1
$\begingroup$

Suppose $\Big(\sum_{k =1}^n x_k^2\Big)^{1/2} \leq \eta_2$.

$\eta_2$ is the bound on the l_2 norm of a matrix. I want to upper bound $\eta_\infty = \max_{i}|x_i|$ using $\eta_2$ the bound on the l_2 norm of the matrix.

Is there a standard approach to do this?

$\endgroup$
2
  • $\begingroup$ Defining all the quantities in the question would be helpful. $\endgroup$
    – Kavi Rama Murthy
    Commented Oct 8, 2018 at 6:11
  • $\begingroup$ made the formula simpler to reflect my question. Is it clearer? $\endgroup$
    – user77005
    Commented Oct 8, 2018 at 7:16

1 Answer 1

Reset to default
2
$\begingroup$

For each $i$, $x_i^{2} \leq \sum_{k=1}^{n} x_k^{2}$ so $|x_i| \leq \sqrt {\sum_{k=1}^{n} x_k^{2}} \leq \eta_2$. Taking sup over $i$ you get $\eta_{\infty} \leq \eta_2$.

$\endgroup$
6
  • $\begingroup$ what is $a^2_k$? Is this bound correct if its an integration instead of sum? $\endgroup$
    – user77005
    Commented Oct 8, 2018 at 13:56
  • $\begingroup$ Sorry, I typed $a_k$ for $x_k$. I have corrected it now. The inequality fails for integrals. For example, if $f(x)=x$ in $(0,1)$ then $\max f(x)=1$ and $\int_0^{1} f^{2}(x) \, dx =\frac 1 3$. $\endgroup$
    – Kavi Rama Murthy
    Commented Oct 8, 2018 at 23:21
  • $\begingroup$ You mean $\left(\int_0^1 f(x)^2\,dx\right)^{1/2}$? $\endgroup$
    – Ted Shifrin
    Commented Oct 8, 2018 at 23:23
  • $\begingroup$ Sorry again. Corrected my comment. $\endgroup$
    – Kavi Rama Murthy
    Commented Oct 8, 2018 at 23:23
  • $\begingroup$ LOL, as did I. :) $\endgroup$
    – Ted Shifrin
    Commented Oct 8, 2018 at 23:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .