|
|
en.wikipedia.org/wiki/Symmetry_of_second_derivatives#Theorem_of_Schwarz
In mathematical analysis, Schwarz's theorem (or Clairaut's theorem on equality of mixed partials)[9] named after Alexis Clairaut and Hermann Schwarz, states that for a function ${\displaystyle f\colon \Omega \to \mathbb {R} }$ defined on a set $\Omega \subset \mathbb{R}^n$, if ${\displaystyle \mathbf {p} \in \mathbb {R} ^{n}}$ is a point such that some neighborhood of $\mathbf {p} $ is contained in $\Omega $ and $f$ has continuous second partial derivatives on that neighborhood of $\mathbf {p} $, then for all $i$ and $j$ in ${\displaystyle \{1,2\ldots ,\,n\}.}$
${\displaystyle {\frac {\partial ^{2}}{\partial x_{i}\,\partial x_{j}}}f(\mathbf {p} )={\frac {\partial ^{2}}{\partial x_{j}\,\partial x_{i}}}f(\mathbf {p} ).}$
The partial derivatives of this function commute at that point.
One easy way to establish this theorem (in the case where $n=2$, $i=1$, and $j=2$, which readily entails the result in general) is by applying Green's theorem to the gradient of $f.$
An elementary proof for functions on open subsets of the plane is as follows (by a simple reduction, the general case for the theorem of Schwarz easily reduces to the planar case).[10] Let ${\displaystyle f(x.y)}$ be a differentiable function on an open rectangle Ω containing a point $(a,b)$ and suppose that $df$ is continuous with continuous ${\displaystyle \partial _{x}\partial _{y}f}$ and ${\displaystyle \partial _{y}\partial _{x}f}$ over Ω. Define
${\displaystyle {\begin{aligned}u\left(h,\,k\right)&=f\left(a+h,\,b+k\right)-f\left(a+h,\,b\right),\\v\left(h,\,k\right)&=f\left(a+h,\,b+k\right)-f\left(a,\,b+k\right),\\w\left(h,\,k\right)&=f\left(a+h,\,b+k\right)-f\left(a+h,\,b\right)-f\left(a,\,b+k\right)+f\left(a,\,b\right).\end{aligned}}}$
These functions are defined for ${\displaystyle \left|h\right|,\,\left|k\right|<\varepsilon }$, where $\varepsilon >0$ and ${\displaystyle \left[a-\varepsilon ,\,a+\varepsilon \right]\times \left[b-\varepsilon ,\,b+\varepsilon \right]}$ is contained in Ω.
By the mean value theorem, for fixed h and k non-zero, ${\displaystyle \theta ,\,\theta ^{\prime },\,\,\phi ,\,\,\phi ^{\prime }}$ can be found in the open interval $ (0,1)$ with
${\displaystyle {\begin{aligned}w\left(h,\,k\right)&=u\left(h,\,k\right)-u\left(0,\,k\right)=h\,\partial _{x}u\left(\theta h,\,k\right)\\&=h\,\left[\partial _{x}f\left(a+\theta h,\,b+k\right)-\partial _{x}f\left(a+\theta h,\,b\right)\right]\\&=hk\,\partial _{y}\partial _{x}f\left(a+\theta h,\,b+\theta ^{\prime }k\right)\\w\left(h,\,k\right)&=v\left(h,\,k\right)-v\left(h,\,0\right)=k\,\partial _{y}v\left(h,\,\phi k\right)\\&=k\left[\partial _{y}f\left(a+h,\,b+\phi k\right)-\partial _{y}f\left(a,\,b+\phi k\right)\right]\\&=hk\,\partial _{x}\partial _{y}f\left(a+\phi ^{\prime }h,\,b+\phi k\right).\end{aligned}}}$
Since ${\displaystyle h,\,k\neq 0}$, the first equality below can be divided by $hk$:
${\displaystyle {\begin{aligned}hk\,\partial _{y}\partial _{x}f\left(a+\theta h,\,b+\theta ^{\prime }k\right)&=hk\,\partial _{x}\partial _{y}f\left(a+\phi ^{\prime }h,\,b+\phi k\right),\\\partial _{y}\partial _{x}f\left(a+\theta h,\,b+\theta ^{\prime }k\right)&=\partial _{x}\partial _{y}f\left(a+\phi ^{\prime }h,\,b+\phi k\right).\end{aligned}}}$
Letting ${\displaystyle h,\,k}$ tend to zero in the last equality, the continuity assumptions on ${\displaystyle \partial _{y}\partial _{x}f}$ and ${\displaystyle \partial _{x}\partial _{y}f}$ now imply that
