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For probability measures $\mu ,\nu$ on ${\mathbb{R}}^d$ with finite second moments, the Knott–Smith optimality criterion states that a transport plan $\pi \in \Pi (\mu ,\nu )$ minimizes the quadratic cost $|x-y{|}^2$ if and only if there is a proper convex function $\Phi :{\mathbb{R}}^d\to \mathbb{R}\cup \{+\infty \}$ that is sequentially lower semicontinuous and satisfiesThe subdifferential is characterized bywith $\Phi (x)<\infty$. Thus the transport plan is concentrated on the graph of the subdifferential. No absolute continuity of measures assumption on $\mu$ is needed. Multiplying the cost by $1/2$ leaves the criterion unchanged. The original quadratic optimal mapping result is Knott and Smith, On the optimal mapping of distributions.

Define $T^{†}(0)=0$; the printed quotient is undefined at the origin, but this continuous extension changes no transport cost because $\mu (\{0\})=0$. In polar coordinates, the two probability density functions giveBoth angular distributions are uniform, with independence of angle and radius. The proposed transport map preserves the angle and sends $r$ to $s=\sqrt{r}$. Thereforeand preservation of the angle proves $(T^{†}{)}_{\#}\mu =\nu$. As a local check using the Jacobian determinant, its radial and tangential derivatives for $r>0$ are $1/(2\sqrt{r})$ and $1/\sqrt{r}$, respectively, so $\det D{T}^{†}=1/(2r)$ and $g(T^{†}(x))\det D{T}^{†}(x)=(2r/\pi )/(2r)=f(x)$.

Now use the convex functionIt is convex because the Euclidean norm is convex and $s\mapsto (2/3)s^{3/2}$ is increasing and convex on $[0,\infty )$. It is differentiable, including at zero, andIts graph transport plan lies in the graph of $\partial \Phi$, so the Knott–Smith optimality criterion proves quadratic optimality. Since that plan is induced by a map, part 1(b) proves optimality for the Monge optimal transport problem as well. The minimum provides a useful independent check:

Choose the bounded, nonnegative nondecreasing function $\tau$ from part (d), and defineThe function $h$ is convex and nondecreasing. For $0\le a\le 1$, the triangle inequality, monotonicity, and convexity giveHence $U$ is a convex function. By the gradient of a radial function formula and the fundamental theorem of calculus,for $x\ne 0$ whenever $h^{\prime }(|x|)=\tau (|x|)$. A monotone function has at most countably many discontinuities, so part (d) makes the exceptional radii $\tilde{\mu }$-null. The origin is also $\mu$-null. Define $\mathcal{T}(0)=0$; thus $\mathcal{T}=\nabla U$ $\mu$-almost everywhere.

For every $Q$ in the orthogonal group, $\mathcal{T}(Qx)=Q\mathcal{T}(x)$, including at zero. Consequently ${\mathcal{T}}_{\#}\mu$ is invariant under the orthogonal group. Moreover $|\mathcal{T}(x)|=\tau (|x|)$ away from the null origin, and composition of pushforward measures givesPart (c) therefore gives ${\mathcal{T}}_{\#}\mu =\nu$. The Knott–Smith optimality criterion makes a transport map given by a convex function's gradient optimal for the quadratic transport cost. Uniqueness from Brenier's theorem yieldsSince directions agree, this also identifies the optimal value: