# Abstract Nonsense

## Higher Order Partial Derivatives and the Equality of Mixed Partials (Pt. II)

Point of Post: This is a continuation of this post.

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Remark: For notational convenience we assume that $i. It’s clear the only place this is implied is the writing of the tuples and thus is inconsequential.

Proof: We begin by showing that it suffices to prove this for mappings $f:\mathbb{R}^2\to\mathbb{R}$. We do this in two steps, first showing that our theorem follows if it’s proven for mappings $f:\mathbb{R}^n\to\mathbb{R}$ and then show that this is implied by the result for mappings $f:\mathbb{R}^2\to\mathbb{R}$. To prove this first claim we suppose that the result is true for all mappings $g:U\to\mathbb{R}$ with $U\subseteq\mathbb{R}^n$ and let $f:U\to\mathbb{R}^m$ satisfy the hypotheses. We then note if $f(x)=(f_1(x),\cdots,f_m(x))$ then from a previous result we have that $D_i f=(D_if_1,\cdots,D_if_m)$ and $D_j f=(D_jf_1,\cdots,D_jf_m)$, we have by assumption that $D_if$ is continuous on $U$ that $D_if_k,D_jf_k$ are continuous on $U$ for each $k\in[n]$. Similarly, since $D_{i,j}f$ exists on $U$ and $D_{i,j}f$ is continuous at $a$ we have that $D_{i,j}f_k$ exists on $U$ and is continuous at $a$ for each $k\in [n]$. Thus, by assumption that the theorem holds for $g:U\to\mathbb{R}$ we have that $D_{j,i}f_k(a)$ exists for $k\in[n]$ and moreover $D_{j,i}f_k(a)=D_{i,j}f_k(a)$ and thus by first principles $D_{j,i}f(a)$ exists and $D_{j,i}f(a)=D_{i,j}f(a)$.

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So, now let’s show that the result being true for $g:E\to\mathbb{R}$, where $E\subseteq\mathbb{R}^2$ is open, which satisfies the same hypotheses implies the result for $f:U\to\mathbb{R}$. Indeed, since $U$ is open there exists some open ball $B_{\delta}(a)\subseteq U$ and there of course sits some symmetric $n$-cube $(x_1,y_1)\times\cdots\times(x_i,y_i)\times\cdots\times(x_j,y_j)\times\cdots\times(x_n,y_n)$ with center $a=(a_1,\cdots,a_n)$. We then define $\pi_{i,j}:(x_1,y_1)\times(x_2,y_2)\to U$ given by

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$\pi_{i,j}(x,y)=(a_1,\cdots,\underbrace{x}_{i^{\text{th}}},\cdots,\underbrace{y}_{j^{\text{th}}},\cdots,a_n)$

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and $h:(x_1,y_1)\times(x_2,y_2)\to\mathbb{R}$ to be $h=f\circ\pi_{i,j}$. Fix then some $(x_0,y_0)\in (x_1,y_1)\times(x_2,y_2)$. We note then if we define $g_1:(-\delta_1,\delta_1)\to\mathbb{R}:t\mapsto h(x_0+t,y_0)$ where $(-\delta_1,\delta_1)$ is some interval for which $(x_0-\delta_1,+\delta_1)\subseteq (x_i,y_i)$ and define $g_2$ similarly we have by previous remark that $D_1h(x_0,y_0)$ exists if only if $g_1$ is differentiable at zero. But, noting that

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$g_1(t)=f(a_1,\cdots,\underbrace{x_0+t}_{i^{\text{th}}},\cdots,\underbrace{y_0}_{\text{j}^{\text{th}}},\cdots,a_n)$

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we have by assumption that $D_if$ exists at $(a_1,\cdots,x_0,\cdots,y_0,\cdots,a_n)$ that $g_1(t)$ is differentiable at zero and moreover then we see  that $D_1h(x_0,y_0)=g_1'(0)=D_if(a_1,\cdots,x_0,\cdots,y_0,\cdots,a_n)$. Thus, since $(x_0,y_0)$ were arbitrary we have that $D_1h$ exists everywhere on $(x_i,y_i)\times(x_j,y_j)$ and $D_1h(x,y)=D_if(a_1,\cdots,x,\cdots,y,\cdots,a_n)$ Moreover, it’s clear that $D_1h$ is continuous since $D_1h=D_if\circ\pi_{i,j}$ and $D_if,\pi_{i,j}$ are continuous (the latter because it’s an isometry). Using the same tricks one shows that $D_2h$ is defined everywhere on $(x_i,y_i)\times(x_j,y_j)$, is continuous, and $D_2h=D_jf\circ\pi_{i,j}$. Moreover, using the same idea, defining $g_3(t)=D_1(x_0+t,y_0)$ we see the same method applies and shows that $D_{1,2}h$ exists on $(x_1,y_1)\times(x_2,y_2)$ and moreover that $D_{1,2}(x,y)=D_{i,j}f\circ\pi_{i,j}$ which in particular implies that $D_{1,2}h$ is continuous at $(a_i,a_j)$. Thus, if we assume the theorem is true for mappings $g:E\to\mathbb{R}$ with $\mathbb{R}^2$ we see that $D_{2,1}h(a_i,a_j)$ exists and $D_{2,1}h(a_i,a_j)=D_{1,2}f(a_i,a_j)$. But, this clearly implies that $D_{j,i}f(a)=D_{2,1}h(a_i,a_j)$ exists and

