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Wachipichay (talk | contribs) Modulus of convexity of the Lp spaces |
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.</ref> Namely, there exists {{nowrap|''q'' ≥ 2}} and a constant {{nowrap|''c'' > 0}} such that
::<math>\delta(\varepsilon) \ge c \, \varepsilon^q, \quad \varepsilon \in [0, 2].</math>
==Modulus of convexity of the <math>L^p</math> spaces==
The modulus of convexity is known for the L^p spaces.<ref>{{citation
| last = Hanner
| first = Olof
| title = On the uniform convexity of <math>L^p</math> and <math>\ell^p</math>
| journal = Arkiv för Mathematik
| volume = 3
| year = 1955
| pages = 239-244
}}</ref> If <math>1<p\le2</math>, then it satisfies the following implicit equation:
:<math>\left(1-\delta_p(\varepsilon)+\frac{\varepsilon}{2}\right)^p+\left(1-\delta_p(\varepsilon)-\frac{\varepsilon}{2}\right)^p=2.
</math>
Knowing that <math>\delta_p(\varepsilon+)=0,</math> one can suppose that <math>\delta_p(\varepsilon)=a_0\varepsilon+a_1\varepsilon^2+\cdots</math>. Substituting this into the above, and expanding the left-hand-side as a Taylor series around <math>\varepsilon=0</math>, one can calculate the <math>a_i</math> coefficients:
:<math>\delta_p(\varepsilon)=\frac{p-1}{8}\varepsilon^2+\frac{1}{384}(3-10p+9p^2-2p^3)\varepsilon^4+\cdots.
</math>
For <math>2<p<\infty</math>, one has the explicit expression
:<math>\delta_p(\varepsilon)=1-\left(1-\left(\frac{\varepsilon}{2}\right)^p\right)^{\frac1p}.
</math>
Therefore, <math>\delta_p(\varepsilon)=\frac{1}{p2^p}\varepsilon^p+\cdots</math>.
== See also ==
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