Talk:Lojasiewicz inequality

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About (1): probably, $|f(x)|$ is meant? Also, I guess what is $Z_f$, but it could be said explicitly. Boris Tsirelson (talk) 20:24, 7 April 2014 (CEST)

Indeed! Thanks for pointing it out. Camillo (talk) 08:59, 8 April 2014 (CEST)

About the recent edit by Bark10731: I'm afraid, the former "$V\subset\subset U$" by Camillo was not an error; rather, its meaning was "the closure of $V$ is contained in $U$". Boris Tsirelson (talk) 17:33, 10 February 2015 (CET)

In fact it meant that "the closure of $V$ is compact and contained in $U$". The point is that the exponent $\alpha$ might degenerate as one reaches the boundary of $U$ (or goes to infinity). I will change "$V$ open" to "$K$ compact", so that there is no confusion in the future. Camillo (talk) 20:53, 10 February 2015 (CET)
My apologies for careless edit. Another point, would you consider whether we need a positive constant to multiply $|f(x)|$ in (1)?--Bark10731 (talk) 00:57, 11 February 2015 (CET)
Indeed, I think, we need. Thank you. Boris Tsirelson (talk) 07:21, 11 February 2015 (CET)
Yes, stated as it is, a constant is needed. Actually, the inequality is really useful when you are in a neighborhood of $Z_f$, i.e. when ${\rm dist}\, (x, Z_f)$ is small: under such assumption one can just raise the exponent of the left hand side to get rid of the eventual constant. Camillo (talk) 09:33, 11 February 2015 (CET)
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