- #1

Shirish

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- 24

I'm referring to this result:

But I'm not sure what happens if I apply a linear differential operator to both sides (like a derivation ##D##) - more specifically I'm not sure at what point should each term be evaluated. Acting ##D## on both sides I'll get $$DF=\sum_{i=1}^n\big[(x^i-a^i).DH_i+H_i.Dx^i\big]$$

If I try evaluating the non-differentiated RHS terms at ##a##, like this $$DF=\sum_{i=1}^n\big[(x^i-a^i)|_{x=a}.DH_i+H_i(a).Dx^i\big]$$

I can't just bring the ##x=a## thing out of thin air, can I? Should the LHS also be evaluated at ##x=a##? i.e. $$DF|_{x=a}=\sum_{i=1}^n\big[(x^i-a^i)|_{x=a}.DH_i|_{x=a}+\sum_{i=1}^nH_i(a).Dx^i|_{x=a}=\sum_{i=1}^nH_i(a).Dx^i|_{x=a}$$

And then finally, if ##H_i=X_i(F)## for some operator ##X_i##, can we say that ##H_i(x)=X_i(F)(x)## for all ##x\in\mathbb{R}^n##, and thus conclude from the above equation that $$DF|_{x=a}=\sum_{i=1}^nDx^i|_{x=a}.X_i(F)(a)=\bigg[\sum_{i=1}^nDx^i.X_i(F)\bigg]_{x=a}$$

Now since both sides are equal for arbitrary ##a\in\mathbb{R}^n##, ##D=\sum_{i=1}^nDx^i.X_i##

Is this fine? I'm not sure at all since there are so many steps and I don't know where I might have made a wrong assumption or step.

If ##F:\mathbb{R}^n\to\mathbb{R}## is ##C^{\infty}##, then for each ##a\in\mathbb{R}^n##, there exist ##C^{\infty}## functions ##H_i## such that for all ##x\in\mathbb{R}^n##, $$F(x)=F(a)+\sum_{i=1}^n(x^i-a^i)H_i(x)$$

But I'm not sure what happens if I apply a linear differential operator to both sides (like a derivation ##D##) - more specifically I'm not sure at what point should each term be evaluated. Acting ##D## on both sides I'll get $$DF=\sum_{i=1}^n\big[(x^i-a^i).DH_i+H_i.Dx^i\big]$$

If I try evaluating the non-differentiated RHS terms at ##a##, like this $$DF=\sum_{i=1}^n\big[(x^i-a^i)|_{x=a}.DH_i+H_i(a).Dx^i\big]$$

I can't just bring the ##x=a## thing out of thin air, can I? Should the LHS also be evaluated at ##x=a##? i.e. $$DF|_{x=a}=\sum_{i=1}^n\big[(x^i-a^i)|_{x=a}.DH_i|_{x=a}+\sum_{i=1}^nH_i(a).Dx^i|_{x=a}=\sum_{i=1}^nH_i(a).Dx^i|_{x=a}$$

And then finally, if ##H_i=X_i(F)## for some operator ##X_i##, can we say that ##H_i(x)=X_i(F)(x)## for all ##x\in\mathbb{R}^n##, and thus conclude from the above equation that $$DF|_{x=a}=\sum_{i=1}^nDx^i|_{x=a}.X_i(F)(a)=\bigg[\sum_{i=1}^nDx^i.X_i(F)\bigg]_{x=a}$$

Now since both sides are equal for arbitrary ##a\in\mathbb{R}^n##, ##D=\sum_{i=1}^nDx^i.X_i##

Is this fine? I'm not sure at all since there are so many steps and I don't know where I might have made a wrong assumption or step.

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