Proof of The Free Entropy: the Semicircular Upper Bound, Semicircular Values, Smoothing, the de Bruijn Identity and Upper Semicontinuity

Each result cited is universally quantified over the data in its own statement. We work in Free Products, Semicircular Laws, the Free Heat Flow, Free Fisher Information and Free Entropy: Standing Notation. Throughout, $\mathrm{Leb}$ is Lebesgue measure on the Borel $\sigma$-algebra of $\mathbb{R}$, as in Rays, and Integrability and the Integral of a Real Function on a Borel Subset of the Real Line §integral; open intervals, open rays and one-point sets are Borel sets. For a measurable $v:\mathbb{R}\to\mathbb{R}$ with $v\ge0$, $\int_{\mathbb{R}}v\,d\mathrm{Leb}$ denotes the integral of Lebesgue Integral of a Nonnegative Measurable Function, a member of $[0,\infty]$; for an integrable $v:\mathbb{R}\to\mathbb{R}$ it denotes the real number of Integrable Function and the Lebesgue Integral. Since $d$ is a natural number, $d\ge1>0$. Moreover $\log1=0$, because $\exp(0)=1$ by claim 1 of Basic Properties of the Exponential Function and $\log(\exp(u))=u$ by The Natural Logarithm; and $\log(xy)=\log x+\log y$ for all reals $x,y>0$, by The Natural Logarithm.

Preliminaries. For a Borel set $B\subseteq(0,\infty)$ and a real function $u$ defined at least on $B$, let $E_{B}u:\mathbb{R}\to\mathbb{R}$ be the function equal to $u$ on $B$ and to $0$ off $B$. By Rays, and Integrability and the Integral of a Real Function on a Borel Subset of the Real Line §integral, $u$ is integrable on $B$ exactly when $E_{B}u$ is measurable and integrable with respect to $\mathrm{Leb}$, and then $\int_{B}u(t)\,dt=\int_{\mathbb{R}}E_{B}u\,d\mathrm{Leb}$. We record seven facts.

(P1) If $u$ is defined on $(0,\infty)$ and $E_{(0,\infty)}u$ is measurable, then for every Borel $B\subseteq(0,\infty)$ the function $E_{B}u=(E_{(0,\infty)}u)\,\mathbf{1}_{B}$ is measurable, by claims 1 and 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions.

(P2) If $v:\mathbb{R}\to\mathbb{R}$ is measurable, $w:\mathbb{R}\to\mathbb{R}$ is integrable and $|v(x)|\le|w(x)|$ for every $x\in\mathbb{R}$, then $v$ is integrable. Indeed $|v|$ and $|w|$ are measurable by claim 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions; $\int_{\mathbb{R}}|v|\,d\mathrm{Leb}\le\int_{\mathbb{R}}|w|\,d\mathrm{Leb}$ by claim 1 of Linearity and Monotonicity of the Lebesgue Integral; the right side is finite, hence so is the left, and $v$ is integrable, both by the criterion stated in Integrable Function and the Lebesgue Integral.

(P3) If $v:\mathbb{R}\to\mathbb{R}$ is measurable and $v(x)\ge0$ for every $x$, then $v^{+}=v$ and $v^{-}=0=\mathbf{1}_{\varnothing}$, and $\int_{\mathbb{R}}\mathbf{1}_{\varnothing}\,d\mathrm{Leb}=\mathrm{Leb}(\varnothing)=0$ by The Integral of an Indicator Function is the Measure of the Set. Hence, by Integrable Function and the Lebesgue Integral, $v$ is integrable if and only if its integral in $[0,\infty]$ is finite, and then its Lebesgue integral equals that integral. In particular the zero function is integrable with integral $0$.

(P4) If $u$ is integrable on $(0,\infty)$ and $B\subseteq(0,\infty)$ is Borel, then $u$ is integrable on $B$: $E_{B}u$ is measurable by (P1), and $|E_{B}u(x)|\le|E_{(0,\infty)}u(x)|$ for every $x$, so (P2) applies.

