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Every function $h$ with Lipschitz constant at most $1$ yields a feasible pair $(h,-h)$, and replacing $h$ by $-h$ reverses the objective. HenceConversely, the Lipschitz regularization of transport potentials in part iii dominates every feasible pair by $(h,-h)$ with $\Vert h{\Vert }_L\le 1$, so $m_L\le \gamma$. Together with the preceding parts this proves the Kantorovich–Rubinstein theorem here in all four requested forms:The seminorm convention is essential. If instead $\Vert h{\Vert }_L$ meant $\Vert h{\Vert }_{\infty }+\mathrm{Lip}(h)$, the equality would fail: on a two-point space at distance $2$, with point masses at the two different points, $W=2$ whereas the latter unit-ball supremum is $1$. Indeed, an oscillation $a$ needs supremum norm at least $a/2$ and Lipschitz constant $a/2$, so the sum norm bounds $a$ by $1$, attained by values $-1/2,1/2$. This distinguishes the Lipschitz seminorm used above from a full bounded-Lipschitz norm.

Apply the reciprocal-and-integer-part algorithm for the continued fraction. The successive complete quotients after the integer part two are $(\sqrt{7}+2)/3$, $(\sqrt{7}+1)/2$, $(\sqrt{7}+1)/3$, and $\sqrt{7}+2$. Their integer parts are $1,1,1,4$, and taking the next reciprocal repeats the first complete quotient. ThereforeThe first four continued fraction convergents are $2,3,5/2,8/3$. The last gives $8^2-7\cdot 3^2=1$. Squaring its unit $8+3\sqrt{7}$ gives $127+48\sqrt{7}$, whose norm is again one. Thus two positive solutions of the Pell equation areDirect verification gives $127^2-7\cdot 48^2=16129-16128=1$.

The Arzelà-Ascoli theorem for $C([0,1])$ states that a family has compact closure in the uniform norm exactly when it is uniformly bounded and equicontinuous. To prove sufficiency, choose a countable dense set. From any sequence of functions, successive subsequence extraction at each dense point and a diagonal selection give convergence at all those points. Given $\epsilon >0$, equicontinuity gives a common $\delta$; cover the interval by finitely many $\delta$-balls centered at dense points. On a sufficiently late tail the functions differ by less than $\epsilon /3$ at those finitely many centres, and equicontinuity makes each of the other two differences less than $\epsilon /3$. Thus the selected subsequence is uniformly Cauchy. Completeness of $C([0,1])$ gives a continuous uniform limit. In a metric space this proves compactness of the closure. Conversely, a compact family has finite uniform $\epsilon$-nets. The finitely many net functions are bounded and uniformly continuous; approximation by them gives a common bound and equicontinuity for the family. This proves both directions.

For $f\in S_N$, a positive interior maximum would have $f^{\prime }=0$ and $f^{\prime \prime }\le 0$, contradicting $f^{\prime \prime }=f>0$. A negative interior minimum likewise contradicts $f^{\prime \prime }=f<0$. The endpoint bounds therefore give $|f|\le N$ everywhere.

Now put $w=f^{\prime }$. Differentiating the equation, using the stipulated third derivative, gives $w^{\prime \prime }=w+2w{w}^{\prime }$. At an interior extremum of $w$, $w^{\prime }=0$, so $w^{\prime \prime }=w$. The same maximum/minimum argument and the derivative endpoint bounds imply $|f^{\prime }|=|w|\le N$. Consequently $|f(x)-f(y)|\le N|x-y|$ by the mean value theorem. Thus $S_N$ is uniformly bounded and equicontinuous, and the theorem makes it relatively compact, hence totally bounded in $C([0,1])$. Its closure need not be asserted to remain thrice differentiable.

For $1\le p<\infty$, the Lp norm is $\Vert f{\Vert }_p=({\int }_E|f{|}^{p}d\mu {)}^{1/p}$; for $p=\infty$ it is the essential supremum of $|f|$. The Lp space consists of measurable functions with finite norm, identifying functions equal almost everywhere. This identification makes the quantity a norm rather than only a seminorm. Convergence in $L^p$ means $\Vert f_n-f{\Vert }_p\to 0$, including the essential-supremum norm when $p=\infty$.

The Minkowski inequality states the triangle inequality for every $L^p$ norm with $1\le p\le \infty$. Apply it to $f_n=(f_n-f)+f$ and then to $f=(f-f_n)+f_n$. The two bounds combine toThus norm convergence implies convergence of the norms themselves.

The heat kernel is $H_t(x)=(4\pi t{)}^{-1/2}{e}^{-x^2/(4t)}$, so the homogeneous solution for smooth bounded data is $u(t,x)=(H_t\ast g)(x)$. Differentiating under the integral gives the heat equation, and the approximate-identity property gives the initial value.

For a tempered distribution, its derivative is defined by $⟨T^{\prime },\psi ⟩=-⟨T,{\psi }^{\prime }⟩$ for every Schwartz test function. A fundamental solution of a constant-coefficient operator $P$ is a distribution $E$ satisfying $PE={\delta }_0$. Put $E(x)=e^{-|x|}/2$. Away from zero, $E^{\prime \prime }=E$; its first derivative has jump $-1$ at zero. Distributional differentiation therefore gives $E^{\prime \prime }=E-{\delta }_0$, andFor an ansatz $u=e^{t}v$, the forced equation becomes $(1-D^2)v=\varphi$. With Fourier convention $\hat{v}(k)=\int e^{-ikx}v(x)dx$, its unique solution in the Schwartz space isThe smooth multiplier and its derivatives have polynomial bounds, so this quotient is Schwartz. For uniqueness, multiplication of the difference's Fourier transform by $1+k^2$ gives zero, forcing the difference to vanish.