${\displaystyle {\frac {\partial ^{2}}{\partial x\partial y}}f\left(a,\,b\right)={\frac {\partial ^{2}}{\partial y\partial x}}f\left(a,\,b\right).}$
This account is a straightforward classical method found in many text books, for example in Burkill, Apostol and Rudin.[10][11][12]
Although the derivation above is elementary, the approach can also be viewed from a more conceptual perspective so that the result becomes more apparent.[13][14][15][16][17] Indeed the difference operators ${\displaystyle \Delta _{x}^{t},\,\,\Delta _{y}^{t}}$ commute and ${\displaystyle \Delta _{x}^{t}f,\,\,\Delta _{y}^{t}f}$ tend to ${\displaystyle \partial _{x}f,\,\,\partial _{y}f}$ as $t$ tends to 0, with a similar statement for second order operators.[a] Here, for $z$ a vector in the plane and $u$ a directional vector, the difference operator is defined by
${\displaystyle \Delta _{u}^{t}f(z)={f(z+tu)-f(z) \over t}.}$
By the fundamental theorem of calculus for $C^{1}$ functions $f$ on an open interval $I$ with ${\displaystyle (a,b)\subset I}$
${\displaystyle \int _{a}^{b}f^{\prime }(x)\,dx=f(b)-f(a).}$
Hence
${\displaystyle |f(b)-f(a)|\leq (b-a)\,\sup _{c\in (a,b)}|f^{\prime }(c)|}$.
This is a generalized version of the mean value theorem. Recall that the elementary discussion on maxima or minima for real-valued functions implies that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there is a point $c$ in $(a,b)$ such that
${\displaystyle {f(b)-f(a) \over b-a}=f^{\prime }(c).}$
For vector-valued functions with $V$ a finite-dimensional normed space, there is no analogue of the equality above, indeed it fails. But since ${\displaystyle \inf f^{\prime }\leq f^{\prime }(c)\leq \sup f^{\prime }}$, the inequality above is a useful substitute. Moreover, using the pairing of the dual of $V$ with its dual norm, yields the following inequality:
${\displaystyle \|f(b)-f(a)\|\leq (b-a)\,\sup _{c\in (a,b)}\|f^{\prime }(c)\|}$.
These versions of the mean valued theorem are discussed in Rudin, Hörmander and elsewhere.[19][20]
For $f$ a $C^{2}$ function on an open set in the plane, define ${\displaystyle D_{1}=\partial _{x}}$ and ${\displaystyle D_{2}=\partial _{y}}$. Furthermore for ${\displaystyle t\neq 0}$ set
${\displaystyle \Delta _{1}^{t}f(x,y)=[f(x+t,y)-f(x,y)]/t,\,\,\,\,\,\,\Delta _{2}^{t}f(x,y)=[f(x,y+t)-f(x,y)]/t}$.
Then for $(x_{0},y_{0})$ in the open set, the generalized mean value theorem can be applied twice:
$\displaystyle \left|\Delta _{1}^{t}\Delta _{2}^{t}f(x_{0},y_{0})-D_{1}D_{2}f(x_{0},y_{0})\right|$
$\displaystyle\leq \sup _{0\leq s\leq 1}\left|\Delta _{1}^{t}D_{2}f(x_{0},y_{0}+ts)-D_{1}D_{2}f(x_{0},y_{0})\right|$
$\displaystyle\leq \sup _{0\leq r,s\leq 1}\left|D_{1}D_{2}f(x_{0}+tr,y_{0}+ts)-D_{1}D_{2}f(x_{0},y_{0})\right|.$
Thus ${\displaystyle \Delta _{1}^{t}\Delta _{2}^{t}f(x_{0},y_{0})}$ tends to ${\displaystyle D_{1}D_{2}f(x_{0},y_{0})}$ as $t$ tends to 0. The same argument shows that ${\displaystyle \Delta _{2}^{t}\Delta _{1}^{t}f(x_{0},y_{0})}$ tends to ${\displaystyle D_{2}D_{1}f(x_{0},y_{0})}$. Hence, since the difference operators commute, so do the partial differential operators $D_{1}$ and $D_{2}$, as claimed.[21][22][23][24][25]
Remark. By two applications of the classical mean value theorem,
${\displaystyle \Delta _{1}^{t}\Delta _{2}^{t}f(x_{0},y_{0})=D_{1}D_{2}f(x_{0}+t\theta ,y_{0}+t\theta ^{\prime })}$
for some $\theta $ and $\theta^\prime$ in $(0,1)$. Thus the first elementary proof can be reinterpreted using difference operators. Conversely, instead of using the generalized mean value theorem in the second proof, the classical mean valued theorem could be used. |
|
