斜率作为二元函数是连续的

hbghlyj

Suppose $U$ is an open subset of $ℝ$ and $f: U→ℝ$ if differentiable on $U$. Show that $f$ is continuously differentiable if and only if the function $s: U×U→ℝ$
$$s(x, y)=\left\{\begin{array}{cc}
\frac{f(x)-f(y)}{x-y}, & x≠y, \\
f'(x), & x=y .
\end{array}\right.
$$is continuous.

Czhang271828

直接看左推右. 选定 $(x_0,y_0)$, 以及 $s(x_0,y_0)=s_0$, 下证明 $s$ 在 $(x_0,y_0)$ 附近连续.

由于 $f$ 连续可导, 从而对任意 $\delta >0$, $\left|\dfrac{f(x)-f(y)}{x-y}\right|$ 在
\[
([x_0-\delta ,x_0+\delta]\times [y_0-\delta ,y_0+\delta ])\setminus\{(x,y)\mid x=y\}
\]
上有一致上界 $\max _{|x-x_0|,|y-y_0|\leq \delta}|f'(x)|<\infty$ 且逐点连续. 由于 $f$ 连续可导, 故 $\{(x,y)\mid x=y\}$ 上的点是可去间断点. 若 $|x_0-y_0|<\delta$, 则存在 $t\in [x_0-\delta ,x_0+\delta]\cap [y_0-\delta ,y_0+\delta ]$ 使得
\begin{align*}
&\lim_{x,y\to t}\dfrac{f(x)-f(y)}{x-y}\\
=\,&\lim_{x,y\to t}\dfrac{f(t)+f'(t+\theta_x)(x-t)-f(t)-f'(t+\eta_y)(y-t)}{x-y}\\
=\,&\lim_{x,y\to t}\dfrac{f'(t+\theta_x)(x-y)+(f'(t+\theta_x)-f'(t+\eta_y))(y-t)}{x-y}\\
=\,&\lim_{x,y\to t}\dfrac{f'(t+\theta_x)(x-y)+L(\theta_x-\eta_y)(y-t)}{x-y}\\
\end{align*}
其中 $\left|\dfrac{\theta_x}{x-t}\right|\leq 1$, $\left|\dfrac{\eta_y}{y-t}\right|\leq 1$, $L$ 为一致连续函数 $f'$ 在 $t$ 邻域内的''一致连续常数''. 从而 $\dfrac{\theta_x-\eta_y}{x-y}$ 在 $x,y\to t$ 时一致有界. 以上极限为 $f'(t)$.

可见 $\dfrac{f(x)-f(y)}{x-y}$ 在可去间断点上的延拓恰好是 $s(x,y)$. 从而 $s(x,y)$ 在 $\mathbb R^2$ 上连续.

hbghlyj

⇐ Suppose $s$ is continuous. $f'$ is the composition of $x\mapsto(x,x)$ and $s$, so $f'$ is continuous, so $f∈𝒞^1$.
⇒ Suppose $f'$ is continuous. $s$ is continuous on $U^2∖\{(x,y):x=y\}$.
$s$ is continuous at $(a,a)∈U^2$ iff $∀ϵ>0∃δ:f(B((a,a),δ))⊂B((a,a),ϵ)$, wlog we use 1-norm in $U^2$. By Mean Value Theorem $∃ξ ∈ ℝ$ between $x,y$ such that $s(x,y)=f'(ξ)$
By uniform continuity of $f',$ $∃δ$ such that $∀z∈B(a,δ):{|f'(z)-f'(a)|}≤ϵ.$
\[\|(x,y)-(a,a)\|≤δ⟹|x-a|+|y-a|≤δ⟹|ξ-a|≤δ⟹{|s(x,y)-s(a,a)|}={|f'(ξ)-f'(a)|}≤ϵ\] 以上有没有问题