Proof of Existence and Equality of the Reversed Mixed Second Partial Derivative

Throughout, $|\cdot|$ is the absolute value on $\mathbb{R}$ and $\lVert\cdot\rVert$ is the Euclidean norm on $\mathbb{R}^{n}$, which by claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ is the unique nonnegative real whose square is $\sum_{l=1}^{n}z_l^{2}$ and by claim 2 of that lemma computes the Euclidean distance between two points as the norm of their difference. For $z\in\mathbb{R}^{n}$, an index $m$ and $u\in\mathbb{R}$, write $z[m{:}u]$ for the point whose $m$th coordinate is $u$ and whose other coordinates are those of $z$. Unqualified claim numbers refer to Elementary Order Arithmetic in an Ordered Field. We use freely that nonnegative reals $\alpha,\beta$ with $\alpha^{2}<\beta^{2}$ satisfy $\alpha<\beta$: otherwise $\beta<\alpha$, whence $\beta^{2}<\alpha^{2}$ by claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, or $\beta=\alpha$, and either alternative contradicts $\alpha^{2}<\beta^{2}$.

If $i=p$, then $\partial_i f=\partial_p f$ and hypothesis 3, read at the single point $a$, is exactly the assertion of the theorem. So assume from now on that $i\ne p$.

Step 0 (a two-dimensional square inside $U$). Since $a\in U$ and $U$ is open, there is a real $r$ with $0<r$ such that every point of $\mathbb{R}^{n}$ at Euclidean distance less than $r$ from $a$ lies in $U$. By claim 8 there is $\rho$ with $0<\rho$ and $\rho+\rho=r$. Let $J_i$ and $J_p$ be the open intervals with endpoints $a_i-\rho,a_i+\rho$ and $a_p-\rho,a_p+\rho$ respectively; by An Open Interval is an Interval All of Whose Points Are Interior each is an interval all of whose points are interior points of it. For $u\in J_i$ and $v\in J_p$ put

the point whose $i$th coordinate is $u$, whose $p$th coordinate is $v$ (the two substitutions concern different coordinates because $i\ne p$) and whose $l$th coordinate is $a_l$ for every other $l$. Thus $P(a_i,a_p)=a$, $P(u,a_p)=a[i{:}u]$ and $P(a_i,v)=a[p{:}v]$.

For such $u$ and $v$, claim 9 of Properties of the Absolute Value in an Ordered Field gives $|u-a_i|<\rho$ and $|v-a_p|<\rho$. Since $P(u,v)$ differs from $a$ only in the $i$th and $p$th coordinates, claim 4 of that lemma, claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field applied to each of the two absolute values, and claim 3 (adding a strict and a non-strict inequality) give

Moreover $(\rho+\rho)^{2}=(\rho^{2}+\rho^{2})+(\rho^{2}+\rho^{2})$ and $0\le\rho^{2}+\rho^{2}$, so $\rho^{2}+\rho^{2}\le(\rho+\rho)^{2}=r^{2}$ by claim 1. Hence $\lVert P(u,v)-a\rVert<r$ and $P(u,v)\in U$.

Step 1 (differentiating along the two coordinate lines).

(a) Let $u\in J_i$ and let $G_u:J_p\to\mathbb{R}$ be given by $G_u(v)=f(P(u,v))$. Then $G_u$ is differentiable at every $v_0\in J_p$, with $G_u'(v_0)=\partial_p f(P(u,v_0))$.

