Technically the introductory part on derivatives on Rn is incorrect, in two different ways.
Firstly the derivative of a map f:Rn→R is a map f′:Rn→L(Rn,R) , that assigns to every point x a linear map sending direction y to a real value (namely the partial derivative of f at x in direction y). Thankfully the space of linear maps from Rn to R is isometrically isomorphic to Rn through the inner product, recovering the expression you gave. Similarly the derivative of a map f:Rn→Rm is a map f′:Rn→L(Rn,Rm) .
Secondly technically the domain of any derivative like the one above is not the vector space we are working with, but the set of directions at point x. This notion is formalised in Manifold theory and called the tangent space. Thankfully for any finite-dimenional vector space the tangent space at any point is canonically isomorphic to the vector space itself (any vector is a direction, that’s what they were invented for). In infinite dimensions this still holds just fine except for the small detail that the notions of manifold and tangent space don’t exist there. The same distinction is necessary in the range. So truly, formally, the derivative of a map f:Rn→R is a map f′:TRn→TR , with TRn=⋃x∈RnTxRn≡Rn×Rn and similarly TR=⋃y∈RTyR≡R×R , with the condition that f′ is simply f on the first coordinate. This coincides with the map above: for every x∈Rn we get a linear map f′(x):TxRn→Tf(x)R.
The above may seem very confusing for n=1, since I claim that the derivative in that case is a map f′:R→L(R,R) instead of simply a real-valued function. This is resolved by noting that each linear map from R to R can be represented with a number, similar to the top bullet point above (the inner product on R is just multiplication). I think lecturers are quite justified in not exploring the details of this when first introducing derivatives or partial derivatives, but unfortunately in possibly infinite-dimensional abstract vector spaces the distinctions are necessary, if only to avoid type errors.
In the definition of the partial derivative of M at f with respect to g (so with a range inside a vector space Y) we do not take the norm or absolute value of that expression, it should be the straight up limit limλ→0M(f+λg)−M(f)λ. The claim that the limit exists does depend on the topology of Y and therefore on the norm, though.
Also there are a lot of discontinuous linear maps out there. A textbook example is considering the vector space P[0,1] of polynomials interpreted as functions on the closed interval [0,1], equipped with supremum norm. The derivative map ddx:P[0,1]→P[0,1] is not continuous, and you can verify this directly by searching for a sequence of functions that converges to 0 whose image does not converge to 0.
Very nice! Two mistakes though:
Technically the introductory part on derivatives on Rn is incorrect, in two different ways.
Firstly the derivative of a map f:Rn→R is a map f′:Rn→L(Rn,R) , that assigns to every point x a linear map sending direction y to a real value (namely the partial derivative of f at x in direction y). Thankfully the space of linear maps from Rn to R is isometrically isomorphic to Rn through the inner product, recovering the expression you gave. Similarly the derivative of a map f:Rn→Rm is a map f′:Rn→L(Rn,Rm) .
Secondly technically the domain of any derivative like the one above is not the vector space we are working with, but the set of directions at point x. This notion is formalised in Manifold theory and called the tangent space. Thankfully for any finite-dimenional vector space the tangent space at any point is canonically isomorphic to the vector space itself (any vector is a direction, that’s what they were invented for). In infinite dimensions this still holds just fine except for the small detail that the notions of manifold and tangent space don’t exist there. The same distinction is necessary in the range. So truly, formally, the derivative of a map f:Rn→R is a map f′:TRn→TR , with TRn=⋃x∈RnTxRn≡Rn×Rn and similarly TR=⋃y∈RTyR≡R×R , with the condition that f′ is simply f on the first coordinate. This coincides with the map above: for every x∈Rn we get a linear map f′(x):TxRn→Tf(x)R.
The above may seem very confusing for n=1, since I claim that the derivative in that case is a map f′:R→L(R,R) instead of simply a real-valued function. This is resolved by noting that each linear map from R to R can be represented with a number, similar to the top bullet point above (the inner product on R is just multiplication). I think lecturers are quite justified in not exploring the details of this when first introducing derivatives or partial derivatives, but unfortunately in possibly infinite-dimensional abstract vector spaces the distinctions are necessary, if only to avoid type errors.
In the definition of the partial derivative of M at f with respect to g (so with a range inside a vector space Y) we do not take the norm or absolute value of that expression, it should be the straight up limit limλ→0M(f+λg)−M(f)λ. The claim that the limit exists does depend on the topology of Y and therefore on the norm, though.
Also there are a lot of discontinuous linear maps out there. A textbook example is considering the vector space P[0,1] of polynomials interpreted as functions on the closed interval [0,1], equipped with supremum norm. The derivative map ddx:P[0,1]→P[0,1] is not continuous, and you can verify this directly by searching for a sequence of functions that converges to 0 whose image does not converge to 0.