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For following problem, either prove it or give a counterexample by finding functions and variables for which it does not hold. Assume $f :\mathbb{R}^2 \to \mathbb{R}$ is differentiable. Assume $x ,y$ be in $\mathbb{R}^2$.

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Uh-forgive me for sounding like a complete idiot, but doesn't that follow from the fact the "del" function is a differential operator whose range is a vector field in Euclidean space and therefore a linear operator? So wouldn't this follow from the definition of a linear transformation? – Mathemagician1234 Mar 17 '13 at 17:37
@PeterTamaroff Yes i did. But i m so confused that x and y are vectors. grad f (x+y) = grad f (x1+y1, x2+y2) but how do i go from from there since f is function from R2 to R. how do i even compute gradient of that?? I sound like a moron. i Know thats why i came to get some help. – Christopher Mar 17 '13 at 17:47
saying x,y are in $R^2$ already implies x,y (hence x+y) are in the domain of $f$ and $g$. – Theta33 Mar 17 '13 at 17:49

I don't know if I understood the point, but you want to proove or give a counterexample that $\nabla f(x+y) = \nabla f(x) + \nabla f(y)$?

Well, that's not always true. Pick $f(x,y) = xy^3$, then $\nabla f(x,y)=(y^3, 3xy^2)$. Now, it's obvious that if we take $u =(1,0)$ and $v=(0,1)$, them $\nabla f(u+v)=\nabla f(1,1) =(1,3)$.

However, we have $\nabla f(1,0) = (0,0)$ and $\nabla f(0,1) = (1, 0)$. Hence, $\nabla f(1,1) \neq \nabla f(1,0) + \nabla f(0,1)$ for this function, and hence we disproove the proposition.

EDIT: On the other hand $\nabla(f+g) = \nabla f+\nabla g$ since by the definition of differentiability, the gradient (and more general, the total derivative) should be linear.

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But f and g are functions right. I n my problems x and y are vectors. Will that still be true? – Christopher Mar 17 '13 at 18:16
Look, if $f$ and $g$ are functions, then the gradient of the sum is the sum of the gradients. However, in your problem you've got two vectors, and one function. You want to know if the gradient of the function applied on the sum of vectors is equal to the sum of the gradient applied on each vector separetely. This is not always true as I've shown with a counterexample. For the $f$ I've build, you can see tha the relation doesn't hold. In other words, $\nabla f(u+v) \neq \nabla f(u) + \nabla f(v)$. Take a look on the counterexample and see if you can note that now. – user1620696 Mar 17 '13 at 18:18

Suppose you have differentiable function that $\nabla f(x+y) = \nabla f(x) + \nabla f(y)$. Than by Gradient theorem you know that $f(x+y) = f(x)+f(y)+c$. So $f$ has to be affine.

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