The compression Jensen inequality for the negative logarithm

For $t>0$ and $R>0$, set $F_t(s)=\ln (t+s)-\ln (1+s)$ on $s>-\min (t,1)$. Translated difference quotients in [F2] give $F_t^{\mathrm{\prime }}(s)=(t+s{)}^{-1}-(1+s{)}^{-1}$, including the one-sided derivatives on $[0,R]$. For $s,u\in [0,R]$, $\mathrm{\mid }(t+s{)}^{-1}-(t+u{)}^{-1}\mathrm{\mid }\le \mathrm{\mid }s-u\mathrm{\mid }\mathrm{/}{t}^2$ and $\mathrm{\mid }(1+s{)}^{-1}-(1+u{)}^{-1}\mathrm{\mid }\le \mathrm{\mid }s-u\mathrm{\mid }$, so the derivative is continuous and integrable by [F5]. Thus [F6] and [F4] give ${\int }_0^R((t+s{)}^{-1}-(1+s{)}^{-1})ds=\ln ((t+R)\mathrm{/}(1+R))-\ln t$. As $R\to \mathrm{\infty }$, the quotient tends to $1$, so [F4] gives the limit $-\ln t$ (also when $t=1$). Applying this scalar identity to each of the finitely many positive eigenvalues in [F7] yields the corresponding operator integral identity in norm, since the maximum of finitely many scalar errors tends to zero. On a zero-dimensional space the empty spectral formula is the unique endomorphism, so the same assertion is vacuous and valid.

Since $V^{\ast }V=I_H$ and $T$ is positive definite, $V^{\ast }TV$ is positive definite on nonzero $H$: for $x\mathrm{\ne }0$, $Vx\mathrm{\ne }0$ and $⟨V^{\ast }TVx,x⟩=⟨TVx,Vx⟩>0$. Apply [F1] at each $s\ge 0$. Subtracting $(1+s{)}^{-1}{I}_H$ preserves the order inequality. For every $x$, integrating the resulting quadratic-form inequality over $[0,R]$ gives a non-negative quadratic form; the operator norm limits as $R\to \mathrm{\infty }$ exist by step 1.1, and the limit remains non-negative because quadratic forms are continuous. The limit is $-\ln (V^{\ast }TV)⪯V^{\ast }(-\ln T)V$, as required.