Gaussian squared-error rate-distortion function

Suppose ${\sigma }^2>0$ and $0<D<{\sigma }^2$. By [F2], every feasible joint law has mutual information in nats at least $\frac{1}{2}\ln ({\sigma }^2\mathrm{/}D)$; division by $\ln 2>0$ and taking the infimum in [F1] gives $R(D)\ge \frac{1}{2}{\log }_2({\sigma }^2\mathrm{/}D)$.

If ${\sigma }^2>0$ and $D\ge {\sigma }^2$, the constant reconstruction $Y=0$ has distortion ${\sigma }^2$ and mutual information zero; [F5] makes zero the minimum, so $R(D)=0$.

Suppose ${\sigma }^2>0$ and $D=0$. Every feasible coupling satisfies $X=Y$ almost surely because the nonnegative random variable $(X-Y{)}^2$ has expectation zero. The diagonal $\{(x,y):x=y\}$ has joint probability one. By [F7] the normal density is bounded by $(2\pi {\sigma }^2{)}^{-1\mathrm{/}2}$, so under the product law, for every $h>0$, the event $\mathrm{\mid }X-Y\mathrm{\mid }\le h$ has probability at most $2h(2\pi {\sigma }^2{)}^{-1\mathrm{/}2}$. For $h=1\mathrm{/}(n+1)$, $n\ge 0$, these measurable events decrease to the diagonal; [F8] therefore makes its product probability zero. Thus the joint law is not absolutely continuous with respect to the product of its marginals and its KL, hence its mutual information, is infinite. Every feasible coupling has this value, proving $R(0)=+\mathrm{\infty }$.

Put $a=\sqrt{{\sigma }^2-D}>0$, $b=\sqrt{D}>0$, and let $U,V$ be independent standard normals. Set $Y=aU$, $X=aU+bV$. The orthogonal matrix ${\sigma }^{-1}\left(\begin{array}{cc}a& b\\ -b& a\end{array}\right)$ and [F3] show $X\sim N(0,{\sigma }^2)$; also $\mathbb{E}(X-Y{)}^2=b^2=D$. By [F7], $U,V$ have product density; the invertible map $(u,v)\mapsto (x,y)=(au+bv,au)$ has determinant $-ab$, so [F4] gives joint density $(2\pi ab{)}^{-1}\exp [-y^2\mathrm{/}(2a^2)-(x-y{)}^2\mathrm{/}(2b^2)]$. The product of the $N(0,{\sigma }^2)$ and $N(0,a^2)$ marginals has density $(2\pi \sigma a{)}^{-1}\exp [-x^2\mathrm{/}(2{\sigma }^2)-y^2\mathrm{/}(2a^2)]$. Their log-density ratio has expectation $\ln (\sigma \mathrm{/}b)-\frac{1}{2}+\frac{1}{2}=\frac{1}{2}\ln ({\sigma }^2\mathrm{/}D)$; [F1] therefore gives a feasible joint law attaining the lower bound in step 1.1.

If ${\sigma }^2=0$, then $X=0$ almost surely. The constant reconstruction $Y=0$ has zero distortion and zero mutual information, while [F5] gives a nonnegative lower bound for every feasible coupling. Therefore $R(D)=0$ for every $D\ge 0$. These branches prove the statement.