(1)

\(C^{1}\)級\(\Rightarrow\)全微分可能

2変数関数\(f\left(x,y\right)\)で示す。
\begin{align*} & \left|\lim_{\left(h_{1},h_{2}\right)\rightarrow\left(0,0\right)}\frac{f\left(a_{1}+h_{1},a_{2}+h_{2}\right)-f\left(a_{1},a_{2}\right)-\left(A_{1}h_{1}+A_{2}h_{2}\right)}{\sqrt{h_{1}^{2}+h_{2}^{2}}}\right|\\ & =\lim_{\left(h_{1},h_{2}\right)\rightarrow\left(0,0\right)}\left|\frac{f\left(a_{1}+h_{1},a_{2}+h_{2}\right)-f\left(a_{1},a_{2}+h_{2}\right)+f\left(a_{1},a_{2}+h_{2}\right)-f\left(a_{1},a_{2}\right)-\left(A_{1}h_{1}+A_{2}h_{2}\right)}{\sqrt{h_{1}^{2}+h_{2}^{2}}}\right|\\ & =\lim_{\left(h_{1},h_{2}\right)\rightarrow\left(0,0\right)}\left|\frac{f_{x}\left(a_{1}+\theta_{1}h_{1},a_{2}+h_{2}\right)h_{1}+f_{y}\left(a_{1},a_{2}+\theta_{2}h_{2}\right)h_{2}-\left(A_{1}h_{1}+A_{2}h_{2}\right)}{\sqrt{h_{1}^{2}+h_{2}^{2}}}\right|\cmt{\text{平均値の定理:}0<\theta_{1}<1,0<\theta_{2}<1}\\ & =\lim_{\left(h_{1},h_{2}\right)\rightarrow\left(0,0\right)}\left|\frac{\left\{ f_{x}\left(a_{1}+\theta_{1}h_{1},a_{2}+h_{2}\right)-A_{1}\right\} h_{1}+\left\{ f_{y}\left(a_{1},a_{2}+\theta_{2}h_{2}\right)-A_{2}\right\} h_{2}}{\sqrt{h_{1}^{2}+h_{2}^{2}}}\right| \end{align*} となるので、
\begin{align*} A_{1} & =\lim_{\left(h_{1},h_{2}\right)\rightarrow\left(0,0\right)}f_{x}\left(a_{1}+\theta_{1}h_{1},a_{2}+h_{2}\right)\\ & =f_{x}\left(a_{1},a_{2}\right) \end{align*} \begin{align*} A_{2} & =\lim_{\left(h_{1},h_{2}\right)\rightarrow\left(0,0\right)}f_{y}\left(a_{1},a_{2}+\theta_{2}h_{2}\right)-A_{2}\\ & =f_{y}\left(a_{1},a_{2}\right) \end{align*} ととれば、
\begin{align*} & \left|\lim_{\left(h_{1},h_{2}\right)\rightarrow\left(0,0\right)}\frac{f\left(a_{1}+h_{1},a_{2}+h_{2}\right)-f\left(a_{1},a_{2}\right)-\left(A_{1}h_{1}+A_{2}h_{2}\right)}{\sqrt{h_{1}^{2}+h_{2}^{2}}}\right|\\ & =\lim_{\left(h_{1},h_{2}\right)\rightarrow\left(0,0\right)}\left|\frac{\left\{ f_{x}\left(a_{1}+\theta_{1}h_{1},a_{2}+h_{2}\right)-f_{x}\left(a_{1},a_{2}\right)\right\} h_{1}+\left\{ f_{y}\left(a_{1},a_{2}+\theta_{2}h_{2}\right)-f_{y}\left(a_{1},a_{2}\right)\right\} h_{2}}{\sqrt{h_{1}^{2}+h_{2}^{2}}}\right|\\ & =\lim_{\left(h_{1},h_{2}\right)\rightarrow\left(0,0\right)}\left|\frac{\left\{ f_{x}\left(a_{1}+\theta_{1}h_{1},a_{2}+h_{2}\right)-f_{x}\left(a_{1},a_{2}\right)\right\} h_{1}}{\sqrt{h_{1}^{2}+h_{2}^{2}}}+\frac{\left\{ f_{y}\left(a_{1},a_{2}+\theta_{2}h_{2}\right)-f_{y}\left(a_{1},a_{2}\right)\right\} h_{2}}{\sqrt{h_{1}^{2}+h_{2}^{2}}}\right|\\ & \leq\lim_{\left(h_{1},h_{2}\right)\rightarrow\left(0,0\right)}\left(\left|\frac{\left\{ f_{x}\left(a_{1}+\theta_{1}h_{1},a_{2}+h_{2}\right)-f_{x}\left(a_{1},a_{2}\right)\right\} h_{1}}{\sqrt{h_{1}^{2}+h_{2}^{2}}}\right|+\left|\frac{\left\{ f_{y}\left(a_{1},a_{2}+\theta_{2}h_{2}\right)-f_{y}\left(a_{1},a_{2}\right)\right\} h_{2}}{\sqrt{h_{1}^{2}+h_{2}^{2}}}\right|\right)\\ & \leq\lim_{\left(h_{1},h_{2}\right)\rightarrow\left(0,0\right)}\left(\left|\frac{\left\{ f_{x}\left(a_{1}+\theta_{1}h_{1},a_{2}+h_{2}\right)-f_{x}\left(a_{1},a_{2}\right)\right\} h_{1}}{h_{1}}\right|+\left|\frac{\left\{ f_{y}\left(a_{1},a_{2}+\theta_{2}h_{2}\right)-f_{y}\left(a_{1},a_{2}\right)\right\} h_{2}}{h_{2}}\right|\right)\\ & =\lim_{\left(h_{1},h_{2}\right)\rightarrow\left(0,0\right)}\left(\left|f_{x}\left(a_{1}+\theta_{1}h_{1},a_{2}+h_{2}\right)-f_{x}\left(a_{1},a_{2}\right)\right|+\left|f_{y}\left(a_{1},a_{2}+\theta_{2}h_{2}\right)-f_{y}\left(a_{1},a_{2}\right)\right|\right)\\ & =0 \end{align*} となるので、
\[ \lim_{\left(h_{1},h_{2}\right)\rightarrow\left(0,0\right)}\frac{f\left(a_{1}+h_{1},a_{2}+h_{2}\right)-f\left(a_{1},a_{2}\right)-\left(A_{1}h_{1}+A_{2}h_{2}\right)}{\sqrt{h_{1}^{2}+h_{2}^{2}}}=0 \] となり全微分可能となる。
従って題意は成り立つ。

