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Corollary 13.5.1143C93EF28279

Corollary 13.5.11.43C93EF28279 Under the hypotheses of corollary 13.5.10, for every $0\leq t\leq 1$,

\[ \left\langle \big ((1-t)T+tS\big )^ n\right\rangle \geq \big ((1-t)f^{1/n}+tg^{1/n}\big )^ n\nu . \]

Proof. For $t=0$ and $t=1$ this is the hypothesis. Otherwise, multilinearity (proposition 13.2.6) expands the left side as $\sum _{k=0}^ n\binom nk(1-t)^ kt^{n-k}\langle T^ k\wedge S^{n-k}\rangle $. Bounding the terms with $1\leq k\leq n-1$ by corollary 13.5.10 and the terms with $k=0,n$ by the hypotheses, we obtain

\[ \left\langle \big ((1-t)T+tS\big )^ n\right\rangle \geq \sum _{k=0}^ n\binom nk(1-t)^ kt^{n-k}f^{k/n}g^{(n-k)/n}\, \nu =\big ((1-t)f^{1/n}+tg^{1/n}\big )^ n\nu \]

by the binomial formula. The measure on the right is finite, because $f$ and $g$ are $\nu $-integrable by the hypotheses and each density $f^{k/n}g^{(n-k)/n}$ is at most $\frac knf+\frac{n-k}ng\leq f+g$: for $k=0$ and $k=n$ this is an equality, and for $1\leq k\leq n-1$ it is (13.5.9). $\square$

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