(P5) Let $\kappa$ be real. If $a<r$ and $b$ are reals with $a+b>0$, then $t\mapsto\kappa/(b+t)$ is integrable on $(a,r)$ and $\int_{(a,r)}\kappa/(b+t)\,dt=\kappa\bigl(\log(b+r)-\log(b+a)\bigr)$; if $a,b,c$ are reals with $a+b>0$ and $a+c>0$, then $t\mapsto\kappa/(b+t)-\kappa/(c+t)$ is integrable on $(a,\infty)$ with integral $\kappa\bigl(\log(a+c)-\log(a+b)\bigr)$. This follows from Logarithmic Integrals of Reciprocal Linear Functions over Bounded and Unbounded Intervals §bounded and Logarithmic Integrals of Reciprocal Linear Functions over Bounded and Unbounded Intervals §half-line (with the letter $s$ there taken to be $r$), since $E_{B}(\kappa u)=\kappa\,E_{B}u$, together with claim 2 of Linearity and Monotonicity of the Lebesgue Integral.

(P6) Let $\mu\in\Sigma_{d}$. For every real $t>0$, $M(\mu)+td>0$ and $d^{2}/(M(\mu)+td)\le f_{\mu}(t)\le d/t$ by The Free Fisher Information along the Free Heat Flow: Bounds, Lower Semicontinuity in Time and Integrability §bounds; the left side is positive by claims 5 and 7 of Elementary Order Arithmetic in an Ordered Field, so $f_{\mu}(t)>0$. The functions $E_{(0,\infty)}f_{\mu}$ and $E_{(0,\infty)}g_{\mu}$ are measurable by The Free Fisher Information along the Free Heat Flow: Bounds, Lower Semicontinuity in Time and Integrability §semicontinuous. If $\mu\in\mathcal{D}_{\chi}$, then $g_{\mu}$ is integrable on $(0,\infty)$ and $\chi^{*}(\mu)=\frac{1}{2}\int_{(0,\infty)}g_{\mu}(t)\,dt+\frac{d}{2}\log(2\pi e)$, by Voiculescu's Non-Microstates Free Entropy of a Noncommutative Law §entropy; and $\mu\in\mathcal{D}_{\chi}$ means that $f_{\mu}$ is integrable on $(0,1)$, by Voiculescu's Non-Microstates Free Entropy of a Noncommutative Law §domain.

(P7) Let $\mu\in\Sigma_{d}$ and let $r\ge0$ be real. For every real $t>0$, $(\mu\boxplus\mathrm{sc}_{d,r})\boxplus\mathrm{sc}_{d,t}=\mu\boxplus\mathrm{sc}_{d,r+t}$ by The Free Heat Flow: Realisation, Norm Bound, Moments, Distance to the Initial Law, Wasserstein Contraction, Semigroup Property and Continuity §semigroup, so, as $r+t>0$, Voiculescu's Non-Microstates Free Entropy of a Noncommutative Law §integrand gives

$$f_{\mu\boxplus\mathrm{sc}_{d,r}}(t)=\Phi^{*}(\mu\boxplus\mathrm{sc}_{d,r+t})=f_{\mu}(r+t).$$

Clause 1 (semicircular upper bound). Let $\lambda\in\mathcal{D}_{\chi}$ and write $M=M(\lambda)=\sum_{j=1}^{d}\lambda(x_{j}^{2})$. Each $x_{j}$ is self-adjoint by Noncommutative Polynomials Form a Unital Complex Algebra with Involution: Linear Extension from Monomials, Products, Adjoints and Self-Adjoint Parts §adjoint, so $\lambda(x_{j}^{2})=\lambda(x_{j}^{*}x_{j})$ is a real number $\ge0$ by condition (b) of a tracial state; hence $M\ge0$ by claim 2 of Elementary Arithmetic in an Ordered Field.

Suppose, for a contradiction, that $M=0$. Then for $t\in(0,1)$ we have $d^{2}/(M+td)=d/t$, so $f_{\lambda}(t)\ge d/t$ by (P6). Put $F=E_{(0,1)}f_{\lambda}$, which is integrable since $\lambda\in\mathcal{D}_{\chi}$ (P6), and let $I=\int_{\mathbb{R}}F\,d\mathrm{Leb}$, a real number. Let $K=(|I|+1)/d>0$ and $\varepsilon=\exp(-K)$. Then $0<\varepsilon<1$, since $\exp(-K)>0$ by claim 2 and $\exp(-K)<\exp(0)=1$ by claims 4 and 1 of Basic Properties of the Exponential Function; and $\log\varepsilon=-K$. By (P5) with $(a,r,b)=(\varepsilon,1,0)$ and $\kappa=d$ (note $a+b=\varepsilon>0$), the function $v=E_{(\varepsilon,1)}(t\mapsto d/t)$ is integrable with $\int_{\mathbb{R}}v\,d\mathrm{Leb}=d(\log1-\log\varepsilon)=dK$. Moreover $v(x)\le F(x)$ for every $x\in\mathbb{R}$: for $x\in(\varepsilon,1)$ this is $d/x\le f_{\lambda}(x)$; for $x\in(0,\varepsilon]$ it is $0<f_{\lambda}(x)$ by (P6); elsewhere both sides are $0$. By claim 2 of Linearity and Monotonicity of the Lebesgue Integral, $dK\le I$, that is $|I|+1\le I$, which is impossible since $I\le|I|$ by claim 3 of Properties of the Absolute Value in an Ordered Field and $|I|<|I|+1$ by claims 6 and 1 of Elementary Order Arithmetic in an Ordered Field. Hence $M\ne0$, and so $M>0$.