逆は一般的に成り立たない

反例で示す。
\[ f\left(x,y\right)=\begin{cases} xy\sin\frac{1}{\sqrt{x^{2}+y^{2}}} & \left(x,y\right)\ne\left(0,0\right)\\ 0 & \left(x,y\right)=\left(0,0\right) \end{cases} \] \begin{align*} f_{x}\left(0,0\right) & =\lim_{h\rightarrow0}\frac{f\left(h,0\right)-f\left(0,0\right)}{h}\\ & =0 \end{align*} \begin{align*} f_{y}\left(0,0\right) & =\lim_{h\rightarrow0}\frac{f\left(0,h\right)-f\left(0,0\right)}{h}\\ & =0 \end{align*} \(h_{1}=r\cos\theta,h_{2}=r\sin\theta\)とおけば、
\begin{align*} \left|\lim_{\left(h_{1},h_{2}\right)\rightarrow\left(0,0\right)}\frac{f\left(0+h_{1},0+h_{2}\right)-f\left(0,0\right)-\left(0h_{1}+0h_{2}\right)}{\sqrt{h_{1}^{2}+h_{2}^{2}}}\right| & =\lim_{\left(h_{1},h_{2}\right)\rightarrow\left(0,0\right)}\left|\frac{f\left(h_{1},h_{2}\right)}{\sqrt{h_{1}^{2}+h_{2}^{2}}}\right|\\ & =\lim_{\left(h_{1},h_{2}\right)\rightarrow\left(0,0\right)}\left|\frac{h_{1}h_{2}}{\sqrt{h_{1}^{2}+h_{2}^{2}}}\sin\frac{1}{\sqrt{h_{1}^{2}+h_{2}^{2}}}\right|\\ & =\lim_{r\rightarrow0^{+}}\left|\frac{r^{2}\cos\theta\sin\theta}{r}\sin\frac{1}{r}\right|\\ & =\lim_{r\rightarrow0^{+}}\left|r\cos\theta\sin\theta\sin\frac{1}{r}\right|\\ & \leq\lim_{r\rightarrow0^{+}}\left|r\cos\theta\sin\theta\right|\\ & =0 \end{align*} となるので、
\[ \lim_{\left(h_{1},h_{2}\right)\rightarrow\left(0,0\right)}\frac{f\left(0+h_{1},0+h_{2}\right)-f\left(0,0\right)-\left(0h_{1}+0h_{2}\right)}{\sqrt{h_{1}^{2}+h_{2}^{2}}}=0 \] となり原点\(\left(0,0\right)\)で全微分可能となる。
次に原点\(\left(0,0\right)\)で\(C^{1}\)級でないことを示す。
\begin{align*} f_{x}\left(x,y\right) & =y\sin\frac{1}{\sqrt{x^{2}+y^{2}}}-\frac{1}{2}\cdot\frac{2x^{2}y}{\left(x^{2}+y^{2}\right)^{\frac{3}{2}}}\cos\frac{1}{\sqrt{x^{2}+y^{2}}}\\ & =y\sin\frac{1}{\sqrt{x^{2}+y^{2}}}-\frac{x^{2}y}{\left(x^{2}+y^{2}\right)^{\frac{3}{2}}}\cos\frac{1}{\sqrt{x^{2}+y^{2}}} \end{align*} となり、\(0<x=y\) として\(x\rightarrow0^{+}\)とすると、
\begin{align*} \lim_{x\rightarrow0^{+}}f_{x}\left(x,x\right) & =\lim_{x\rightarrow0^{+}}\left(y\sin\frac{1}{\sqrt{2x^{2}}}-\frac{x^{3}}{\left(2x^{2}\right)^{\frac{3}{2}}}\cos\frac{1}{\sqrt{2x^{2}}}\right)\\ & =\lim_{x\rightarrow0^{+}}\left(y\sin\frac{1}{\sqrt{2}x}-\frac{1}{2\sqrt{2}}\cos\frac{1}{\sqrt{2}x}\right) \end{align*} となり振動するので原点\(\left(0,0\right)\)では不連続となり、\(C^{1}\)級ではない。
従って原点\(\left(0,0\right)\)では全微分可能であるが\(C^{1}\)級ではない。

全微分可能\(\Rightarrow\)偏微分可能

\(a=\left(a_{1},a_{2},\cdots,a_{n}\right)\)で全微分可能なので
\[ \lim_{\left(h_{1},h_{2}\cdots,h_{n}\right)\rightarrow\left(0,0,\cdots,0\right)}\frac{f\left(a_{1}+h_{1},a_{2}+h_{2},\cdots,a_{n}+h_{n}\right)-f\left(a_{1},a_{2},\cdots,a_{n}\right)-\left(A_{1}h_{1}+A_{2}h_{2}+\cdots+A_{n}h_{n}\right)}{\sqrt{h_{1}^{2}+h_{2}^{2}+\cdots+h_{n}^{2}}}=0 \] とある定数\(A_{1},A_{2},\cdots,A_{n}\)が存在する。
このとき、\(j\ne i\rightarrow h_{j}=0\)として\(h_{i}\rightarrow0\)とすれば、
\[ \lim_{h_{i}\rightarrow0}\frac{f\left(a_{1},a_{2},\cdots a_{i}+h_{i},\cdots,a_{n}\right)-f\left(a_{1},a_{2},\cdots,a_{n}\right)-\left(A_{i}h_{i}\right)}{\left|h_{i}\right|}=0 \] となるので、
\begin{align*} A_{i} & =\lim_{h_{i}\rightarrow0}\frac{f\left(a_{1},a_{2},\cdots a_{i}+h_{i},\cdots,a_{n}\right)-f\left(a_{1},a_{2},\cdots,a_{n}\right)}{h_{i}}\\ & =\frac{\partial f\left(a\right)}{\partial x_{i}} \end{align*} となるので偏微分可能である。

