So, I have trouble proving the following, I'd be grateful if somebody helps me with this.

Let $z$ be a given point in $\mathbb{R}^m$. Then, $x\in \mathbb{R}^m$ is a dual vector of $z$ with respect to $\|.\|$ if it satisfies $\|x\|=1$ and $z^Tx=\|z\|'$.
A norm $\|.\|$ is said to be strictly convex if the unit sphere $\{x:\|x\|=1\}$ contains no line segment.

Now, how does one prove that

The norm $\|.\|$ is strictly convex if and only if each $z\in \mathbb{R}^m$ has a unique dual vector.

Here is my attempt: Suppose that $\|.\|$ is strictly convex, that is, $\{x:\|x\|=1\}$ does not contain any line segment.
Let us take $x$ such that $\|x\|=1$. Then,
Then, can I say $\|z\|'=z^Tx$ since $\|x\|=1$ and hence $x$ is a dual vector of $z$?
For uniqueness, let $x_0$ be another dual vector of $z$, hence $\|x_0\|=1$.

Now, how do I use the strict convexity of $\|.\|$ to show that $x_0=x$?
Conversely, suppose that $z$ has a unique dual vector $x$. So, $\|z\|'=z^Tx$ and $\|x\|=1$. We know that $\|.\|$ is strictly convex when the sphere $\{x:\|x\|=1\}$ does not contain any line segments.
How do I connect these two ideas?

  • $\begingroup$ To confirm, $\|z\|'$ is the dual norm of $z$? Which definition of dual norm do you use? $\endgroup$ – Theo Bendit May 30 at 5:43
  • $\begingroup$ I am downvoting your post and opting to close it, because just as in your previous lengthy post on the definition of a norm of a convex body, you put a lot of effort in copying the problem, but no effort whatsoever to solve it yourself. $\endgroup$ – uniquesolution May 30 at 5:46
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    $\begingroup$ @TheoBendit $\|z\|'=\underset{\|x\|\le 1}{\max} z^Tx$. $\endgroup$ – Octagonal Monk May 30 at 6:03
  • $\begingroup$ @uniquesolution Aww, you put so much effort into writing this lengthy comment but put no effort into helping me? I didn't solve it because I have no idea where to start it. $\endgroup$ – Octagonal Monk May 30 at 6:04
  • $\begingroup$ I'm not sure this is true. Should we instead be proving $\|\cdot\|'$ is strictly convex? $\endgroup$ – Theo Bendit May 30 at 6:16

Sorry, I got it backwards. Yes, this indeed holds.

Fix $z \in \Bbb{R}^m$, and consider the set $$C_z =\left\{x \in \Bbb{R}^m : \|x\| = 1 \text{ and } z \cdot x = \|z\|' = \max\limits_{\|y\| \le 1} z \cdot y\right\}.$$ Note that $C_z$ is a subset of the unit sphere (of $\| \cdot \|$). I claim that $C_z$ is convex. Take two $x_1, x_2 \in C_z$ and $\lambda \in [0, 1]$. Then $$z \cdot (\lambda x_1 + (1 - \lambda)x_2) = \lambda \|z\|' + (1 - \lambda)\|z\|' = \|z\|'.$$ Also, $$\|\lambda x_1 + (1 - \lambda)x_2\| \le \lambda \|x_1\| + (1 - \lambda)\|x_2\| = 1.$$ We now have to show that the above inequality is not strict. Let $$y = \frac{\lambda x_1 + (1 - \lambda)x_2}{\|\lambda x_1 + (1 - \lambda)x_2\|}.$$ Note that $y$ is in the unit sphere (of $\| \cdot \|$), and $$z \cdot y = \frac{z \cdot (\lambda x_1 + (1 - \lambda)x_2)}{\|\lambda x_1 + (1 - \lambda)x_2\|} = \frac{\|z\|'}{\|\lambda x_1 + (1 - \lambda)x_2\|} \ge \|z\|' \ge z \cdot y.$$ Hence $\|\lambda x_1 + (1 - \lambda)x_2\| = 1$, so $$\lambda x_1 + (1 - \lambda)x_2 \in C_z$$ and $C_z$ is convex.

Therefore, if $\|\cdot\|$ is strictly convex, then the sphere contains no line segments. That is, the only non-empty convex subsets of the sphere are singletons. So, $|C_z| = 1$ in such a case, i.e. there is a unique dual vector $x \in C_z$.

On the other hand, suppose $\| \cdot \|$ is not strictly convex. Pick a line segment $L = \operatorname{conv} \{x_1, x_2\}$ in the unit sphere, and let $x$ be their midpoint. Find a supporting hyperplane $H = \{y \in \Bbb{R}^m : z \cdot y = z \cdot x\}$, where $z \in \Bbb{R}^m$ is some fixed non-zero vector. In particular, we choose the direction of $z$ so that the halfspace $S = \{y \in \Bbb{R}^m : z \cdot y \le z \cdot x\}$ contains the unit sphere, and hence its convex hull, the unit ball.

In particular, this means that $z \cdot y$ achieves its maximum over $y$ in the unit ball at $x$, and, indeed, any other point in the intersection of $H$ and the unit sphere. Such points are dual to $z$, by definition!

Now, I claim that $x_1$ and $x_2$ are two such points. We have, \begin{align*} z \cdot x_1 \le z \cdot x &= z \cdot \frac{x_1 + x_2}{2} \\ z \cdot x_2 \le z \cdot x &= z \cdot \frac{x_1 + x_2}{2}. \end{align*} Assume either of the inequalities above was strict. Then, adding these inequalities together, $$z \cdot x_1 + z \cdot x_2 < 2 z \cdot \frac{x_1 + x_2}{2},$$ which is absurd. Thus, equalities hold above, and $x_1, x_2 \in H$, as claimed. Thus, $z$ is dual to (more than) two points $x_1, x_2$.

  • $\begingroup$ $x=\frac{x_1+x_2}{2}$, how? And, $z$ is dual to more than two points $x_1, x_2$? But we have to prove that $z$ has a unique dual. $\endgroup$ – Octagonal Monk May 30 at 7:42
  • $\begingroup$ That was my definition of $x$: the midpoint between $x_1$ and $x_2$. The thrust of that part of the proof is to show that, if $\|\cdot\|$ is not strictly convex, then dual points are not unique. Contrapositively, this shows that unique dual points implies strict convexity. $\endgroup$ – Theo Bendit May 30 at 7:45
  • $\begingroup$ Oh, ok. Is that why put $z.x_1\le z.x$? $\endgroup$ – Octagonal Monk May 30 at 7:47
  • $\begingroup$ I put $z \cdot x_1 \le z \cdot x$ because $x_1$ lies in the unit sphere, and hence lies in the halfspace $S$. $\endgroup$ – Theo Bendit May 30 at 7:48
  • $\begingroup$ Ok, thank you so much. I now fully understand it. $\endgroup$ – Octagonal Monk May 30 at 7:50

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