Differentia of function-valued mapping

[yifli]

I have difficulty in understanding why the differential $$d\varphi$$ of a function $$\varphi: \eta \rightarrow F(\cdot, \eta)$$ on $$\Re$$ can be written as:

$$[\varphi'(b)](\xi)=\frac{\partial F}{\partial y}(\xi,b)$$ (F is defined on RxR)
according to the following theorem.

the differential $$d\varphi_{b}$$ is a linear mapping, and the skeleton (1x1 matrix) of this linear mapping is its derivative at b, which is $$\varphi'(b)$$. So the differential $$d\varphi_{b}(\eta)=\varphi'(b)\eta$$, why does $$\eta$$ disappear from the above formula?

I think the above formula should be written as:
$$[\varphi'(b)\eta](\xi)=[\frac{\partial F}{\partial y}(\xi,b)](\eta)$$

Related Theorem:
If F is a bounded continuos mapping from an open product set MxN of a normed linear space VxW to a normed linear space X, and if $$dF{^2}{_<\alpha,\beta>}$$ exists and is a bounded uniformly continuous function of $$<\alpha,\beta>$$, then $$\varphi:\eta \rightarrow F(\cdot,\eta)$$ is a differentiable mapping and $$[d{\varphi_\beta}(\eta)](\xi)=dF{^2}_{<\xi,\beta>}(\eta)$$

 

[jvicens]

Thank you for bringing up this question. The formula you have suggested, [\varphi'(b)\eta](\xi)=[\frac{\partial F}{\partial y}(\xi,b)](\eta), is indeed correct and equivalent to the one stated in the theorem. Let me explain why.

First, let's clarify some notation. In the theorem, \varphi_{b} refers to the differential of \varphi at the point b, which is a linear mapping from the domain of \varphi (in this case, \Re) to the range of \varphi (in this case, F(\cdot, b)). This mapping is defined as d\varphi_{b}(\eta) = \varphi'(b)\eta, where \varphi'(b) is the derivative of \varphi at b and \eta is an element of the domain of \varphi.

Now, let's look at the formula in question, [\varphi'(b)](\xi)=\frac{\partial F}{\partial y}(\xi,b). The left side of the equation is the value of the linear mapping \varphi'(b) at the point \xi, which is a function of \xi. On the right side, we have the partial derivative of the function F with respect to its second argument y, evaluated at the point (\xi, b). This partial derivative is also a function of \xi.

So, we can rewrite this formula as [\varphi'(b)](\xi)=\frac{\partial F(\xi, b)}{\partial y}. Now, if we plug in the definition of \varphi'(b), we get [\varphi'(b)](\xi) = d\varphi_{b}(\xi). This means that the left side of the equation is the value of the differential of \varphi at the point b, evaluated at the point \xi. On the right side, we have the partial derivative of the function F with respect to its second argument y, evaluated at the point (\xi, b). This is exactly what we have in the formula you suggested, [\frac{\partial F}{\partial y}(\xi,b)](\eta).

In summary, the formula you suggested is indeed equivalent to the one stated in the theorem. The reason why \eta disappears in the original formula is because it is already contained in the definition of the differential d\var