I'm trying clarify my understanding of the following expression

$\displaystyle \frac{\partial (-f(z))}{\partial (-z)}$

I believe I can treat the negative sign in the operand as a constant $-1$ and pull it out in front of the expression to give

$\displaystyle -\frac{\partial f(z)}{\partial (-z)}$

What I'm not sure about is the negative sign in the operator. Can I also pull a negative sign out there? In other words, does my initial expression simplify to

$\displaystyle \frac{\partial (-f(z))}{\partial (-z)} = \frac{\partial f(z)}{\partial z}$

or do I have to keep the operator as is? If $z$ is complex, does that place any restrictions on my ability to pull the negative sign out (assuming I can) or does it not matter if $z$ is real or complex?

If $f$ is Fréchet-differentiable at $z$ then

$$\frac{\partial[ -f(z)]}{\partial(-z)}=-\partial[-f(z)]z=\partial_z f(z)$$

where $\partial$ means Fréchet derivative and $\frac{\partial}{\partial x}\equiv \partial_x$. The above manipulation is the result of the linearity of $\partial$ and $\partial[-f(z)]$.

If $f$ is not Fréchet-differentiable at $z$ we need to justify the result using directional derivatives, that is

$$\frac{\partial [-f(z)]}{\partial(-z)}=\lim_{h\to 0}\frac{-f(z-hz)+f(z)}h=\lim_{r\to 0}\frac{f(z+rz)-f(z)}r=\partial_z f(z)$$

where we had set $r:=-h$ and the final step is just the result of the definition of directional derivative.

Note: if $z$ belongs to a complex vector space then $h\in\Bbb C$. If $z$ belongs to a real vector space then $h\in\Bbb R$.

• Kinda a new world to me...can you tell me how I would "read" your second expression $-\partial[-f(z)]z$. I get the first one and the last would be "the partial of $f(z)$ with respect to $z$", but the middle one is a bit foreign to me. I looked up Fréchet derivative and as I said, it's a new world to me. – ThatsRightJack Nov 18 '17 at 2:27
• In the expression you can rename $g(z):=-f(z)$ and $y:=-z$ to have an easy reading as $\partial g(z)y$, where $\partial g(z)$ is the Fréchet derivative of $g$ at $z$ (assuming that it exists). When the Fréchet derivative exists at some point (in this case in the point $z$) then $\partial g(z)y=\partial_y g(z)$ where $\partial_y$ means partial or directional derivative, ... – Masacroso Nov 18 '17 at 3:01
• ... where the concept of partial derivative is more general than the directional derivative, because the first doesnt necessarily rely in a "direction" of a vector space but the second yes. I assumed that in this case the domain of $f$ is a subset of some vector space (then $-z$ is a direction) so $\partial_y$ is a directional derivative also (not just a partial derivative). – Masacroso Nov 18 '17 at 3:01
• Sorry, I need to says also that the concept of "partial derivative" depends on the context, so dont follow too much what I said above about it generality. Just follow what you learn in your multivariable calculus course. – Masacroso Nov 18 '17 at 3:20
• yes, that correct. But in general, if you read some book about linear functions the usual notation is $Ax$ instead of $A(x)$, by example if $A$ is a matrix (it is also a costume to name linear functions with a capital letter but the case of a Fréchet derivative). Take a look at the wikipedia article (or some book) of Fréchet derivative. – Masacroso Nov 18 '17 at 3:43

I would imagine that there is no problem with doing that method! After all, even though its complex analysis, I imagine issues only arise when starting to deal with derivatives with respect to the conjugate...but aside from that, you have a single variable complex function, so its derivative would be the same as in the real case ie. $$\frac{\partial(-f(z))}{\partial (-z)}=-\frac{df(z)}{d(-z)}=-\frac{df(z)}{dz}\frac{dz}{d(-z)}=\frac{df}{dz}$$

• O and just one more simple question with regards to your notation. You dropped the partial differential notation, yet $z$ can be written as $x+iy$. I ask because I always get directed back to Wirtinger derivative operators which are expressed as partials. Can you just make a comment on this. – ThatsRightJack Nov 18 '17 at 1:49
• I suppose it may be a bad habit of mine to drop the partial but as I see it, let $\sigma=f^2+g-x^4+sin(y)$...if you have $f(\sigma)$ its derivative with respect to $\sigma$ is just $\frac{df}{d\sigma}$ – Keith Afas Nov 18 '17 at 1:56