Denotarem
f
x
i
=
∂
f
∂
x
i
{\displaystyle f_{x_{i}}={\frac {\partial f}{\partial x_{i}}}}
i
f
x
i
x
j
:=
∂
2
f
∂
x
i
∂
x
j
{\displaystyle f_{x_{i}x_{j}}:={\frac {\partial ^{2}f}{\partial x_{i}\partial x_{j}}}}
i demostrarem que si existeixen
f
x
i
,
f
x
j
i
f
x
i
x
j
{\displaystyle f_{x_{i}},f_{x_{j}}{\text{ i }}f_{x_{i}x_{j}}}
en tot l'obert
Ω
{\displaystyle \Omega }
i
f
x
i
x
j
{\displaystyle f_{x_{i}x_{j}}}
és contínua en un punt
a
∈
Ω
{\displaystyle a\in \Omega }
, aleshores
∃
f
x
j
x
i
(
a
)
=
f
x
i
x
j
(
a
)
{\displaystyle \exists f_{x_{j}x_{i}}(a)=f_{x_{i}x_{j}}(a)}
.
Sigui
α
:=
f
x
i
x
j
(
a
)
{\displaystyle \alpha :=f_{x_{i}x_{j}}(a)}
. Per continuïtat de
f
x
i
x
j
{\displaystyle f_{x_{i}x_{j}}}
en
a
{\displaystyle a}
tenim que donat
ε
>
0
,
∃
r
>
0
{\displaystyle \varepsilon >0,\ \ \exists r>0}
tal que
B
(
a
,
r
)
⊆
Ω
{\displaystyle B(a,r)\subseteq \Omega }
(per ser
Ω
{\displaystyle \Omega }
obert) i
|
f
x
i
x
j
(
x
)
−
α
|
≤
ε
∀
x
∈
B
(
a
,
r
)
{\displaystyle |f_{x_{i}x_{j}}(x)-\alpha |\leq \varepsilon \ \ \ \forall x\in B(a,r)}
.
(
1
)
{\displaystyle (1)}
Considerem
δ
:=
r
2
{\displaystyle \delta :={\frac {r}{\sqrt {2}}}}
. Aleshores, denotant per
e
i
{\displaystyle e_{i}}
l'
i
{\displaystyle i}
-èsim vector de la base canònica de
R
n
{\displaystyle \mathbb {R} ^{n}}
, per a tot
h
,
k
∈
(
−
δ
,
δ
)
{\displaystyle h,k\in (-\delta ,\delta )}
, tenim que
|
|
a
+
h
e
i
+
k
e
j
−
a
|
|
=
|
|
h
e
i
+
k
e
j
|
|
=
h
2
+
k
2
<
2
δ
2
=
2
r
2
2
=
|
r
|
=
r
⇒
a
+
h
e
i
+
k
e
j
∈
B
(
a
,
r
)
(
2
)
{\displaystyle ||a+he_{i}+ke_{j}-a||=||he_{i}+ke_{j}||={\sqrt {h^{2}+k^{2}}}<{\sqrt {2\delta ^{2}}}={\sqrt {2{\frac {r^{2}}{2}}}}=|r|=r\Rightarrow a+he_{i}+ke_{j}\in B(a,r)\quad \quad (2)}
En particular, com que, per
(
1
)
{\displaystyle (1)}
,
B
(
a
,
r
)
⊆
Ω
{\displaystyle B(a,r)\subseteq \Omega }
, podem definir la següent funció:
A
:
(
−
δ
,
δ
)
2
⟶
R
{\displaystyle A:(-\delta ,\delta )^{2}\longrightarrow \mathbb {R} }
(
h
,
k
)
⟼
A
(
h
,
k
)
=
f
(
a
+
h
e
i
+
k
e
j
)
−
f
(
a
+
h
e
i
)
−
f
(
a
+
k
e
j
)
+
f
(
a
)
{\displaystyle \quad \quad \ \ (h,k)\longmapsto A(h,k)=f(a+he_{i}+ke_{j})-f(a+he_{i})-f(a+ke_{j})+f(a)}