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$D_{j,i}f(a)=D_{2,1}h(a_i,a_j)=D_{1,2}h(a_i,a_j)=D_{i,j}h(a)$

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Thus, we now assume without loss of generality that $U\subseteq\mathbb{R}^2$ and let $f:U\to\mathbb{R}$ satisfy the hypotheses of the theorem with $a=(a_1,a_2)$. Since the following argument (as you shall see) is independent of whether $(i,j)=(1,2)$ or $(i,j)=(2,1)$ we assume the latter.So, what we want to show is that

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$\displaystyle \lim_{s\to0}\frac{D_2f(a_1+s,a_2)-D_2f(a_1,a_2)}{s}=\lim_{s\to0}\frac{\displaystyle \lim_{t\to0}\frac{f(a_1+s,a_2+t)-f(a_1+s,a_2)}{t}-\lim_{t\to0}\frac{f(a_1,a_2+t)-f(a_1,a_2)}{t}}{s}\quad\bold{(1)}$

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exists and equals $D_{1,2}f(a_1,a_2)$. Now, define $g_t(z)=f(a_1+z,a_2+t)-f(a_1+z,a_2)$ for $z\in(-\gamma,\gamma)$ where $\delta$ is chosen to be sufficiently large so that $(a_1+z,a_2+t)\in U$ for all $z\in(-\gamma,\gamma)$ (this can be done for sufficiently small $t$). We note then by our favorite observation that $g_t(z)$ is differentiable on $(-\gamma,\gamma)$ and $g'_t(z)=D_1f(a_1+z,a_2+t)-D_1f(a_1+z,a_2)$. We note then that $\bold{(1)}$ may be rewritten as

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$\displaystyle \lim_{s\to0}\frac{\displaystyle \lim_{t\to0}\frac{g_t(s)-g_t(0)}{t}}{s}=\lim_{s\to0}\lim_{t\to0}\frac{g'_t(\xi_s)}{t}=\lim_{s\to0}\lim_{t\to0}\frac{D_1f(a_1+\xi_s,a_2+t)-D_1f(a_1+\xi_s,a_2)}{t}\quad\bold{(2)}$

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where we’ve used the mean value theorem on $g_t'(s)-g_t'(0)$ (where we know $s\in(-\gamma,\gamma)$ for sufficiently small $s$). Using a similar approach we see that $\bold{(2)}$ may be rewritten as

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$\displaystyle \lim_{s\to0}\lim_{t\to0}D_{1,2}f(a_1+\xi_s,\xi_t)$

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It’s not quite as easy as it looks “hey! We know that $\xi_s\to 0$ and $\xi_t\to a_2$ and so the result follows from continuity”, the problem with this is that $\xi_s$ actually depends on $t$ as well–so we don’t get this totally for free (although the idea is right). Well, from continuity at $(a_1,a_2)$ we know that for any $\varepsilon>0$ there exists some $2\delta>0$ such that $\|(x,y)-(a_1,a_2)\|<2\delta$ implies $\|D_{1,2}f(x,y)-D_{1,2}f(a_1,a_2)\|<\varepsilon$. So, choosing $|s|,|t|<\delta$ enables us to conclude that $\left\|D_{1,2}f(a_1+\xi_s,\xi_t)-D_{1,2}f(a_1,a_2)\right\|<\varepsilon$. In particular, if we choose $|s|<\delta$ and let $t$ tend to zero we get that

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$\displaystyle \left\|\lim_{t\to0}D_{1,2}f(a_1+\xi_s,\xi_t)-D_1f(a_1,a_2)\right\|\leqslant \varepsilon$

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Thus, since $\varepsilon>0$ was arbitrary we have the desired result. $\blacksquare$

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References:

1.  Spivak, Michael. Calculus on Manifolds; a Modern Approach to Classical Theorems of Advanced Calculus. New York: W.A. Benjamin, 1965. Print.

2. Apostol, Tom M. Mathematical Analysis. Reading, MA: Addison-Wesley Pub., 1974. Print.