Abstract Nonsense

Crushing one theorem at a time

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


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

\text{ }

Remark: For notational convenience we assume that i<j. 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).

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

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)

\text{ }

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

\text{ }

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)

\text{ }

\text{ }

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

\text{ }

\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)}

\text{ }

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

\text{ }

\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)}

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

\text{ }

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.

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June 1, 2011 - Posted by | Analysis, Uncategorized | , , , , , , ,

2 Comments »

  1. […] just telescopes). But, using logic similar to that in the proof of Clairaut’s theorem we may apply the mean value theorem to find with such that   is […]

    Pingback by Relationship Between the Notions of Directional and Total Derivatives (Pt.II) « Abstract Nonsense | June 2, 2011 | Reply

  2. […] We know that by the prior theorem and thus we may use Clairaut’s theorem to make successive switches, and so the conclusion […]

    Pingback by Functions of Class C^k « Abstract Nonsense | June 4, 2011 | Reply


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