Ara, donats
h
,
k
{\displaystyle h,k}
amb
0
<
|
h
|
,
|
k
|
<
δ
{\displaystyle 0<|h|,|k|<\delta }
, definim la funció
u
:
(
−
δ
,
δ
)
⟶
R
{\displaystyle u:(-\delta ,\delta )\longrightarrow \mathbb {R} }
t
⟼
u
(
t
)
=
f
(
a
+
h
e
i
+
t
e
j
)
−
f
(
a
+
t
e
j
)
{\displaystyle \quad \quad \ \ t\longmapsto u(t)=f(a+he_{i}+te_{j})-f(a+te_{j})}
Per
(
2
)
{\displaystyle (2)}
i
(
1
)
{\displaystyle (1)}
, tenim que
a
+
h
e
i
+
t
e
j
,
a
+
t
e
j
∈
B
(
a
,
r
)
⊆
Ω
{\displaystyle a+he_{i}+te_{j},a+te_{j}\in B(a,r)\subseteq \Omega }
, d'on, com que existeix
f
x
j
{\displaystyle f_{x_{j}}}
per hipòtesi,
u
{\displaystyle u}
és derivable i
u
′
=
f
x
j
(
a
+
h
e
i
+
t
e
j
)
−
f
x
j
(
a
+
t
e
j
)
{\displaystyle u'=f_{x_{j}}(a+he_{i}+te_{j})-f_{x_{j}}(a+te_{j})}
. Com que
0
,
k
∈
(
−
δ
,
δ
)
,
{\displaystyle 0,k\in (-\delta ,\delta ),}
podem aplicar el teorema del valor mitjà de Lagrange d'una variable a
u
{\displaystyle u}
a l'interval amb extrems
0
{\displaystyle 0}
i
k
{\displaystyle k}
. Així,
∃
ξ
∈
⟨
0
,
k
⟩
tal que
A
(
h
,
k
)
=
u
(
k
)
−
u
(
0
)
=
u
′
(
ξ
)
(
k
−
0
)
=
[
f
x
j
(
a
+
h
e
i
+
ξ
e
j
)
−
f
(
a
+
ξ
e
j
)
]
k
(
∗
)
{\displaystyle \exists \xi \in \left\langle 0,k\right\rangle {\text{ tal que }}A(h,k)=u(k)-u(0)=u'(\xi )(k-0)=\left[f_{x_{j}}(a+he_{i}+\xi e_{j})-f(a+\xi e_{j})\right]k\quad \quad (*)}
Considerem ara
v
:
(
−
δ
,
δ
)
⟶
R
{\displaystyle v:(-\delta ,\delta )\longrightarrow \mathbb {R} }
t
⟼
v
(
t
)
=
f
x
j
(
a
+
t
e
i
+
ξ
e
j
)
{\displaystyle \quad \quad \ \ t\longmapsto v(t)=f_{x_{j}}(a+te_{i}+\xi e_{j})}
Com que
t
∈
(
−
δ
,
δ
)
i
ξ
∈
⟨
0
,
k
⟩
⊆
(
−
δ
,
δ
)
{\displaystyle t\in (-\delta ,\delta ){\text{ i }}\xi \in \left\langle 0,k\right\rangle \subseteq (-\delta ,\delta )}
, per
(
2
)
i
(
1
)
{\displaystyle (2){\text{ i }}(1)}
tenim que
a
+
t
e
i
+
ξ
e
j
∈
B
(
a
,
r
)
⊆
Ω
{\displaystyle a+te_{i}+\xi e_{j}\in B(a,r)\subseteq \Omega }
, d'on, com que existeix
f
x
i
x
j
{\displaystyle f_{x_{i}x_{j}}}
per hipòtesi,
v
{\displaystyle v}
és derivable i
v
′
(
t
)
=
f
x
i
x
j
(
a
+
t
e
i
+
ξ
e
j
)
{\displaystyle v'(t)=f_{x_{i}x_{j}}(a+te_{i}+\xi e_{j})}
.
