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 $i=1:n$.

  • $\begingroup$ 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

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.

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

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