Lemma 14.19.43C3049F567279

Proof. Each $f_{\infty ,i}$ is an increasing limit of positive concave functions, hence concave with values in $(0,+\infty ]$. Suppose it is infinite at one point and finite at another point of $\Gamma $. Extend the segment joining them slightly beyond the second point, inside the open cone. The second point is then an interior point of the longer segment, so concavity forces the value there to be infinite, a contradiction. Thus $f_{\infty ,i}$ is either finite everywhere, and then continuous, or infinite everywhere. By symmetry, the same alternative holds for every $i$.

Take $\delta =\frac18\min \{ \delta _0,\min _ Xf(\mu )\} >0$. In the finite case set $H=\min \{ H_0,f_\infty (\mu )-4\delta \} $; otherwise set $H=H_0$. Since $f_\infty (\mu )\geq f(\mu )\geq 8\delta $, this function is positive. The conditions on the family give $h\leq H$ and $f_\infty (\mu )\geq H+4\delta $. Since $\mu (X)$ is a compact subset of $\Gamma $ and the functions $f_{\infty ,i}$ are continuous (or identically infinite), a sufficiently small $r>0$ gives $\mu -2r\mathbf1\in \Gamma $ and $f_\infty (\mu -2r\mathbf1)\geq H+3\delta $ on $X$.

If the matrix bound failed, there would be points $x_ j\to x_*$ and $P_ j\geq 0$ with unbounded norms and $F(A_{\underline u}(x_ j)-r\operatorname {Id}+P_ j)\leq H(x_ j)+2\delta $. By monotonicity of eigenvalues, the ordered eigenvalues $\lambda ^{(j)}$ of $A_{\underline u}(x_ j)-r\operatorname {Id}+P_ j$ satisfy $\lambda _ i^{(j)}\geq \mu _ i(x_ j)-r$. Their sum tends to infinity, so some component $k$ tends to infinity along a subsequence. For every fixed $t>0$, eventually

Since $f$ is increasing in each variable, letting $t\to \infty $ shows that the lower limit of $F(A_{\underline u}(x_ j)-r\operatorname {Id}+P_ j)$ is at least $f_\infty (\mu (x_*)-2r\mathbf1)\geq H(x_*)+3\delta $, which contradicts the upper bound $H(x_ j)+2\delta $. Adding positive semidefinite matrices preserves the cone, so all these comparisons take place in $\Gamma $. $\square$