Com que
0
,
h
∈
(
−
δ
,
δ
)
,
{\displaystyle 0,h\in (-\delta ,\delta ),}
podem aplicar el teorema del valor mitjà de Lagrange d'una variable a
v
{\displaystyle v}
a l'interval amb extrems
0
{\displaystyle 0}
i
h
{\displaystyle h}
. Així,
∃
η
∈
⟨
0
,
h
⟩
tal que
f
x
j
(
a
+
h
e
i
+
ξ
e
j
)
−
f
x
j
(
a
+
ξ
e
j
)
=
v
(
h
)
−
v
(
0
)
=
v
′
(
η
)
(
h
−
0
)
=
f
x
i
x
j
(
a
+
η
e
i
+
ξ
e
j
)
h
⇒
⋅
k
{\displaystyle \exists \eta \in \left\langle 0,h\right\rangle {\text{ tal que }}f_{x_{j}}(a+he_{i}+\xi e_{j})-f_{x_{j}}(a+\xi e_{j})=v(h)-v(0)=v'(\eta )(h-0)=f_{x_{i}x_{j}}(a+\eta e_{i}+\xi e_{j})h\quad {\overset {\cdot k}{\Rightarrow }}}
⇒
⋅
k
[
f
x
j
(
a
+
h
e
i
+
ξ
e
j
)
−
f
x
j
(
a
+
ξ
e
j
)
]
k
=
f
x
i
x
j
(
a
+
η
e
i
+
ξ
e
j
)
h
k
⇒
(
∗
)
A
(
h
,
k
)
=
f
x
i
x
j
(
a
+
η
e
i
+
ξ
e
j
)
h
k
{\displaystyle {\overset {\cdot k}{\Rightarrow }}\left[f_{x_{j}}(a+he_{i}+\xi e_{j})-f_{x_{j}}(a+\xi e_{j})\right]k=f_{x_{i}x_{j}}(a+\eta e_{i}+\xi e_{j})hk~~{\overset {(*)}{\Rightarrow }}~~A(h,k)=f_{x_{i}x_{j}}(a+\eta e_{i}+\xi e_{j})hk}
.
Definint
z
:=
a
+
η
e
i
+
ξ
e
j
{\displaystyle z:=a+\eta e_{i}+\xi e_{j}}
, com que
h
,
k
≠
0
⇒
h
k
≠
0
{\displaystyle h,k\neq 0\Rightarrow hk\neq 0}
, tenim que
A
(
h
,
k
)
h
k
=
f
x
i
x
j
(
z
)
{\displaystyle {\frac {A(h,k)}{hk}}=f_{x_{i}x_{j}}(z)}
. Observem que
(
2
)
⇒
z
∈
B
(
a
,
r
)
⇒
(
1
)
|
f
x
i
x
j
−
α
|
≤
ε
{\displaystyle (2)\Rightarrow z\in B(a,r)~{\overset {(1)}{\Rightarrow }}~|f_{x_{i}x_{j}}-\alpha |\leq \varepsilon }
. Així, tenim que
|
A
(
h
,
k
)
h
k
−
α
|
=
|
f
x
i
x
j
(
z
)
−
α
|
≤
ε
(
∗
∗
)
{\displaystyle \left|{\frac {A(h,k)}{hk}}-\alpha \right|=\left|f_{x_{i}x_{j}}(z)-\alpha \right|\leq \varepsilon \quad (**)}
.
Finalment, observem que
A
(
h
,
k
)
h
k
=
1
k
[
f
(
a
+
h
e
i
+
k
e
j
)
−
f
(
a
+
k
e
i
)
h
−
f
(
a
+
h
e
i
)
−
f
(
a
)
h
]
⟶
h
→
0
1
k
[
f
x
i
(
a
+
k
e
j
)
−
f
x
i
(
a
)
]
⇒
{\displaystyle {\frac {A(h,k)}{hk}}={\frac {1}{k}}\left[{\frac {f(a+he_{i}+ke_{j})-f(a+ke_{i})}{h}}-{\frac {f(a+he_{i})-f(a)}{h}}\right]{\overset {h\rightarrow 0}{\longrightarrow }}~~{\frac {1}{k}}\left[f_{x_{i}}(a+ke_{j})-f_{x_{i}}(a)\right]\Rightarrow }
⇒
ε
≥
(
∗
∗
)
|
A
(
h
,
k
)
h
k
−
α
|
⟶
h
→
0
|
f
x
i
(
a
+
k
e
j
)
−
f
x
i
(
a
)
k
−
α
|
≤
ε
⇒
f
x
i
(
a
+
k
e
j
)
−
f
x
i
(
a
)
k
⟶
k
→
0
α
⇒
f
x
j
x
i
(
a
)
=
α
=
def
f
x
i
x
j
(
a
)
◻
{\displaystyle \Rightarrow \varepsilon {\overset {(**)}{\geq }}\left|{\frac {A(h,k)}{hk}}-\alpha \right|~~{\overset {h\rightarrow 0}{\longrightarrow }}~~\left|{\frac {f_{x_{i}}(a+ke_{j})-f_{x_{i}}(a)}{k}}-\alpha \right|\leq \varepsilon \Rightarrow {\frac {f_{x_{i}}(a+ke_{j})-f_{x_{i}}(a)}{k}}~~{\overset {k\rightarrow 0}{\longrightarrow }}~~\alpha \Rightarrow f_{x_{j}x_{i}}(a)=\alpha ~{\overset {\text{def}}{=}}~f_{x_{i}x_{j}}(a)\quad \square }