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By admin 数学代写 2022年2月23日

7.11 Derivatives and Limits of Sequences
When you have a function which is a limit of a sequence of functions, when can you say the derivative of the limit function is the limit of the derivatives of the functions in the sequence? The following theorem seems to be one of the best results available. It is based on the mean value theorem. Thus it is understood that the functions are real valued and defined on an interval of $\mathbb{R}$. First of all, recall Definition $6.9 .6$ on Page 111 listed here for convenience.

Definition 7.11.1 Let $\left{f_{n}\right}$ be a sequence of functions defined on $D$. Then $\left{f_{n}\right}$ is said to converge uniformly to $f$ if it converges pointwise to $f$ and for every $\varepsilon>0$ there exists $N$ such that for all $n \geq N,\left|f(x)-f_{n}(x)\right|<\varepsilon$ for all $x \in D$.
To save on notation, denote by $|k| \equiv \sup {|k(\xi)|: \xi \in D}$. Then
because for each $\xi \in D,|k(\xi)+l(\xi)| \leq|k|+|l|$ and taking sup yields 7.18. From the definition of uniform convergence, you see that $f_{n}$ converges uniformly to $f$ is the same as saying $\lim {n \rightarrow \infty}\left|f{n}-f\right|=0$. Now here is the theorem. Note how the mean value theorem is one of the principal parts of the argument.

Theorem 7.11.2 Let $(a, b)$ be a finite open interval and let $f_{k}:(a, b) \rightarrow \mathbb{R}$ be differentiable and suppose there exists $x_{0} \in(a, b)$ such that
$\left{f_{k}\left(x_{0}\right)\right}$ converges,
$\left{f_{k}^{\prime}\right}$ converges uniformly to a function $g$ on $(a, b)$.
Then there exists a function $f$ defined on $(a, b)$ such that
$$
f_{k} \rightarrow f \text { uniformly }
$$
and $f^{\prime}=g .$ Proof: Let $c \in(a, b)$ and define
Proof: Let $c \in(a, b)$ and define $$ g_{n}(x, c) \equiv\left{\begin{array}{l}\frac{f_{n}(x)-f_{n}(c)}{x-c} \text { if } x \neq c \ f_{n}^{\prime}(c) \text { if } x=c\end{array}\right. $$
Also let $|h| \equiv \sup {|h(x)|: x \in(a, b)}$. Thus $h_{k} \rightarrow h$ uniformly means $\left|h_{k}-h\right| \rightarrow 0$.
Claim 1: For each $c, x \rightarrow g_{n}(x, c)$ converges uniformly to a continuous function $h_{c}$, on $(a, b)$ and $h_{c}(c)=g(c)$.

Proof: First note that each $x \rightarrow g_{n}(x, c)$ is continuous. Next consider the claim about uniform convergence. Let $x \neq c$. Then by the mean value theorem applied to the function $x \rightarrow f_{n}(x)-f_{m}(x)$
$$
\begin{aligned}
&\left|g_{n}(x, c)-g_{m}(x, c)\right| \
=&\left|\frac{f_{n}(x)-f_{m}(x)-\left(f_{n}(c)-f_{m}(c)\right)}{x-c}\right| \
=&\left|f_{n}^{\prime}(\xi)-f_{m}^{\prime}(\xi)\right| \leq\left|f_{n}^{\prime}(\xi)-g(\xi)\right|+\left|g(\xi)-f_{m}^{\prime}(\xi)\right| \
\leq &\left|f_{n}^{\prime}-g\right|+\left|f_{m}^{\prime}-g\right|
\end{aligned}
$$
By the assumption that $\left{f_{n}^{\prime}\right}$ converges uniformly to $g$, it follows each of the last two terms converges to 0 as $n, m \rightarrow \infty$. If $x=c$, then
$$
\left|g_{n}(c, c)-g_{m}(c, c)\right|=\left|f_{n}^{\prime}(c)-f_{m}^{\prime}(c)\right| \leq\left|f_{n}^{\prime}-g\right|+\left|f_{m}^{\prime}-g\right|
$$Thus $x \rightarrow g_{n}(x, c)$ is uniformly Cauchy and must converge uniformly to a continuous function $h_{c}$ by Theorem 6.9.7 and Corollary 6.9.10. Also $h_{c}(c)=g(c)$ by the assumption that $f_{k}^{\prime}$ converges uniformly to $g$. This proves the first claim.
Claim 2: $f_{n}$ converges uniformly to a function $f$.
Proof: From the definition of $g_{n}$ in the case where $c=x_{0}$,
Hence from the above claim, $f_{n}$ converges uniformly because it is given that $f_{n}\left(x_{0}\right)$ converges to $f\left(x_{0}\right)$.
Now to complete the proof of the theorem, for $c$ given and $x \neq c$,
$$
\frac{f(x)-f(c)}{x-c}=\lim {n \rightarrow \infty} \frac{f{n}(x)-f_{n}(c)}{x-c}=\lim {n \rightarrow \infty} g{n}(x, c)=h_{c}(x) .
$$
Since $h_{c}$ is continuous,
$$
f^{\prime}(c)=\lim {x \rightarrow c} \frac{f(x)-f(c)}{x-c}=\lim {x \rightarrow c} h_{c}(x)=h_{c}(c)=g(c)
$$