Now for real $t>0$ we have $M+td=d(M/d+t)$, so $d^{2}/(M+td)=d/(M/d+t)$, and (P6) gives

$$g_{\lambda}(t)=\frac{d}{1+t}-f_{\lambda}(t)\le\frac{d}{1+t}-\frac{d}{M/d+t}.$$

By (P5) with $(a,b,c)=(0,1,M/d)$ and $\kappa=d$ (note $a+b=1>0$ and $a+c=M/d>0$), the function $w=E_{(0,\infty)}\bigl(t\mapsto d/(1+t)-d/(M/d+t)\bigr)$ is integrable with $\int_{\mathbb{R}}w\,d\mathrm{Leb}=d(\log(M/d)-\log1)=d\log(M/d)$. Since $E_{(0,\infty)}g_{\lambda}$ is integrable (P6) and $E_{(0,\infty)}g_{\lambda}\le w$ pointwise on $\mathbb{R}$ (both vanish off $(0,\infty)$), claim 2 of Linearity and Monotonicity of the Lebesgue Integral gives $\int_{(0,\infty)}g_{\lambda}(t)\,dt\le d\log(M/d)$. Multiplying by $\frac{1}{2}\ge0$ (claim 5 of Elementary Arithmetic in an Ordered Field) and adding $\frac{d}{2}\log(2\pi e)$,

$$\chi^{*}(\lambda)\le\frac{d}{2}\log\Bigl(\frac{M}{d}\Bigr)+\frac{d}{2}\log(2\pi e)=\frac{d}{2}\log\Bigl(\frac{2\pi e\,M}{d}\Bigr),$$

where the last step is $\log(xy)=\log x+\log y$ with $x=2\pi e>0$ and $y=M/d>0$. This proves clause 1.

Clause 2 (semicircular laws). First, $\boxplus$ is commutative on $\Sigma_{d}$. Let $\alpha,\beta\in\Sigma_{d}$ and let $\varrho=(x_{d+1},\dots,x_{2d},x_{1},\dots,x_{d})$, a $2d$-tuple in $\mathcal{P}_{2d}$. By The Free Product of Two Noncommutative Laws: Existence, Uniqueness, Norm Bound, Substitutions and Weak-Star Continuity §symmetry (with $m=n=d$, applied to the pair $(\beta,\alpha)$), $(\beta\star\alpha)\circ\sigma_{\varrho}=\alpha\star\beta$. By Free Additive Convolution of Two Noncommutative Laws §convolution, $\alpha\boxplus\beta=(\alpha\star\beta)\circ\sigma_{\mathrm{add}}$, where, by the same definition together with Affine Data and Affine Substitutions of Noncommutative Polynomials §tuple and Affine Data and Affine Substitutions of Noncommutative Polynomials §substitution, $\sigma_{\mathrm{add}}$ is the substitution of the tuple $(x_{1}+x_{d+1},\dots,x_{d}+x_{2d})$ of the affine datum $\mathrm{add}$ introduced in Free Additive Convolution of Two Noncommutative Laws §convolution. By Substitution is the Unique Unital Homomorphism with Prescribed Values on the Variables: Monomials, Products, Adjoints and Composition §composition, $\sigma_{\varrho}\circ\sigma_{\mathrm{add}}$ is the substitution of the tuple whose $i$-th entry is $\sigma_{\varrho}(x_{i}+x_{d+i})=x_{d+i}+x_{i}$ (linearity of $\sigma_{\varrho}$, which is a linear map by Substitution of Noncommutative Polynomials into the Variables §substitution, and Substitution is the Unique Unital Homomorphism with Prescribed Values on the Variables: Monomials, Products, Adjoints and Composition §values), which is the tuple of $\mathrm{add}$; since both maps are substitutions of the same tuple (Affine Data and Affine Substitutions of Noncommutative Polynomials §substitution, and uniqueness in Substitution of Noncommutative Polynomials into the Variables §substitution), we get $\sigma_{\varrho}\circ\sigma_{\mathrm{add}}=\sigma_{\mathrm{add}}$. Hence for every $p\in\mathcal{P}_{d}$,

$$(\alpha\boxplus\beta)(p)=(\beta\star\alpha)\bigl(\sigma_{\varrho}(\sigma_{\mathrm{add}}(p))\bigr)=(\beta\star\alpha)(\sigma_{\mathrm{add}}(p))=(\beta\boxplus\alpha)(p).$$

Next, let $r,t\ge0$ be real. The law $\mathrm{sc}_{d,0}\in\Sigma_{d}$ exists by The Semicircular Law and Its Scalings §scaled, and $\mu\boxplus\mathrm{sc}_{d,0}=\mu$ for every $\mu\in\Sigma_{d}$ by The Free Heat Flow: Realisation, Norm Bound, Moments, Distance to the Initial Law, Wasserstein Contraction, Semigroup Property and Continuity §bound; with commutativity, $\mathrm{sc}_{d,0}\boxplus\mu=\mu$. Applying The Free Heat Flow: Realisation, Norm Bound, Moments, Distance to the Initial Law, Wasserstein Contraction, Semigroup Property and Continuity §semigroup to the law $\mathrm{sc}_{d,0}$,

$$\mathrm{sc}_{d,r}\boxplus\mathrm{sc}_{d,t}=(\mathrm{sc}_{d,0}\boxplus\mathrm{sc}_{d,r})\boxplus\mathrm{sc}_{d,t}=\mathrm{sc}_{d,0}\boxplus\mathrm{sc}_{d,r+t}=\mathrm{sc}_{d,r+t}.$$

Now let $s>0$ be the given real. For real $t>0$, the display with $r=s$ and Voiculescu's Non-Microstates Free Entropy of a Noncommutative Law §integrand give $f_{\mathrm{sc}_{d,s}}(t)=\Phi^{*}(\mathrm{sc}_{d,s+t})=d/(s+t)$ by Free Fisher Information: the Cramer-Rao Inequality, the Semicircular Laws, and Closedness under Weak-Star Limits §semicircular (applied with variance $s+t>0$). Thus $E_{(0,1)}f_{\mathrm{sc}_{d,s}}=E_{(0,1)}(t\mapsto d/(s+t))$, which is integrable by (P5) with $(a,r,b)=(0,1,s)$ and $\kappa=d$ (note $a+b=s>0$); so $\mathrm{sc}_{d,s}\in\mathcal{D}_{\chi}$ by Voiculescu's Non-Microstates Free Entropy of a Noncommutative Law §domain. Further $g_{\mathrm{sc}_{d,s}}(t)=d/(1+t)-d/(s+t)$ for $t>0$, so by (P5) with $(a,b,c)=(0,1,s)$ and $\kappa=d$,

$$\int_{(0,\infty)}g_{\mathrm{sc}_{d,s}}(t)\,dt=d(\log s-\log1)=d\log s,$$

and by Voiculescu's Non-Microstates Free Entropy of a Noncommutative Law §entropy, $\chi^{*}(\mathrm{sc}_{d,s})=\frac{d}{2}\log s+\frac{d}{2}\log(2\pi e)=\frac{d}{2}\log(2\pi e\,s)$. This proves clause 2.

Clause 3 (smoothing). Let $\mu=\lambda\boxplus\mathrm{sc}_{d,s}\in\Sigma_{d}$. For real $t>0$, (P7) with $r=s$ gives $f_{\mu}(t)=\Phi^{*}(\lambda\boxplus\mathrm{sc}_{d,s+t})\le d/(s+t)$ by Conjugate Variables along the Free Heat Flow: the Projected Semicircular Increment and the Bound d/t on the Free Fisher Information §fisher-bound (applied at time $s+t>0$), and $f_{\mu}(t)>0$ by (P6). Hence $0\le E_{(0,1)}f_{\mu}(x)\le E_{(0,1)}(t\mapsto d/(s+t))(x)$ for every $x\in\mathbb{R}$. The left function is measurable by (P6) and (P1), and the right one is integrable by (P5) with $(a,r,b)=(0,1,s)$ and $\kappa=d$. By (P2), $f_{\mu}$ is integrable on $(0,1)$, so $\mu\in\mathcal{D}_{\chi}$ by Voiculescu's Non-Microstates Free Entropy of a Noncommutative Law §domain. This proves clause 3.

Clause 4 (de Bruijn identity). Let $\lambda\in\mathcal{D}_{\chi}$ and put $\lambda_{s}=\lambda\boxplus\mathrm{sc}_{d,s}$, which lies in $\mathcal{D}_{\chi}$ by clause 3.

Integrability of $f_{\lambda}$ on $(0,s)$. The function $E_{(0,s)}f_{\lambda}$ is measurable by (P6) and (P1). We claim that, for every $x\in\mathbb{R}$,

$$0\le E_{(0,s)}f_{\lambda}(x)\le E_{(0,1)}f_{\lambda}(x)+E_{(0,s)}\Bigl(t\mapsto\frac{2d}{1+t}\Bigr)(x).$$

Both terms on the right are $\ge0$, by (P6) and since $1+x>0$ for $x>0$. If $x\notin(0,s)$ the middle term is $0$. If $x\in(0,s)$ and $x<1$, the middle term equals the first term on the right. If $x\in(0,s)$ and $x\ge1$, then $1+x\le2x$, so $d(1+x)\le2dx$ and, dividing by $x(1+x)>0$, $d/x\le2d/(1+x)$; with (P6) the middle term is $f_{\lambda}(x)\le d/x\le2d/(1+x)$. The right side is integrable: $E_{(0,1)}f_{\lambda}$ because $\lambda\in\mathcal{D}_{\chi}$ (P6), the second term by (P5) with $(a,r,b)=(0,s,1)$ and $\kappa=2d$, and their sum by claim 2 of Linearity and Monotonicity of the Lebesgue Integral. By (P2), $f_{\lambda}$ is integrable on $(0,s)$.

Splitting $\int g_{\lambda}$. Let $G=E_{(0,\infty)}g_{\lambda}$, which is integrable (P6). Pointwise on $\mathbb{R}$,

$$G=E_{(0,s)}g_{\lambda}+E_{\{s\}}g_{\lambda}+E_{(s,\infty)}g_{\lambda},$$

and the three terms are integrable by (P4). The one-point set $\{s\}=[s,s]$ has $\mathrm{Leb}(\{s\})=s-s=0$ by claim 4 of Existence of Lebesgue Measure on the Real Line, so it is null, and $E_{\{s\}}g_{\lambda}$ agrees with the zero function off $\{s\}$; by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison and (P3) its integral is $0$. Next, $E_{(0,s)}g_{\lambda}=E_{(0,s)}(t\mapsto d/(1+t))-E_{(0,s)}f_{\lambda}$, both terms being integrable (by (P5) with $(a,r,b)=(0,s,1)$, $\kappa=d$, and by the previous step). By claim 2 of Linearity and Monotonicity of the Lebesgue Integral and (P5),

$$\int_{(0,\infty)}g_{\lambda}(t)\,dt=d\log(1+s)-\int_{(0,s)}f_{\lambda}(t)\,dt+\int_{(s,\infty)}g_{\lambda}(t)\,dt,$$

using $d(\log(1+s)-\log(1+0))=d\log(1+s)$.

Computing $\int g_{\lambda_{s}}$. Let $P$ and $N$ be the positive and negative parts of $E_{(s,\infty)}g_{\lambda}$; they are measurable, nonnegative and real-valued with finite integrals, and $\int_{(s,\infty)}g_{\lambda}(t)\,dt=\int_{\mathbb{R}}P\,d\mathrm{Leb}-\int_{\mathbb{R}}N\,d\mathrm{Leb}$, by Integrable Function and the Lebesgue Integral. Let $\Psi:\mathbb{R}\to\mathbb{R}$, $\Psi(x)=(E_{(s,\infty)}g_{\lambda})(x+s)$. By claim 2 of Translation Invariance of Lebesgue Measure and the Lebesgue Integral (with translation $s$), $x\mapsto P(x+s)$ and $x\mapsto N(x+s)$ are measurable with $\int_{\mathbb{R}}P(x+s)\,d\mathrm{Leb}(x)=\int_{\mathbb{R}}P\,d\mathrm{Leb}$ and likewise for $N$. Since positive and negative parts are taken pointwise, these two functions are $\Psi^{+}$ and $\Psi^{-}$; so $\Psi=\Psi^{+}-\Psi^{-}$ is measurable (claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions) and integrable with $\int_{\mathbb{R}}\Psi\,d\mathrm{Leb}=\int_{(s,\infty)}g_{\lambda}(t)\,dt$, by Integrable Function and the Lebesgue Integral. Let $H=E_{(0,\infty)}\bigl(t\mapsto d/(1+t)-d/(1+s+t)\bigr)$; by (P5) with $(a,b,c)=(0,1,1+s)$ and $\kappa=d$, $H$ is integrable with integral $d(\log(1+s)-\log1)=d\log(1+s)$.