Subtract this particular solution and evolve the remaining initial data by the heat kernel:For smooth bounded $f$, the second term remains uniformly bounded, so $e^{-t}u(t,\cdot )\to v$ uniformly. If additionally $f-v$ is integrable, its uniform norm is $O(t^{-1/2})$, so the remainder itself decays. General bounded data need not have a decaying heat remainder, and no such stronger claim follows without a condition on the data. The homogeneous uniqueness assertion is understood in the bounded or usual tempered-growth solution class; the uniqueness of the particular Schwartz ansatz is unconditional within that stated class.

The oscillator operators satisfy $[a,a^{†}]=1$ and $[a,a]=[a^{†},a^{†}]=0$. Substitution into the kinetic and potential energies givesThe number operator $N=a^{†}a$ is nonnegative because $⟨\psi ,N\psi ⟩=\Vert a\psi {\Vert }^2$. The commutators $[N,a]=-a$ and $[N,a^{†}]=a^{†}$ lower and raise an eigenvalue by one. If an eigenvalue were a noninteger $\nu \ge 0$, repeated lowering would produce a nonzero eigenvector with eigenvalue between zero and one, and one further lowering would have negative eigenvalue; its norm is nonzero because $\Vert a\psi {\Vert }^2=\nu \Vert \psi {\Vert }^2$. This contradicts nonnegativity. Thus only nonnegative integers are allowed. The normalized Gaussian solves $a|0⟩=0$, and raising it supplies every level, withFor the three-dimensional unperturbed oscillator the product states are $|n_1,n_2,n_3⟩=|n_1⟩\otimes |n_2⟩\otimes |n_3⟩$, with energy $\hslash \omega (n_1+n_2+n_3+3/2)$.

Write the perturbation as $\lambda W$, where $W=(\hslash \omega /2){\sum }_{i<j}(a_i+a_i^{†})(a_j+a_j^{†})$. Its ground-state expectation is zero. Acting on the ground state, it produces $|110⟩,|101⟩,|011⟩$, each with coefficient $\hslash \omega /2$. They lie $2\hslash \omega$ above the ground state, so second-order perturbation theory givesIn the first excited subspace with basis $|100⟩,|010⟩,|001⟩$, the first-order perturbation matrix is $(\lambda \hslash \omega /2)\left(\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right)$. Its symmetric eigenvector has eigenvalue $\lambda \hslash \omega$, while the perpendicular two-dimensional space has eigenvalue $-\lambda \hslash \omega /2$. Thus the lowest excited triplet splits intoAs an independent check, diagonalizing the quadratic potential gives normal frequencies $\omega \sqrt{1+2\lambda }$ and $\omega \sqrt{1-\lambda }$ twice; their small-$\lambda$ expansions reproduce both results.

Two norms $\Vert \cdot {\Vert }_a,\Vert \cdot {\Vert }_b$ are Lipschitz equivalent norms when constants $c,C>0$ satisfy $c\Vert x{\Vert }_a\le \Vert x{\Vert }_b\le C\Vert x{\Vert }_a$ for every vector $x$. Applying these inequalities to $x_j-x_k$ shows that a sequence is a Cauchy sequence for one norm exactly when it is so for the other; applying them to $x_j-x$ likewise equates convergence. If one norm is complete, a Cauchy sequence for the other converges using the first norm and hence also the second. Interchanging the norms proves completeness is preserved in both directions.

Write $x={\sum }_i{x}_i{e}_i$ and set $C={\sum }_i\Vert e_i\Vert$. If $\Vert x{\Vert }_2=1$, then $|x_i|\le 1$, and the triangle inequality givesScaling a nonzero $x$ gives $\Vert x\Vert \le C\Vert x{\Vert }_2$ for every $x$. The reverse triangle inequality now impliesso $x\mapsto \Vert x\Vert$ is continuous, indeed Lipschitz, in the Euclidean norm.

Suppose there were no positive lower bound on the Euclidean unit sphere. We could choose $x_j$ with $\Vert x_j{\Vert }_2=1$ but $\Vert x_j\Vert <1/j$. The Bolzano-Weierstrass theorem supplies a subsequence converging in the Euclidean norm to $x$. Continuity of both norms gives $\Vert x{\Vert }_2=1$ and $\Vert x\Vert =0$, contradicting definiteness. Therefore some $m>0$ satisfies $\Vert x\Vert \ge m$ on the unit sphere. Scaling again proves the finite-dimensional equivalence of norms:

Both functionals are finite on $X$: a continuous function on the compact metric space $[0,1]$ is bounded and integrable. Homogeneity follows from $|af|=|a||f|$, and the triangle inequality follows by integrating, or taking the supremum of, $|f+g|\le |f|+|g|$. Each is nonnegative. If $\Vert f{\Vert }_{\infty }=0$, then $f$ vanishes everywhere. If a continuous function $f$ is nonzero at some point, continuity supplies an interval of positive length on which $|f|$ is bounded below by a positive number; hence $\Vert f{\Vert }_1>0$. Thus both are norms.

For the nonnegative $f_n$, direct integration gives the L1 normTherefore $\boxed{f_n⟶0\text{ in }\Vert \cdot {\Vert }_1}$. On the other hand, differentiation places the maximum of $t^n(1-t)$ at $t=n/(n+1)$, giving the supremum normFor $t<1$, exponential decay implies $n{t}^n(1-t)\to 0$, while $f_n(1)=0$. Thus the pointwise convergence is to zero throughout $[0,1]$. Any convergence in the supremum norm would imply uniform convergence, hence the same pointwise limit zero, contradicting the displayed positive limiting norm. Therefore there is no convergence in the supremum norm.