# Bounding the infinity norm using the l_2 norm

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?

• Defining all the quantities in the question would be helpful. Commented Oct 8, 2018 at 6:11
• made the formula simpler to reflect my question. Is it clearer? Commented Oct 8, 2018 at 7:16

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 $$\eta_{\infty} \leq \eta_2$$.
• what is $a^2_k$? Is this bound correct if its an integration instead of sum? Commented Oct 8, 2018 at 13:56
• 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 $\int_0^{1} f^{2}(x) \, dx =\frac 1 3$. Commented Oct 8, 2018 at 23:21
• You mean $\left(\int_0^1 f(x)^2\,dx\right)^{1/2}$? Commented Oct 8, 2018 at 23:23