We claim $E_{(0,\infty)}g_{\lambda_{s}}=H+\Psi$ pointwise on $\mathbb{R}$. For $x\le0$ both sides are $0$, since $x+s\le s$. For $x>0$ we have $x+s>s$, and (P7) with $r=s$ gives $f_{\lambda_{s}}(x)=f_{\lambda}(s+x)$; hence

$$H(x)+\Psi(x)=\frac{d}{1+x}-\frac{d}{1+s+x}+\frac{d}{1+s+x}-f_{\lambda}(s+x)=\frac{d}{1+x}-f_{\lambda_{s}}(x)=g_{\lambda_{s}}(x).$$

By claim 2 of Linearity and Monotonicity of the Lebesgue Integral,

$$\int_{(0,\infty)}g_{\lambda_{s}}(t)\,dt=d\log(1+s)+\int_{(s,\infty)}g_{\lambda}(t)\,dt.$$

Conclusion. Subtracting the two displayed formulas, $\int_{(0,\infty)}g_{\lambda_{s}}(t)\,dt-\int_{(0,\infty)}g_{\lambda}(t)\,dt=\int_{(0,s)}f_{\lambda}(t)\,dt$. Since $\lambda,\lambda_{s}\in\mathcal{D}_{\chi}$, (P6) gives $\chi^{*}(\lambda_{s})-\chi^{*}(\lambda)=\frac{1}{2}\int_{(0,s)}f_{\lambda}(t)\,dt$. This proves clause 4.

Clause 5 (upper semicontinuity). Let $R>0$ and $c$ be real, and let $(\lambda_{k})_{k\in\mathbb{N}}$ be a sequence in $\mathcal{D}_{\chi}\cap\Sigma_{d,R}$ with $\chi^{*}(\lambda_{k})\ge c$ for every $k$ that converges weak-star to $\lambda\in\Sigma_{d}$.

Step 1: second moments. Let $j\in[d]$. The polynomial $x_{j}^{2}=x_{w}$ for the word $w=(j,j)$ of length $2$ is self-adjoint, since $(x_{j}x_{j})^{*}=x_{j}^{*}x_{j}^{*}=x_{j}x_{j}$ by Noncommutative Polynomials Form a Unital Complex Algebra with Involution: Linear Extension from Monomials, Products, Adjoints and Self-Adjoint Parts §adjoint; so for every $\nu\in\Sigma_{d}$ the number $\nu(x_{j}^{2})$ is real by Basic Properties of Noncommutative Laws: Adjoints, the Self-Adjoint Pairing, Cauchy-Schwarz, Monotonicity in the Bound, and Laws of Constant Tuples §adjoint. For each $k$, $\lambda_{k}(x_{j}^{2})=\operatorname{Re}\lambda_{k}(x_{j}^{2})\le|\lambda_{k}(x_{w})|\le R^{2}$ by claim 6 of Properties of Complex Conjugation and Modulus and Noncommutative Laws of Finitely Many Self-Adjoint Variables with a Norm Bound §norm-bound. By Weak-Star Convergence of Noncommutative Laws §weak-star, $\operatorname{Re}\lambda_{k}(x_{j}^{2})\to\operatorname{Re}\lambda(x_{j}^{2})=\lambda(x_{j}^{2})$, so $\lambda(x_{j}^{2})\le R^{2}$ by claim 1 of Order Properties of Limits of Real Sequences (compared with the constant sequence $R^{2}$). Summing over $j$, $M(\lambda_{k})\le dR^{2}$ for every $k$ and $M(\lambda)\le dR^{2}$.

