Differential operator in multivariable fundamental theorem

[Shirish]

I'm referring to this result:

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.

 

[PeroK]

But I'm not sure what happens if I apply a 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 $$DF=\sum_{i=1}^n\big[(x^i-a^i).DH_i+H_i.Dx^i\big]$$

I don't understand what you are doing here? You have dropped some function arguments but not others, so formally the equation cannot be right.

 

[Shirish]

I don't understand what you are doing here? You have dropped some function arguments but not others, so formally the equation cannot be right.

Since $F(a)$ is constant, acting $D$ on it will give zero. The remaining stuff is applying the product rule to the 2nd RHS term in the very first equation (in quote block).

 

[PeroK]

Since $F(a)$ is constant, acting $D$ on it will give zero. The remaining stuff is applying the product rule to the 2nd RHS term in the very first equation (in quote block).

My point is not that you misapplied the product rule, but you played fast and loose with functional arguments. So, it's not suprising that your calculations thereafter go from bad to worse. Your next equation is effectively a total mish-mash of notation:

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

 

[Shirish]

My point is not that you misapplied the product rule, but you played fast and loose with functional arguments. So, it's not suprising that your calculations thereafter go from bad to worse. Your next equation is effectively a total mish-mash of notation:

The thread's intention is for me to get a better understanding, hence I stated

But I'm not sure what happens if I apply a linear differential operator to both sides (like a derivation $D$)

I'm certain I've made mistakes in several steps. Could you help with those? (The motivation for the thread is to try and understand Theorem 2.2.1 from Wald's book)

Also you're right about this equation being wrong $$DF=\sum_{i=1}^n\big[(x^i-a^i)|_{x=a}.DH_i+H_i(a).Dx^i\big]$$
But immediately after that I acknowledge:

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. $$DF|_{x=a}=\sum_{i=1}^n\big[(x^i-a^i)|_{x=a}.DH_i|_{x=a}+\sum_{i=1}^nH_i(a).Dx^i|_{x=a}=\sum_{i=1}^nH_i(a).Dx^i|_{x=a}$$

Is the new equation correct? If not, could you help me understand the proper way to apply the $D$ operator and evaluate at a specific point $a$?

 

[PeroK]

The thread's intention is for me to get a better understanding, hence I statedI'm certain I've made mistakes in several steps. Could you help with those? (The motivation for the thread is to try and understand Theorem 2.2.1 from Wald's book)

Also you're right about this equation being wrong $$DF=\sum_{i=1}^n\big[(x^i-a^i)|_{x=a}.DH_i+H_i(a).Dx^i\big]$$
But immediately after that I acknowledge:

Is the new equation correct? If not, could you help me understand the proper way to apply the $D$ operator and evaluate at a specific point $a$?

The first step is to apply the linear operator $D$ to both sides:
$$(DF)(x)=\sum_{i=1}^n\big[(x^i-a^i).(DH_i)(x)+H_i(x).D(x^i-a^i)\big]$$Edit: I missed out $a^i$ in the second term. Fixed now.

Although, to really clarify what is going on we could introduce coordinate functions $X^i(x) = x^i$. That allows us to avoid something like $Dx^i$ and have:
$$(DF)(x)=\sum_{i=1}^n\big[X^i(x-a).(DH_i)(x)+H_i(x).(DX^i)(x-a)\big]$$

 

[Shirish]

The first step is to apply the linear operator $D$ to both sides:
$$(DF)(x)=\sum_{i=1}^n\big[(x^i-a^i).(DH_i)(x)+H_i(x).Dx^i\big]$$Although, to really clarify what is going on we could introduce coordinate functions $X^i(x) = x^i$. That allows us to avoid something like $Dx^i$ and have:
$$(DF)(x)=\sum_{i=1}^n\big[X^i(x-a).(DH_i)(x)+H_i(x).(DX^i)(x-a)\big]$$

Thanks! Understood so far. How does the rest of the treatment after that look? (i.e. evaluating both sides at some point $x=a$ and reaching the conclusion $D=\sum_{i=1}^n dx^i.X_i$)

 

[PeroK]

Thanks! Understood so far. How does the rest of the treatment after that look? (i.e. evaluating both sides at some point $x=a$ and reaching the conclusion $D=\sum_{i=1}^n dx^i.X_i$)

$a$ is already the point about which the expansion was taken. I.e. $x-a = 0$. The only question is how the differential operator $D$ acts on the coordinate functions. That depends on the details of $D$. In any case, we have:
$$(DF)(a)=\sum_{i=1}^n\big[H_i(a).(DX^i)(0)\big]$$If you are asking about another point that is not $a$, then $x$ does just as well, although you might want to use $x = b$.
$$(DF)(b)=\sum_{i=1}^n\big[X^i(b-a).(DH_i)(b)+H_i(b).(DX^i)(b-a)\big]$$Note these notations are equivalent:$$F(a) \equiv F(x) \big |_{x = a}$$

 

[Shirish]

The first step is to apply the linear operator $D$ to both sides:
$$(DF)(x)=\sum_{i=1}^n\big[(x^i-a^i).(DH_i)(x)+H_i(x).Dx^i\big]$$Although, to really clarify what is going on we could introduce coordinate functions $X^i(x) = x^i$. That allows us to avoid something like $Dx^i$ and have:
$$(DF)(x)=\sum_{i=1}^n\big[X^i(x-a).(DH_i)(x)+H_i(x).(DX^i)(x-a)\big]$$

Sorry I didn't notice before, but why is there $x-a$ in the very last term in the 2nd equation? I'm guessing it should be $DX^i(x)$?

