# Strict Convexity and Uniqueness of Dual norm

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,
$$\|z\|'=\underset{\|x\|=1}{\max}\frac{z^Tx}{\|x\|}.$$
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?

• To confirm, $\|z\|'$ is the dual norm of $z$? Which definition of dual norm do you use? – Theo Bendit May 30 at 5:43
• 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. – uniquesolution May 30 at 5:46
• @TheoBendit $\|z\|'=\underset{\|x\|\le 1}{\max} z^Tx$. – Octagonal Monk May 30 at 6:03
• @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. – Octagonal Monk May 30 at 6:04
• I'm not sure this is true. Should we instead be proving $\|\cdot\|'$ is strictly convex? – 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 $$x_1, x_2$$.

• $x=\frac{x_1+x_2}{2}$, 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. – Octagonal Monk May 30 at 7:42
• 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. – Theo Bendit May 30 at 7:45
• Oh, ok. Is that why put $z.x_1\le z.x$? – Octagonal Monk May 30 at 7:47
• I put $z \cdot x_1 \le z \cdot x$ because $x_1$ lies in the unit sphere, and hence lies in the halfspace $S$. – Theo Bendit May 30 at 7:48
• Ok, thank you so much. I now fully understand it. – Octagonal Monk May 30 at 7:50