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Total Differentiability (Real Scalar Fields)#

TODO

Theorem: Total Differentiability via Bachmann-Landau

Let \(f: \mathcal{D} \subseteq \mathbb{R}^n \to \mathbb{R}\) be a real scalar field and let \(\boldsymbol{p} \in \operatorname{int} \mathcal{D}\).

Then \(f\) is totally differentiable at \(\boldsymbol{p}\) if and only if there exists some real vector \(\boldsymbol{v}\) such that \(f(\boldsymbol{p} + \boldsymbol{h}) - f(\boldsymbol{p}) - \boldsymbol{v}^{\mathsf{T}}\boldsymbol{h}\) is little o of \(||\boldsymbol{h}||\) for \(\boldsymbol{h} \to \boldsymbol{0}\):

\[f(\boldsymbol{p} + \boldsymbol{h}) - f(\boldsymbol{p}) - \boldsymbol{v}^{\mathsf{T}}\boldsymbol{h} = o(||\boldsymbol{h}||) \qquad \text{for} \qquad \boldsymbol{h} \to \boldsymbol{0}\]

In this case, \(\boldsymbol{v}\) is the gradient \(\nabla f(\boldsymbol{p})\).

Proof

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Theorem: Continuous Partial Differentiability \(\implies\) Total Differentiability

Let \(f: \mathcal{D} \subseteq \mathbb{R}^n \to \mathbb{R}\) be a real scalar field and let \(\boldsymbol{p}\) be an interior point.

If \(f\) is partially differentiable on an open neighborhood of \(\boldsymbol{p}\) and is continuously partially differentiable at \(\boldsymbol{p}\), then \(f\) is totally differentiable at \(\boldsymbol{p}\).

Proof

TODO

Theorem: Mean Value Theorem via Total Differential (Real Scalar Fields)

Let \(f: \mathcal{D} \subseteq \mathbb{R}^n \to \mathbb{R}\) be a real scalar field and let \(\boldsymbol{a}, \boldsymbol{b} \in \mathcal{D}\) such that \(L = \{\boldsymbol{a} + t(\boldsymbol{b} - \boldsymbol{a}) \mid t \in [0,1]\} \subseteq \mathcal{D}\).

If \(f\) is continuous on \(L\) and totally differentiable on \(\operatorname{int} L\), then there exists some \(\boldsymbol{\xi} \in \operatorname{int} L\) such that \(f(\boldsymbol{b}) - f(\boldsymbol{a})\) is equal to \(f\)'s total differential at \(\boldsymbol{\xi}\) applied to \(\boldsymbol{b} - \boldsymbol{a}\):

\[f(\boldsymbol{b}) - f(\boldsymbol{a}) = \mathrm{d}f_{\boldsymbol{\xi}}(\boldsymbol{b} - \boldsymbol{a})\]
Proof

We define a real function \(g: [0,1] \to \mathbb{R}\) as follows:

\[g(t) = f(\boldsymbol{a} + t(\boldsymbol{b} - \boldsymbol{a}))\]

Since \(f\) is continuous on \(L\) and the curve \(\gamma(t) = \boldsymbol{a} + t(\boldsymbol{b} - \boldsymbol{a})\) is continuous on \([0,1]\), we know that \(g\) must be continuous on \([0,1]\). Since \(f\) is totally differentiable on \(\operatorname{int} L\) and \(\gamma(t)\) is differentiable on \((0,1)\), we know that \(g\) must be differentiable on \((0,1)\) with the following "total differential" due to the chain rule:

\[\mathrm{d}_t g = \mathrm{d}_{\gamma(t)}f \circ \mathrm{d}_t \gamma\]

The total differentials are functions, specifically, \(\mathrm{d}_t g: \mathbb{R} \to \mathbb{R}\), \(\mathrm{d}_{\gamma(t)}f: \mathbb{R}^n \to \mathbb{R}\) and \(\mathrm{d}_t \gamma: \mathbb{R} \to \mathbb{R}^n\). Therefore, we have

\[\mathrm{d}_t g(h) = \mathrm{d}_{\gamma(t)}f (\mathrm{d}_t \gamma(h))\]

for all \(h \in \mathbb{R}\).

The left-hand side can be expressed using \(g\)'s derivative as \(\mathrm{d}_t g(h) = g'(t) h\). The total differential \(\mathrm{d}_t \gamma\) can be express using \(\gamma\)'s derivative as \(\mathrm{d}_t \gamma(h) = \gamma'(t) h\). Therefore:

\[g'(t) h = \mathrm{d}_{\gamma(t)}f (\gamma'(t)h)\]

Since \(\gamma'(t) = \boldsymbol{b} - \boldsymbol{a}\), we have:

\[g'(t) h = \mathrm{d}_{\gamma(t)}f ((\boldsymbol{b} - \boldsymbol{a})h)\]

This must hold for \(h = 1\) and we get:

\[g'(t) = \mathrm{d}_{\gamma(t)}f (\boldsymbol{b} - \boldsymbol{a})\]

Since \(g\) is continuous on \([0,1]\) and differentiable on \((0,1)\), we can use the mean value theorem for \(g\). Specifically, there exists some \(\xi \in (0,1)\) such that

\[g(1) - g(0) = g'(\xi)(1 - 0) = g'(\xi)\]
\[g(1) - g(0) = g'(\xi)\]

Evaluating \(g\) at \(1\) and \(0\) gives us \(g(1) = f(\boldsymbol{b})\) and \(g(0) = f(\boldsymbol{a})\). Evaluating \(g'(\xi)\) gives us \(\mathrm{d}_{\gamma(\xi)}f (\boldsymbol{b} - \boldsymbol{a})\) We thus have:

\[f(\boldsymbol{b}) - f(\boldsymbol{a}) = \mathrm{d}_{\gamma(\xi)}f (\boldsymbol{b} - \boldsymbol{a}) = \mathrm{d}f_{\boldsymbol{\xi}} (\boldsymbol{b} - \boldsymbol{a})\]

with \(\boldsymbol{\xi} = \gamma(\xi)\). Finally, since \(\xi \in (0,1)\) and since \(\gamma\) is injective, we know that \(\boldsymbol{\xi} \in \operatorname{int} L\).