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What do we mean when we say $Df$ is continuous? What is the definition?
In general $J_f $ is $n \times m $ matrix. What is the definition of limit for such a matrix?
How do we check its continuity?

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  • $\begingroup$ Actually, $J_f$ isn't a matrix, it is a map whose values are matrices. What actually is a matrix is $J_f(\mathbf{x})$, for all $\mathbf{x}$ such that $f$ is is differentiable at $\mathbf x$. $\endgroup$ – Git Gud Mar 19 '17 at 11:25
  • $\begingroup$ if $f:E\to F$, then $Df : E\to \mathcal L_c(E,F)$. $\endgroup$ – Gabriel Romon Mar 19 '17 at 11:28
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$Df $ is a map that associate a point $x_0$ of the domain of $f$ with his Jacobian matrix $ J_f(x_0) $. You can say that this map is continuos because we are working in metric spaces. In any case, I suppose you mean $Df(x_0)$ , the Jacobian matrix at a point. So, a matrix is continuos if you consider it as a linear map between, in our case, $\Bbb R^m$ and $\Bbb R^n$.

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The correct way of viewing the derivative of a function in a vector variable is by means of normed spaces. Thus, consider two normed spaces $\mathrm{V}$ and $\mathrm{W},$ an open subset $\mathrm{E}$ of the former and a function $f:\mathrm{E} \to \mathrm{W}.$ The function $f$ is said to be differentiable at a point $v \in \mathrm{E}$ if there exists a continuous linear function $\mathrm{T}:\mathrm{V} \to \mathrm{W}$ such that $f$ and $\mathrm{T}$ are "tangent of first ordre" at $v,$ meaning this that $$\lim_{h \to 0} \dfrac{\|f(v + h) - f(v) - \mathrm{T}v\|}{\|h\|} = 0;$$ the latter limit is also written as $f(v + h) = f(v) + \mathrm{T}v + o(h).$ A consequence follows: if there exists one such $\mathrm{T}$, then it is unique and thus, it is written as $\mathbf{D}f(v)$. If $f$ is differentiable at every $v \in \mathrm{E}$ then, the derived mapping $\mathbf{D}f$ defined on all of $\mathrm{E}$ and such that for $v \in \mathrm{E}$ associates the unique continuous linear function $\mathbf{D}f(v)$ satisfying being tangent to $f$ at $v$ of ordre one is then up on table. Such mapping $\mathbf{D}f$ sends $v$ to $\mathbf{D}f(v)$ which is a continuous linear function $\mathrm{V} \to \mathrm{W}.$ By defining $\mathscr{L}_\mathrm{W}(\mathrm{V})$ as the set of all continuous linear mappings $\mathrm{V} \to \mathrm{W},$ one can then write $\mathbf{D}f:\mathrm{E} \to \mathscr{L}_\mathrm{W}(\mathrm{V})$. The set $\mathrm{X} = \mathscr{L}_\mathrm{W}(\mathrm{V})$ is endowed with the obvious structure of linear space (pointwise sum and pointwise multiplication by scalars in the field; either the complex numeric field or the real numeric field); however, what is not obvious is that $\mathrm{X}$ is also a normed space with a natural norm; said norm is given according to the rule $$\| x \| = \sup_{\|v\| \leq 1} \|x(v)\|, \quad x \in \mathrm{X}.$$ This makes $\mathrm{X}$ a normed vector space and hence, the function $\mathbf{D}f:\mathrm{E} \to \mathrm{X}$ is a function between normed vector spaces. Naturally, being "differentiable with continuity of the first ordre" means that (1) the derived function $\mathbf{D}f$ exists and that (2) said function is continuous.

This definition is quite a nuissance and one can then recall the following useful theorem.

Theorem. For a function $f$ defined on an open subset of $\Bbb R^d$ with values in some normed space $\mathrm{W}$ to be differentiable with continuity of first ordre it is necessary and suficient that all the partial derivatives $\mathbf{D}_k f$ exists (with same domain as $f$) and be continuous (from $\Bbb R^d \to \mathrm{W}$).

The previous theorem is specially useful when $\mathrm{W} = \Bbb R$ as it reduces tho check the standard definition of continuity between $\Bbb R^d$ and $\Bbb R.$

Hope this clarifies the issue.

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