0
$\begingroup$

The norm $||\cdot||$ on $\mathbb{R}^n$ is absolute if || |$x$| || = ||$x$||. Let $x = [x_1, x_2, ..., x_n]^T$. I want to show that $||[x_1, x_2, ..., x_{k-1},\alpha x_k, x_{k+1},...,x_n]^T|| \le ||x||$, where $\alpha \in [0,1]$. Let $e_k \in \mathbb{R}^n$ where $k$-th entry is 1, otherwise 0. We can verify that

$[x_1, x_2, ..., x_{k-1},\alpha x_k, x_{k+1},...,x_n]^T = \frac{1}{2}(1-\alpha)(x-2x_ke_k) + \frac{1}{2}(1-\alpha)x + \alpha x$.

Using this equation, it follows that

$||\frac{1}{2}(1-\alpha)(x-2x_ke_k) + \frac{1}{2}(1-\alpha)x + \alpha x||$

$\le||\frac{1}{2}(1-\alpha)(x-2x_ke_k +x)|| + ||\alpha x||$ by triangle inequality

$=\frac{1}{2}(1-\alpha)||2x-2x_ke_k|| + \alpha||x||$

$=(1-\alpha)||x-x_ke_k|| + \alpha||x||$

This is as far as I could possibly get. I would probably have to use the definition of absolute norm here, but I'm a little stumped on how to apply here. It would be nice if $||x-x_ke_k|| \le ||x_k||$, because that would finish the proof, but how can I finish this last line?

Edit: To clarify, if $x \in \mathbb{R}^n$, |x| means you apply absolute value for every entry of $x$, so it is still a vector, not a scalar. Also, $|x| \le |y|$ means $|x_i| \le |y_i|$ for $i=1:n$.

$\endgroup$
2
  • $\begingroup$ What does $|x|$ mean for $x\in\mathbb R^n$? $\endgroup$ – Jason Jan 23 '18 at 3:43
  • $\begingroup$ Just edited it. $\endgroup$ – Ted Jan 23 '18 at 3:48
1
$\begingroup$

Put $y=x-x_ke_k$ and $z=x-2x_ke_k$. Observe that $y$ and $z$ are just $x$ with the $k^\text{th}$ coordinate replaced by $0$ and $-x_k$, respectively. In particular, by assumption of the norm being absolute, $\|z\|=\|x\|$. Writing $y=\frac12 (x+z)$ and applying the triangle inequality allows you to conclude.

$\endgroup$
1
  • $\begingroup$ Nice answer! Thanks! $\endgroup$ – Ted Jan 23 '18 at 4:04

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.