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given $f$ - a smooth function, $f\colon\mathbb{R}^2\to \mathbb{R}$.

I have a differential operator that takes $f$ to $\frac{\partial f}{\partial x}x$, but I am unsure what this is.

If, for example, the operator tooked $f$ to $\frac{\partial f}{\partial x}$ then I understand that this operator derives by $x$, $\frac{\partial f}{\partial x}x$ derives by $x$ and then what ?

Can someone please help with this ? (I'm confused...)

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$f \mapsto \frac{\partial f}{\partial x}x$ derives by $x$ and multiplies by $x$ afterwards. –  martini Apr 23 '12 at 19:09
    
For example, if $f(x,y) = x^2+2xy$, then $\frac{\partial f}{\partial x} = 2x+2y$, and $\frac{\partial f}{\partial x}x = (2x+2y)x = 2x^2+2xy$. –  Arturo Magidin Apr 23 '12 at 19:10

1 Answer 1

up vote 2 down vote accepted

Let $D$ be a (first order) differential operator (i.e. a derivation) and let $h$ be a function. (In your example $h$ is the function $x$ and $D$ is $\frac{\partial}{\partial x}$). I claim that there is a differential operator $hD$ given by $$ f \mapsto hDf. $$ (This means multiply $h$ by $Df$ pointwise.) For this claim to hold, we just need it to be true that the Leibniz (product) rule holds, i.e. $$ hD(fg) = f(hD)g + g(hD)f. $$ Since $D$ itself satisfies the Leibniz rule, this is true. (We also need the operator $hD$ to take sums to sums, which it does.)

Put less formally, to compute $x \frac{\partial}{\partial x}$ on a function $f$, take the partial with respect to $x$, then multiply what you get by $x$.

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Thank you for the help, but why are we checking these conditions ? –  Belgi Apr 23 '12 at 19:17
    
I was verifying that you have a first order differential operator and not just an arbitrary operator. –  user29743 Apr 23 '12 at 19:32
    
Is it needed for this to be well defined ? –  Belgi Apr 23 '12 at 19:36
    
No, as something that eats a function and takes it to another function, it's well-defined without checking anything. But my assumption was that you were working with vector fields in coordinates on some manifold, in which case you want to see that this is a vector field and not any old operator on functions. –  user29743 Apr 23 '12 at 20:20

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