$11\mathrm{.}2$ Theorem. Suppose that $V$ is open in $R^2,$ that $(a,b)$

If $f$ is $C^1$ on $V,$ and if one of the mixed second partial derivauves of $/$ exsus on $v$ and is continuous at the point $(a,b),$ then the other mixed second partial derivative exists at $(a,b)$ and

$\frac{de{l}^{2}f}{\text{d}\text{e}\text{l}\text{y}\text{d}\text{e}\text{l}\text{x}}(a,b)=\frac{de{l}^{2}f}{\text{d}\text{e}\text{l}\text{x}\text{d}\text{e}\text{l}\text{y}}(a,b)$

NOTE: These hypotheses are met if $fin{C}^2(V).$

Proof. Suppose that $f_{yx}$ exists on $V$ and is continuous at the point $(a,b).$ Consider $(h,k)$:$=f(ah,bk)-f(ah,b)-f(a,bk)f(a,b),$ defined for $r>0B_r(a,b)subVs,$tin$(0,1)(h,k)=k\frac{\text{d}\text{e}\text{l}\text{f}}{\text{d}\text{e}\text{l}\text{y}}(ah,btk)-k\frac{\text{d}\text{e}\text{l}\text{f}}{\text{d}\text{e}\text{l}\text{y}}(a,btk)=hk\frac{de{l}^{2}f}{\text{d}\text{e}\text{l}\text{x}\text{d}\text{e}\text{l}\text{y}}(ash,btk).R^{n}a,b\underset{k0}{lim}\underset{h0}{lim}\frac{(h,k)}{hk}=\frac{de{l}^{2}f}{\text{d}\text{e}\text{l}\text{x}\text{d}\text{e}\text{l}\text{y}}(a,b)uin(0,1)(h,k)=f(ah,bk)-f(a,bk)-f(ah,b)f(a,b)$

$=h\frac{\text{d}\text{e}\text{l}\text{f}}{\text{d}\text{e}\text{l}\text{x}}(auh,bk)-h\frac{\text{d}\text{e}\text{l}\text{f}}{\text{d}\text{e}\text{l}\text{x}}(auh,b)\underset{k0h0}{lim}\underset{k}{lim}\frac{1}{k}(\frac{\text{d}\text{e}\text{l}\text{f}}{\text{d}\text{e}\text{l}\text{x}}(auh,bk)-\frac{\text{d}\text{e}\text{l}\text{f}}{\text{d}\text{e}\text{l}\text{x}}(auh,b))$

$=\underset{k0}{lim}\underset{h0}{lim}\frac{(h,k)}{hk}=\frac{de{l}^{2}f}{\text{d}\text{e}\text{l}\text{x}\text{d}\text{e}\text{l}\text{y}}(a,b)f_x{B}_r(a,b)h=0\frac{de{l}^{2}f}{\text{d}\text{e}\text{l}\text{y}\text{d}\text{e}\text{l}\text{x}}(a,b)=\underset{k0}{lim}\frac{1}{k}(\frac{\text{d}\text{e}\text{l}\text{f}}{\text{d}\text{e}\text{l}\text{x}}(a,bk)-\frac{\text{d}\text{e}\text{l}\text{f}}{\text{d}\text{e}\text{l}\text{x}}(a,b))=\frac{de{l}^{2}f}{\text{d}\text{e}\text{l}\text{x}\text{d}\text{e}\text{l}\text{y}}(a,b)|h|,|k|,$ where $r>0isso$ small that $B_r(a,b)subV.$ Apply the Mean Value Theorem twice $to$ choose scalars $s,$tin$(0,1)$ such that

$(h,k)=k\frac{\text{d}\text{e}\text{l}\text{f}}{\text{d}\text{e}\text{l}\text{y}}(ah,btk)-k\frac{\text{d}\text{e}\text{l}\text{f}}{\text{d}\text{e}\text{l}\text{y}}(a,btk)=hk\frac{de{l}^{2}f}{\text{d}\text{e}\text{l}\text{x}\text{d}\text{e}\text{l}\text{y}}(ash,btk).$

$385$

$386$ Chapter $11$ Differentiability $on{R}^n$

Since this last mixed partial derivative $is$ continuous $at$ the point $(a,b),we$ have

$\underset{k0}{lim}\underset{h0}{lim}\frac{(h,k)}{hk}=\frac{de{l}^{2}f}{\text{d}\text{e}\text{l}\text{x}\text{d}\text{e}\text{l}\text{y}}(a,b)$

$On$ the other hand, the Mean Value Theorem also implies that there $is$ a scalar uin$(0,1)$ such that

$(h,k)=f(ah,bk)-f(a,bk)-f(ah,b)f(a,b)$

$=h\frac{\text{d}\text{e}\text{l}\text{f}}{\text{d}\text{e}\text{l}\text{x}}(auh,bk)-h\frac{\text{d}\text{e}\text{l}\text{f}}{\text{d}\text{e}\text{l}\text{x}}(auh,b)$

Hence, $it$ follows from $(1)$ that

$\underset{k0h0}{lim}\underset{k}{lim}\frac{1}{k}(\frac{\text{d}\text{e}\text{l}\text{f}}{\text{d}\text{e}\text{l}\text{x}}(auh,bk)-\frac{\text{d}\text{e}\text{l}\text{f}}{\text{d}\text{e}\text{l}\text{x}}(auh,b))$

$=\underset{k0}{lim}\underset{h0}{lim}\frac{(h,k)}{hk}=\frac{de{l}^{2}f}{\text{d}\text{e}\text{l}\text{x}\text{d}\text{e}\text{l}\text{y}}(a,b)$

Since $f_{x}is$ continuous $on{B}_r(a,b),we$ can let $h=0in$ the first expression. $We$ conclude $by$ definition that

$\frac{de{l}^{2}f}{\text{d}\text{e}\text{l}\text{y}\text{d}\text{e}\text{l}\text{x}}(a,b)=\underset{k0}{lim}\frac{1}{k}(\frac{\text{d}\text{e}\text{l}\text{f}}{\text{d}\text{e}\text{l}\text{x}}(a,bk)-\frac{\text{d}\text{e}\text{l}\text{f}}{\text{d}\text{e}\text{l}\text{x}}(a,b))=\frac{de{l}^{2}f}{\text{d}\text{e}\text{l}\text{x}\text{d}\text{e}\text{l}\text{y}}(a,b)$