Step 2: a uniform lower bound. Let $\nu\in\Sigma_{d}$ with $M(\nu)\le dR^{2}$, and let $t>0$ be real. By (P6), $0<M(\nu)+td\le dR^{2}+td$; multiplying this inequality by the positive number $d^{2}(M(\nu)+td)^{-1}(dR^{2}+td)^{-1}$ (claims 5 and 7 of Elementary Order Arithmetic in an Ordered Field and claim 5 of Elementary Arithmetic in an Ordered Field) gives $d^{2}/(dR^{2}+td)\le d^{2}/(M(\nu)+td)$, so by (P6)

$$f_{\nu}(t)\ge\frac{d^{2}}{dR^{2}+td}=\frac{d}{R^{2}+t}.$$

By Step 1 this applies to $\nu=\lambda$ and to every $\nu=\lambda_{k}$.

Step 3: the functions $h_{k}$ and $h$. Let $q=E_{(0,\infty)}\bigl(t\mapsto d/(1+t)-d/(R^{2}+t)\bigr)$; by (P5) with $(a,b,c)=(0,1,R^{2})$ and $\kappa=d$, $q$ is integrable with $\int_{\mathbb{R}}q\,d\mathrm{Leb}=d(\log(R^{2})-\log1)=d\log(R^{2})$. For $k\in\mathbb{N}$ let $h_{k}=q-E_{(0,\infty)}g_{\lambda_{k}}$ and let $h=q-E_{(0,\infty)}g_{\lambda}$. For $x>0$ we have $h_{k}(x)=f_{\lambda_{k}}(x)-d/(R^{2}+x)$ and $h(x)=f_{\lambda}(x)-d/(R^{2}+x)$, and $h_{k}(x)=h(x)=0$ for $x\le0$; by Step 2, $h_{k}\ge0$ and $h\ge0$ on $\mathbb{R}$. The function $h$ is measurable by (P6) and claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. Each $h_{k}$ is integrable, as $\lambda_{k}\in\mathcal{D}_{\chi}$ (P6) and by claim 2 of Linearity and Monotonicity of the Lebesgue Integral; in particular $h_{k}$ is measurable, since integrability in Integrable Function and the Lebesgue Integral is defined for measurable functions. By (P6)

$$\int_{\mathbb{R}}h_{k}\,d\mathrm{Leb}=d\log(R^{2})-\int_{(0,\infty)}g_{\lambda_{k}}(t)\,dt=d\log(R^{2})-2\chi^{*}(\lambda_{k})+d\log(2\pi e)\le d\log(2\pi e\,R^{2})-2c,$$

using $\chi^{*}(\lambda_{k})\ge c$ and $\log(2\pi e)+\log(R^{2})=\log(2\pi e\,R^{2})$. Put $K_{0}=d\log(2\pi e\,R^{2})-2c$. By (P3), the integral of $h_{k}$ in $[0,\infty]$ is this same real number, so it is at most $K_{0}$.

Step 4: pointwise comparison. Let $\ell:\mathbb{R}\to[0,\infty]$ be the function $\liminf_{k}h_{k}$ of Fatou's Lemma, $\ell(x)=\sup_{n\in\mathbb{N}}\inf_{k\ge n}h_{k}(x)$. We show $h(x)\le\ell(x)$ for every $x\in\mathbb{R}$. For $x\le0$ both are $0$. Fix $x>0$. For every $k$, $0\le h_{k}(x)\le f_{\lambda_{k}}(x)\le d/x$ by (P6), so each $\inf_{k\ge n}h_{k}(x)$ lies in $[0,d/x]$, and hence $\ell(x)$ is a real number in $[0,d/x]$. The laws $\lambda_{k}$ lie in $\Sigma_{d,R}$ and converge weak-star to $\lambda$, so $\lambda_{k}\boxplus\mathrm{sc}_{d,x}\to\lambda\boxplus\mathrm{sc}_{d,x}$ weak-star by The Free Heat Flow: Realisation, Norm Bound, Moments, Distance to the Initial Law, Wasserstein Contraction, Semigroup Property and Continuity §continuity. Now let $\varepsilon>0$ be real. For every $n\in\mathbb{N}$, $\inf_{k\ge n}h_{k}(x)\le\ell(x)<\ell(x)+\varepsilon$, so $\ell(x)+\varepsilon$ is not a lower bound of $\{h_{k}(x):k\ge n\}$ and there is $k\ge n$ with $h_{k}(x)<\ell(x)+\varepsilon$. Choose recursively $k_{1}$ with $h_{k_{1}}(x)<\ell(x)+\varepsilon$ and, given $k_{n}$, an index $k_{n+1}\ge k_{n}+1$ with $h_{k_{n+1}}(x)<\ell(x)+\varepsilon$. Then $(k_{n})_{n\in\mathbb{N}}$ is strictly increasing, and $(\lambda_{k_{n}}\boxplus\mathrm{sc}_{d,x})_{n\in\mathbb{N}}$ converges weak-star to $\lambda\boxplus\mathrm{sc}_{d,x}$, because for each $p\in\mathcal{P}_{d}$ the real sequences of real and imaginary parts are subsequences of convergent real sequences (A Subsequence of a Convergent Sequence Has the Same Limit, for the absolute value metric). Each $\lambda_{k_{n}}\boxplus\mathrm{sc}_{d,x}$ has conjugate variables by Conjugate Variables along the Free Heat Flow: the Projected Semicircular Increment and the Bound d/t on the Free Fisher Information §conjugate, and

