\beta_1+a\beta_2=\eta,\\ \max\{|\beta_1|, |\beta_2|\}=1,\end{cases} \end{align*} you can determine $F$.

Observe that this system does not have unique solutions. For instance, $(\beta_1, \beta_2)=(0, 1)$ and $(\beta_1, \beta_2)=(1, 0)$ solves the system, and each solution is associated to a different linear extension. The condition to have a unique Hahn-Banach extension (preserving the norm) for a linear functional $f: M\leq X\to \mathbb{R}$, is that the dual space $X^*$ is strictly convex.

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  • $\begingroup$ Could you please explain what the induced 1-norm of $F$ is ? where does it come from ? $\endgroup$
    – Physor
    Nov 8, 2021 at 21:44
  • $\begingroup$ It is the best constant satisfying $|F(x)|\leq C\|x\|_1$ for all $x\in \mathbb{R}^2$. There are some characterizations for such number (I assume that those are well known for you from the definitions you gave), and, as I assert, in this case such number is given by $\max\{|\beta_1|, |\beta_2|\}$. It is a good exercise to prove that claim. $\endgroup$
    – rebo79
    Nov 8, 2021 at 21:54
  • $\begingroup$ Could you remind me what characterisations you mean, perhaps I know what you mean $\endgroup$
    – Physor
    Nov 8, 2021 at 21:56
  • $\begingroup$ The second definition you gave is exactly the induced $1$-norm of a linear functional $\endgroup$
    – rebo79
    Nov 8, 2021 at 21:58
  • $\begingroup$ The whole difficulty was to arrive at that max, but is this an implication of sth like the spaces $L^p$ and $L^q$ are dual if $\frac{1}{p} + \frac{1}{q} = 1$ ? I say that because the max norm is $(q = \infty)$-norm dual to $(p = 1)$-norm of the space $\endgroup$
    – Physor
    Nov 8, 2021 at 22:02

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