# why restricted isometry property constant $\delta_{2k}<\sqrt{2}-1$?

It's said that $$\delta_{2k} < \sqrt{2} -1$$ , the solution of the $$l_{1}$$ problem is that of $$l_{0}$$ problem.

I checked the proof of $$||x^{*}-x||_{l_{2}}\leq C_{0}s^{-1/2}||x-x_{s}||_{l_{1}}+C_{1}\epsilon$$

in this file but I haven't found out where the $$\sqrt{2}-1$$ works.

So it will be very helpful if anyone could tell me if I missed something in that paper or direct me to some relevant literature or give me some tips.

• Welcome. Could you please share how far you did check the proof? Jun 4 '19 at 22:19
• Thanks very much, I tried to write the proof briefly and found that the inequality implicitly requires $1-\frac{\sqrt{2}\delta_{2k}}{1-\delta_{2k}}>0$ which is excatly $\delta_{2k}<\sqrt{2}-1$. Jun 5 '19 at 6:29