Fréchet differentiation as a uniform linear approximation

Try

For f(x,y)=xy near (a,b), a displacement (h,k) changes the output by bh+ak+hk. The first two terms form a linear map of the displacement. The remaining product is at most (h²+k²)/2 in magnitude, so divided by the displacement length it tends to zero. This exact expansion explains total differentiability without relying only on coordinate slopes.

Predict

Predict why separate directional derivatives might be insufficient. A derivative map must be linear in direction, and its error must shrink relative to displacement uniformly across directions. Taking a limit along each fixed line can miss both requirements. Different line limits may exist but fail to assemble into a linear map; even compatible directional data can require further uniform control.

Experiment

Compare the objects.

Object	Input	Required structure
Partial derivative	One coordinate direction	A scalar limit
Directional derivative	One fixed direction	A one-variable limit
Fréchet derivative	Every small displacement	One linear map with relative error tending to zero

For the product function, evaluate the linear map on two vectors and their sum to check linearity. Then divide the exact remainder by √(h²+k²) and bound it without choosing a preferred path.

Understand

Let F map an open subset of R^n into R^m. It is Fréchet differentiable at a if a linear map A satisfies F(a+h)=F(a)+Ah+r(h), with ||r(h)||/||h||→0 as h→0 through nonzero displacements. The derivative is A=DF(a). The open-domain hypothesis ensures all sufficiently small directions are admissible. Equivalent finite-dimensional norms give the same differentiability property, although estimates use different constants.

The derivative is unique. If both A and B work, subtracting the expansions gives (A−B)h=o(||h||). For a fixed vector v, put h=tv and divide by |t|. The quantity ||(A−B)v|| must be zero. Since this holds for every v, A=B. Differentiability also implies continuity because the linear term is bounded by ||A||||h|| and the remainder is smaller than first order.

Apply

For a counterexample, define q(x,y)=x³/(x²+y²) away from zero and q(0,0)=0. Along h=tv with v≠0, q(tv)/t=v₁³/(v₁²+v₂²), so every two-sided directional derivative exists. The axis values would force A(v)=v₁, but direction (1,1) gives 1/2 rather than 1. Thus no linear derivative can match all directions.

Continuous first partials on a neighborhood are sufficient. Decompose an increment into coordinate moves and apply one-variable MVT to each component of F along each move. Subtract the partial derivatives at a. Their continuity makes all finitely many coefficient differences small throughout a small neighborhood, and the sum of coordinate magnitudes is bounded by √n||h||₂. Hence the remainder is o(||h||).

Continue

Optional depth distinguishes existence of a derivative at one point from continuity of the derivative as a function of the point. The latter is the C¹ condition needed by the local inverse theorem later. A function may be differentiable everywhere yet have a discontinuous derivative.

Save a direct remainder proof and a directional counterexample. State the dimensions of the derivative map and explain why its matrix is a representation of a linear transformation, not the definition of differentiability. Next, composing such maps requires tracking the inner displacement’s size so that both remainder terms remain negligible.