Problem about strictly normed spaces.

Question

A normed vector space $(V,\Vert \cdot \Vert)$ is strictly normed if $$\Vert x + y\Vert = \Vert x \Vert + \Vert y \Vert$$ with $x,y\neq 0$ only if $y = \lambda x$ where $\lambda >0$.

(a) Prove that $V$ is strictly normed if only if the sphere $$\sigma_{1}(0) = \{x \in V \mid \Vert x \Vert = 1\}$$ contains no segments.

(b) Give examples of strictly normed spaces and not strictly normed spaces.

My attempt.

(a) Suppose that $V$ is strictly normed. Take $x, y \in \sigma_{1}(0)$ with $x \neq y$. If $y = \alpha x$ with $\alpha > 0$, then $$1 = \Vert y \Vert = \alpha \Vert x \Vert = \alpha \Longrightarrow y = x.$$ Moreover $$\Vert \lambda x + (1-\lambda)y\Vert = \Vert \lambda x \Vert + \Vert(1-\lambda)y\Vert$$ only if $(1-\lambda)y = \alpha \lambda x$, that is, $\displaystyle y = \left(\frac{\lambda\alpha}{1-\lambda}\right)x$ (we can take $\lambda \in (0,1)$), then $$\Vert \lambda x + (1-\lambda)y\Vert < \lambda\Vert x \Vert + (1-\lambda)\Vert y \Vert = 1$$ and if $\lambda x + (1-\lambda)y \in \sigma_{1}(0)$, $\Vert \lambda x + (1-\lambda)y \Vert = 1$ and so, $1<1$, an absurd!

For converse, I take $x,y \in V$ with $x \neq y$. So, $\displaystyle \frac{x}{\Vert x \Vert},\frac{y}{\Vert y \Vert} \in \sigma_{1}(0)$. But I dont know how to use the hypothesis. Can someone help me?

Edit. $$\left\Vert \lambda \frac{x}{\Vert x \Vert} + (1-\lambda)\frac{y}{\Vert y \Vert}\right\Vert \leq \lambda \frac{\Vert x \Vert}{\Vert x \Vert} + (1-\lambda)\frac{\Vert y \Vert}{\Vert y \Vert} = 1,$$ but, if $\lambda \in (0,1)$, $$\left\Vert \lambda \frac{x}{\Vert x \Vert} + (1-\lambda)\frac{y}{\Vert y \Vert}\right\Vert \not\in \sigma_{1}(0),$$ then $$\left\Vert \lambda \frac{x}{\Vert x \Vert} + (1-\lambda)\frac{y}{\Vert y \Vert}\right\Vert < 1.$$

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(b) Consider the Euclidean norm $\Vert \cdot \Vert_{E}$ and the sum norm $\Vert \cdot \Vert_{\infty}$. So, $(\mathbb{R}^{n},\Vert \cdot \Vert_{E})$ is strictly normed and $(\mathbb{R}^{n}, \Vert \cdot \Vert_{\infty})$ is not strictly normed.

Here, I didnt write proof of this, I'm just using the previous equivalence. Please, correct me if Im wrong.

Can someone knows another examples of not strictly normed?

Answer 1

$$ \frac{x+y}{|x|+|y|}=\frac{|x|}{|x|+|y|}\Big(\frac{x}{|x|}\Big)+\frac{|y|}{|x|+|y|}\Big(\frac{y}{|y|}\Big) $$ This shows that $v:=(x+y)/(|x|+|y|)$ is on the segment connecting $x/|x|$ to $y/|y|$. Since the unit sphere has no segments, $|v|$ cannot be $1$. By the triangle inequality, $|v|\le 1$.

Answer 2

Suppose $V$'s unit sphere contains no line segments, and $x, y \in V$ such that $$\|x + y\| = \|x\| + \|y\|.$$ Let $z$ be the point on the line segment $[0, x + y]$ that you would expect to be distance $\|x\|$ from $0$ and distance $\|y\|$ from $x + y$. Working this out, you'll get $$z = \frac{\|x\|(x + y)}{\|x + y\|}.$$ Note that $z$ lies on the spheres $S[0; \|x\|]$ and $S[x + y; \|y\|]$.

Also note the same is true for $x$. That is, $x$ and $z$ lie in both spheres. Let's suppose they're different points. Since they both lie in $S[0; \|x\|]$, it follows from the convexity of the ball that $\frac{x + z}{2}$ must lie in the open ball $B(0; \|x\|)$, which is to say $\left\|\frac{x + z}{2}\right\| < \|x\|$. On the same token, we have $\frac{x + z}{2} \in B(x + y, \|y\|)$. Hence,

$$\|x + y\| \le \left\|\frac{x + z}{2}\right\| + \left\|x + y - \frac{x + z}{2}\right\| < \|x\| + \|y\| = \|x + y\|,$$

which is a contradiction. Thus, $x = z$, and from this it's easy to see that $x$ and $y$ are parallel.

As for your other question, you can form a norm from a unit ball. The eligible unit balls are precisely the non-empty symmetric, closed, bounded, convex subsets of $\mathbb{R}^n$. This gives you a lot of scope to find norms that are strict or non-strict.