2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $1$. $\endgroup$ – Mike Earnest Jan 5 at 2:48
  • $\begingroup$ @MikeEarnest I had a similar idea, but it was not enough (for me). I'll write it. $\endgroup$ – Corrêa Jan 5 at 2:52
  • 1
    $\begingroup$ 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 $\lambda=\|x\|/(\|x\|+\|y\|)$. $\endgroup$ – Mike Earnest Jan 5 at 3:01
  • $\begingroup$ Also, the $L_1$ norm is not strict. $\endgroup$ – Mike Earnest Jan 5 at 15:34
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.