# convergence in higher order mean implies convergence in lower order mean

Why does convergence in higher order mean implies convergence in lower order mean?

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@user957: Homework? –  Shai Covo Mar 30 '11 at 16:13
Hint. For $0<r<s$, write $r=s(r/s)$; note that $r/s < 1$. There is a well-known inequality concerning concave functions...