 

[PeroK]

Sorry I didn't notice before, but why is there $x-a$ in the very last term in the 2nd equation? I'm guessing it should be $DX^i(x)$?

It right as it is. It's true that for a differential operator the two will (probably) be equal. But, I was keeping the action of $D$ as general as possible until we have a specific operator in mind. The point of the exercise was to emphasise the formal structure of the equation of functions, acted on by an operator. You had a number of (often confusing) short cuts in your OP. I was trying to avoid that scenario.

 

[PeroK]

PS the assumption you want to make is that
$$D(X^i(x - a)) = D(X^i(x))$$If, for example, $D \equiv \frac{\partial}{\partial x^j}$, then:
$$D(X^i(x - a)) = D(X^i(x)) = \frac{\partial x^i}{\partial x^j}= \delta^i_j$$

 

[Shirish]

It right as it is. It's true that for a differential operator the two will (probably) be equal. But, I was keeping the action of $D$ as general as possible until we have a specific operator in mind. The point of the exercise was to emphasise the formal structure of the equation of functions, acted on by an operator. You had a number of (often confusing) short cuts in your OP. I was trying to avoid that scenario.

So a problematic case could be if $D$ is non-linear? If $D$ is linear, then $$D(X^i(x-a))=D(X^i(x))-D(X^i(a))=D(X^i(x))$$ since $X^i$ is also linear, and $X^i(a)=a^i$ is just a constant, so $D(\text{constant})=0$. It didn't occur to me that there can be operators $D$ such that $Dx^i\neq DX^i(x)$

 

[PeroK]

So a problematic case could be if $D$ is non-linear?

We don't need a non-linear operator. Only an operator that does not map a constant function to the zero function. If $D$ were the identity operator, for example. Ironically, then it could be seen as the zeroth derivative!

If $D$ is linear, then $D(X^i(x-a))=D(X^i(x))-D(X^i(a))=D(X^i(x))$ is what I think.

This does not hold for all operators. See above.

It didn't occur to me that there can be operators $D$ such that $Dx^i\neq DX^i(x)$

That's not what we are saying. We are saying that for some operators $D(X^i(x)) \ne D(X^i(x - a))$.

 

[Shirish]

I'm sure I'm probably trying your patience by now, but please bear with me since I'm not very bright. You mentioned this:

The first step is to apply the linear operator $D$ to both sides:
$$(DF)(x)=\sum_{i=1}^n\big[(x^i-a^i).(DH_i)(x)+H_i(x).Dx^i\big]$$

Call the above equation 1. All perfectly fine so far. Then you mention:

Although, to really clarify what is going on we could introduce coordinate functions $X^i(x) = x^i$. That allows us to avoid something like $Dx^i$ and have:
$$(DF)(x)=\sum_{i=1}^n\big[X^i(x-a).(DH_i)(x)+H_i(x).(DX^i)(x-a)\big]$$

Call this equation 2. What I'm trying to understand is - what's the pitfall of writing equation 1 in terms of coordinate functions like this: $$(DF)(x)=\sum_{i=1}^n\big[X^i(x-a).(DH_i)(x)+H_i(x).(DX^i)(x)\big]$$ Call the above equation 3. Equation 3 will be the wrong way to write only if $(DX^i)(x)$ is not the correct way to write $Dx^i$, right? i.e. there is some operator $D$ for which $Dx^i\neq DX^i(x)$

 

[PeroK]

I'm sure I'm probably trying your patience by now, but please bear with me since I'm not very bright. You mentioned this:

Call the above equation 1. All perfectly fine so far. Then you mention:

Call this equation 2. What I'm trying to understand is - what's the pitfall of writing equation 1 in terms of coordinate functions like this: $$(DF)(x)=\sum_{i=1}^n\big[X^i(x-a).(DH_i)(x)+H_i(x).(DX^i)(x)\big]$$ Call the above equation 3. Equation 3 will be the wrong way to write only if $(DX^i)(x)$ is not the correct way to write $Dx^i$, right? i.e. there is some operator $D$ for which $Dx^i\neq DX^i(x)$

In post #6, I missed out the $a^i$ in the second term. I've fixed that now. The only question is whether we simplify the equation using $D(x-a) = D(x)$ or $D(x^i - a^i) = D(x^i)$ or $D(X^i(x-a)) = D(X^i(x))$.