26.9 Derivatives on $\mathbb{R}^{n}$

Proposition 26.9.1.label Let $E$ be a separated topological vector space and $\sigma \subset \mathfrak{B}(\real)$ be a covering ideal, then

  1. (1)

    $\mathcal{R}_{\sigma}(\real; E) = \mathcal{R}_{\mathfrak{B}(\real)}(\real; E)$. Hence, all forms of $\sigma$-differentiability on $\real$ are equivalent.

  2. (2)

    For any $U \subset \real$ open, $f: U \to E$, and $x_{0} \in U$, $f$ is differentiable at $x_{0}$ if and only if

    \[\lim_{t \to 0}\frac{f(x + t) - f(x)}{t}\]

    exists. In which case, the above limit is identified with the derivative of $f$ at $0$.

  3. (3)

    For any $U \subset \real$ open, $f: U \to E$, and $x_{0} \in U$, if $f$ is differentiable at $x_{0}$, then $f$ is continuous at $x_{0}$.

Proof. (1): Let $r \in \mathcal{R}_{\sigma}(\real; E)$. For any $R > 0$ and $U \in \cn_{E}(0)$, there exists $\delta > 0$ such that $t^{-1}r(tR), t^{-1}r(-tR) \in U$ for all $t \in (0, \delta)$. Thus $t^{-1}r(tB(0, R)) \subset U$, and $r \in \mathcal{R}_{\mathfrak{B}(\real)}(\real; E)$.

(2): Suppose that $f$ is differentiable at $x_{0}$, then there exists $r \in \mathcal{R}_{\sigma}$ such that for any $t \in \real$ with $x_{0} + t \in U$,

\begin{align*}f(x_{0} + t) - f(x_{0})&= Df(x_{0})(t) + r(t) \\ \frac{f(x_{0} + t) - f(x_{0})}{t}&= Df(x_{0})(1) + t^{-1}r(t) \\ \lim_{t \to 0}\frac{f(x_{0} + t) - f(x_{0})}{t}&= Df(x_{0})(1)\end{align*}

Now suppose that $v = \lim_{t \to 0}\frac{f(x + t) - f(x)}{t}$ exists. Let $T: \real \to E$ be defined by $t \mapsto tv$, then

\[\lim_{t \to 0}\frac{f(x_{0} + t) - f(x_{0}) - Tt}{t}= \lim_{t \to 0}\frac{f(x_{0} + t) - f(x_{0})}{t}- v = 0\]

and $Df(x_{0}) = T$.$\square$

Proposition 26.9.2.label Let $E$ be a separated locally convex space over $K \in \RC$, $U \subset K$ be open, $Y$ be a Hausdorff space, and $f: U \times Y \to E$. If $f$ is differentiable in the first variable and $\frac{df}{dx}\in C(U \times Y; E)$, then

\[\frac{f(x + h, y) - f(x, y)}{h}\to \frac{df}{dx}(x, y)\]

as $h \to 0$, uniformly on compact sets.

Proof. Let $A \subset U$ and $B \subset Y$ be compact, then by the Mean Value Theorem, for any $(x, y) \in A \times B$ and $h \in \real$ with $x + h$,

\begin{align*}&\frac{f(x + h, y) - f(x, y)}{h}- \frac{df}{dx}(x, y) \\&\in \overline{\text{Conv}}\bracs{\frac{df}{dx}(x + k, y) - \frac{df}{dx}(x, y) \bigg | k \in B_K(0, |h|)}\end{align*}

Let $\eps > 0$ such that $A + B_{K}(0, |\eps|) \subset U$, then since $\frac{df}{dx}\in C(U \times Y; E)$, $\frac{df}{dx}|_{(A + B_K(0, |\eps|)) \times B}$ is uniformly continuous[1]. Since $E$ is locally convex,

\[\frac{f(x + h, y) - f(x, y)}{h}- \frac{df}{dx}(x, y) \to 0\]

uniformly on $A \times B$.$\square$

  1. $K$ is a compact Hausdorff space, which comes with a unique uniform structure. keyboard_return

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