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 $$\left\Vert \lambda \frac{x}{\Vert x \Vert} + (1-\lambda)\frac{y}{\Vert y \Vert}\right\Vert < 1.$$

(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?

• The idea is to express some scalar multiple of $x+y$ as a convex combination of $x/\|x\|$ and $y/\|y\|$, and since the sphere has no segments conclude that the norm of that multiple is strictly less than $1$. Jan 5, 2019 at 2:48
• @MikeEarnest I had a similar idea, but it was not enough (for me). I'll write it. Jan 5, 2019 at 2:52
• Your work was correct, you just had to choose $\lambda$ so that $\|x+y\|$ somehow entered the picture. It turns out the correct choice is $\lambda=\|x\|/(\|x\|+\|y\|)$. Jan 5, 2019 at 3:01
• Also, the $L_1$ norm is not strict. Jan 5, 2019 at 15:34

$$\frac{x+y}{|x|+|y|}=\frac{|x|}{|x|+|y|}\Big(\frac{x}{|x|}\Big)+\frac{|y|}{|x|+|y|}\Big(\frac{y}{|y|}\Big)$$ 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, $$|v|\le 1$$.
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.