$$\Phi^{*}(\lambda_{k_{n}}\boxplus\mathrm{sc}_{d,x})=f_{\lambda_{k_{n}}}(x)=h_{k_{n}}(x)+\frac{d}{R^{2}+x}\le\ell(x)+\varepsilon+\frac{d}{R^{2}+x},$$

a real number $\ge0$. By Free Fisher Information: the Cramer-Rao Inequality, the Semicircular Laws, and Closedness under Weak-Star Limits §closed, $f_{\lambda}(x)=\Phi^{*}(\lambda\boxplus\mathrm{sc}_{d,x})\le\ell(x)+\varepsilon+d/(R^{2}+x)$, that is $h(x)\le\ell(x)+\varepsilon$. As $\varepsilon>0$ was arbitrary, $h(x)\le\ell(x)$: otherwise $\varepsilon=(h(x)-\ell(x))/2>0$ would give $h(x)\le\ell(x)+\varepsilon<h(x)$ (claim 8 of Elementary Order Arithmetic in an Ordered Field).

Step 5: Fatou. By Fatou's Lemma, $\ell$ is measurable and $\int_{\mathbb{R}}\ell\,d\mathrm{Leb}\le\liminf_{k}\int_{\mathbb{R}}h_{k}\,d\mathrm{Leb}$, both sides in $[0,\infty]$. For each $n$, $\inf_{k\ge n}\int_{\mathbb{R}}h_{k}\,d\mathrm{Leb}\le\int_{\mathbb{R}}h_{n}\,d\mathrm{Leb}\le K_{0}$ by Step 3, so the right side is at most $K_{0}$. By Step 4 and claim 1 of Linearity and Monotonicity of the Lebesgue Integral, $\int_{\mathbb{R}}h\,d\mathrm{Leb}\le\int_{\mathbb{R}}\ell\,d\mathrm{Leb}\le K_{0}$, integrals in $[0,\infty]$. By (P3), $h$ is integrable and its Lebesgue integral is at most $K_{0}$.

Step 6: conclusion. Pointwise on $\mathbb{R}$, $E_{(0,1)}f_{\lambda}=h\,\mathbf{1}_{(0,1)}+E_{(0,1)}\bigl(t\mapsto d/(R^{2}+t)\bigr)$, since for $x\in(0,1)$ the right side is $f_{\lambda}(x)-d/(R^{2}+x)+d/(R^{2}+x)$ and both sides vanish elsewhere. The first term is measurable (claims 1 and 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions) with $|h(x)\mathbf{1}_{(0,1)}(x)|\le|h(x)|$, so it is integrable by (P2); the second is integrable by (P5) with $(a,r,b)=(0,1,R^{2})$ and $\kappa=d$. By claim 2 of Linearity and Monotonicity of the Lebesgue Integral, $f_{\lambda}$ is integrable on $(0,1)$, so $\lambda\in\mathcal{D}_{\chi}$ by Voiculescu's Non-Microstates Free Entropy of a Noncommutative Law §domain. Finally $E_{(0,\infty)}g_{\lambda}=q-h$, so by claim 2 of Linearity and Monotonicity of the Lebesgue Integral,

$$\int_{(0,\infty)}g_{\lambda}(t)\,dt=d\log(R^{2})-\int_{\mathbb{R}}h\,d\mathrm{Leb}\ge d\log(R^{2})-K_{0}=2c-d\log(2\pi e),$$

and by (P6), $\chi^{*}(\lambda)=\frac{1}{2}\int_{(0,\infty)}g_{\lambda}(t)\,dt+\frac{d}{2}\log(2\pi e)\ge c$. This proves clause 5.