Indeed, let $y=P(u,v_0)$, a point of $U$ whose $p$th coordinate is $v_0$. Claim 1 of Slice Function and the Partial Derivative, applied to $f$ at $y$ in the $p$th variable, provides $\rho_0$ with $0<\rho_0$ such that $y[p{:}w]\in U$ whenever $|w-v_0|<\rho_0$, and claim 2 of that lemma states that the slice function $w\mapsto f(y[p{:}w])$, defined on the open interval with endpoints $v_0-\rho_0$ and $v_0+\rho_0$, is differentiable at $v_0$ with derivative $\partial_p f(y)$, the partial derivative existing by hypothesis 2. Since $y[p{:}w]=P(u,w)$, that slice function agrees with $G_u$ on the overlap of the two intervals. By An Open Interval is an Interval All of Whose Points Are Interior the point $v_0$ is an interior point of each of the two open intervals, so each contains every real within some positive distance of $v_0$; taking the least of the two radii by claim 9, the overlap does as well. Shrinking the $\delta$ in the defining condition of Derivative at an Interior Point so that the increments stay in the overlap therefore shows that $G_u$ is differentiable at $v_0$ with the same derivative.

(b) Let $v\in J_p$ and let $H_v:J_i\to\mathbb{R}$ be given by $H_v(u)=\partial_p f(P(u,v))$. Then $H_v$ is differentiable at every $u_0\in J_i$, with $H_v'(u_0)=\partial_i\partial_p f(P(u_0,v))$. The argument is the one just given, applied to the function $\partial_p f:U\to\mathbb{R}$ of hypothesis 2 at the point $P(u_0,v)$ of $U$ in the $i$th variable, the relevant partial derivative existing by hypothesis 3.

Step 2 (a double difference and two mean values). Let $u\in J_i$ with $u\ne a_i$ and let $v\in J_p$ with $v\ne a_p$, and set

Let $\psi:J_p\to\mathbb{R}$ be the function $G_u+(-1)G_{a_i}$, that is $\psi(w)=f(P(u,w))-f(P(a_i,w))$, so that $\Delta(u,v)=\psi(v)-\psi(a_p)$. By part (a) and claim 2 of Sum, Constant Multiple, and Product Rules for One-Dimensional Derivatives, whose hypothesis that the base point be an interior point of the interval holds by An Open Interval is an Interval All of Whose Points Are Interior, the function $\psi$ is differentiable at every $w\in J_p$ with

Applying Mean Value Theorem on an Open Interval to $\psi$ on $J_p$ with the two distinct points $a_p$ and $v$, taken in increasing order, yields $w^{*}$ strictly between them with

the identity is the same for either order of the two points, since interchanging them negates both sides. As $w^{*}$ lies strictly between $a_p$ and $v$ we have $|w^{*}-a_p|<|v-a_p|$, so in particular $w^{*}\in J_p$.

By part (b) and Mean Value Theorem on an Open Interval applied in the same way to $H_{w^{*}}$ on $J_i$ with the two distinct points $a_i$ and $u$, there is $u^{*}$ strictly between them, so with $|u^{*}-a_i|<|u-a_i|$ and $u^{*}\in J_i$, such that

Combining the two displays,

Step 3 (using continuity at $a$). Let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. By claim 8 there is $\varepsilon_1$ with $0<\varepsilon_1$ and $\varepsilon_1+\varepsilon_1=\varepsilon$. Since $\partial_i\partial_p f$ is continuous at $a$, that definition, read for a map into $\mathbb{R}^{1}$ with single coordinate function $\partial_i\partial_p f$, provides $\theta$ with $0<\theta$ such that every $z\in U$ at Euclidean distance less than $\theta$ from $a$ satisfies $\bigl(\partial_i\partial_p f(z)-\partial_i\partial_p f(a)\bigr)^{2}<\varepsilon_1^{2}$, hence, by claim 4 of Properties of the Absolute Value in an Ordered Field and the monotonicity of squares recalled above, $|\partial_i\partial_p f(z)-\partial_i\partial_p f(a)|<\varepsilon_1$.

