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Tuesday, May 30, 2023

The Mixed Derivative Theorem and the Ring Structure of Differential Operators

Mixed derivative theorem

Let E is an open subset of Rn, f:E→R∈C2 , ∀1≤i,j≤n and ∀x∈E, ∂i∂j(f)|x=∂j∂i(f)|x

Proof.

Let a=∂i∂j(f)|x0,a′=∂j∂i(f)x0

We need to prove that a=a′

Because of f∈C2,∀ϵ>0,∃δ, when |x−x0|≤δ|,|∂i∂j(f)|x0−a|≤ϵ

Similarly, ∀ϵ>0,∃δ, when |x−x0|≤δ|,|∂j∂i(f)|x0−a′|≤ϵ

Now, consider F:=f(δei+δej+x0)−f(δej+x0)−(f(δei+x0)−f(x0))

According to FTC, F=∫0δ∂xif(x0+δej+xiei)−∂xif(x0+xiei)dxi

According to the Mean value theorem, ∀xi,∃0≤xj≤δ,∂xif(x0+δej+xiei)−∂xif(x0+xiei)=δ∂xj∂xif(x0+xiei+xjej)

Observe that xi,xj≤δ

Thus |∂xif(x0+δej+xiei)−∂xif(x0+xiei)−δa|≤δϵ

Therefore ∫0δ|∂xif(x0+δej+xiei)−∂xif(x0+xiei)−δa|dxi=|F−δ2a|≤δ2ϵ

And change the order of i,j, we have |F−δ2a′|≤δ2ϵ

According to triangle inequality, |δ2a−F+F−δ2a′|≤|F−δ2a|+|F−δ2a′|≤2δ2ϵ

Thus |a−a′|≤2ϵ

Then we can generalize it to f:E→Rm

And consider C∞:E⊆Rn→R

Observe that R[∂1,∂2,...,∂n]≅R[x1,x2,...,xn]

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