By claim 8 let $\theta_1$ satisfy $0<\theta_1$ and $\theta_1+\theta_1=\theta$, and by claim 9 let $\delta_1$ be the least of $\rho$ and $\theta_1$, so that $0<\delta_1$. Suppose $u,v\in\mathbb{R}$ satisfy $0<|u-a_i|<\delta_1$ and $0<|v-a_p|<\delta_1$. Since $\delta_1\le\rho$, claim 9 of Properties of the Absolute Value in an Ordered Field gives $u\in J_i$ and $v\in J_p$, so Step 2 applies. For the points $u^{*}$ and $w^{*}$ produced there, Step 2 gives $|u^{*}-a_i|<|u-a_i|<\delta_1\le\theta_1$ and $|w^{*}-a_p|<|v-a_p|<\delta_1\le\theta_1$. Hence, exactly as in the computation of Step 0 with $\theta_1$ in place of $\rho$, the point $z=P(u^{*},w^{*})$ satisfies

so $\lVert z-a\rVert<\theta$ and $z\in U$. As $(u-a_i)(v-a_p)\ne 0$, the last display of Step 2 gives

Step 4 (letting the increment in the $i$th variable go to zero). Keep $\varepsilon$, $\varepsilon_1$ and $\delta_1$ as in Step 3, fix $v$ with $0<|v-a_p|<\delta_1$, and set

which is defined because $\partial_i f$ exists at every point of $U$ by hypothesis 1 and $a[p{:}v]=P(a_i,v)\in U$ by Step 0.

Grouping the four terms of $\Delta(u,v)$ as $\bigl(f(P(u,v))-f(P(a_i,v))\bigr)-\bigl(f(P(u,a_p))-f(a)\bigr)$, dividing by $(u-a_i)(v-a_p)$ and subtracting $D(v)$ gives, for every $u\in J_i$ with $u\ne a_i$, the identity

By claim 8 let $\varepsilon_2$ satisfy $0<\varepsilon_2$ and $\varepsilon_2+\varepsilon_2=\varepsilon_1$, and put $\eta=\varepsilon_2|v-a_p|$, which is positive by claim 5. By hypothesis 1 and Partial Derivative on a Euclidean Open Set applied at the point $a[p{:}v]$ in the $i$th variable, there is $\delta_2$ with $0<\delta_2$ such that every real $h$ with $0<|h|<\delta_2$ satisfies $a[p{:}v][i{:}a_i+h]\in U$ and

since $a[p{:}v][i{:}a_i+h]=P(a_i+h,v)$. Applied at $a$ in the same way, there is $\delta_3$ with $0<\delta_3$ such that every real $h$ with $0<|h|<\delta_3$ satisfies

By claim 9 choose a real $h$ with $0<|h|$ and $|h|$ less than each of $\delta_1,\delta_2,\delta_3$, and set $u=a_i+h$, so that $0<|u-a_i|<\delta_1$ and $u\in J_i$. The identity above, the triangle inequality (claim 5 of Properties of the Absolute Value in an Ordered Field), multiplicativity of the absolute value (claim 4 of that lemma) and division of a strict inequality by the positive number $|v-a_p|$ (claims 7 and 10) give

Combining this with the last display of Step 3 for these $u$ and $v$, and using the triangle inequality once more,

Step 5 (conclusion). Let $k\in\mathbb{R}$ with $0<|k|<\delta_1$ and put $v=a_p+k$, so that $0<|v-a_p|<\delta_1$. By Step 0 the point $a[p{:}a_p+k]=P(a_i,v)$ lies in $U$, and by Step 4

Thus for every real $\varepsilon$ with $0<\varepsilon$ there is a real $\delta_1$ with $0<\delta_1$ such that every real $k$ with $0<|k|<\delta_1$ satisfies $a[p{:}a_p+k]\in U$ and the displayed inequality. By Partial Derivative on a Euclidean Open Set this says exactly that the partial derivative of $\partial_i f$ with respect to the $p$th variable exists at $a$ with value $\partial_i\partial_p f(a)$; that is, $\partial_p\partial_i f(a)=\partial_i\partial_p f(a)$.