逆は一般的に成り立たない

反例で示す。
\[ f\left(x,y\right)=\begin{cases} 0 & xy=0\\ 1 & xy\ne0 \end{cases} \] は
\begin{align*} \frac{\partial f\left(0,0\right)}{\partial x} & =\lim_{h\rightarrow0}\frac{f\left(0+h,0\right)-f\left(0,0\right)}{h}\\ & =\lim_{h\rightarrow0}\frac{0}{h}\\ & =0 \end{align*} \begin{align*} \frac{\partial f\left(0,0\right)}{\partial y} & =\lim_{h\rightarrow0}\frac{f\left(0,0+h\right)-f\left(0,0\right)}{h}\\ & =\lim_{h\rightarrow0}\frac{0}{h}\\ & =0 \end{align*} であるので偏微分可能である。
もし、全微分可能であれば、
\[ \lim_{\left(h_{1},h_{2}\right)\rightarrow\left(0,0\right)}\frac{f\left(0+h_{1},0+h_{2}\right)-f\left(0,0\right)-\left(A_{1}h_{1}+A_{2}h_{2}\right)}{\sqrt{h_{1}^{2}+h_{2}^{2}}}=0 \] となる定数\(A_{1},A_{2}\)が存在する。
しかし\(h_{1}=h_{2}=\epsilon\)として\(\epsilon\rightarrow0\)とすれば、
\begin{align*} \lim_{\epsilon\rightarrow0}\frac{f\left(0+\epsilon,0+\epsilon\right)-f\left(0,0\right)-\left(A_{1}+h_{2}\right)\epsilon}{\sqrt{2\epsilon^{2}}} & =\lim_{\epsilon\rightarrow0}\frac{1-0-\left(A_{1}+A_{2}\right)\epsilon}{\sqrt{2}\left|\epsilon\right|}\\ & =\lim_{\epsilon\rightarrow0}\left(\frac{1}{\sqrt{2}\left|\epsilon\right|}-\frac{\left(A_{1}+A_{2}\right)}{\sqrt{2}\sgn\left(\epsilon\right)}\right) \end{align*} となり第1項は発散するので定数\(A_{1},A_{2}\)は存在しない。
従って、全微分可能可能ではないので、偏微分可能であっても全微分可能であるとは限らない。

(2)

全微分可能\(\Rightarrow\)連続

全微分可能であるので
\[ \lim_{\left(h_{1},h_{2}\cdots,h_{n}\right)\rightarrow\left(0,0,\cdots,0\right)}\frac{f\left(a_{1}+h_{1},a_{2}+h_{2},\cdots,a_{n}+h_{n}\right)-f\left(a_{1},a_{2},\cdots,a_{n}\right)-\left(A_{1}h_{1}+A_{2}h_{2}+\cdots+A_{n}h_{n}\right)}{\sqrt{h_{1}^{2}+h_{2}^{2}+\cdots+h_{n}^{2}}}=0 \] となる定数\(A_{1},A_{2}\)が存在する。
このとき、
\begin{align*} & \lim_{\left(h_{1},h_{2}\cdots,h_{n}\right)\rightarrow\left(0,0,\cdots,0\right)}\left\{ f\left(a_{1}+h_{1},a_{2}+h_{2},\cdots,a_{n}+h_{n}\right)-f\left(a_{1},a_{2},\cdots,a_{n}\right)\right\} \\ & =\lim_{\left(h_{1},h_{2}\cdots,h_{n}\right)\rightarrow\left(0,0,\cdots,0\right)}\left\{ f\left(a_{1}+h_{1},a_{2}+h_{2},\cdots,a_{n}+h_{n}\right)-f\left(a_{1},a_{2},\cdots,a_{n}\right)-\left(A_{1}h_{1}+A_{2}h_{2}+\cdots+A_{n}h_{n}\right)+\left(A_{1}h_{1}+A_{2}h_{2}+\cdots+A_{n}h_{n}\right)\right\} \\ & =\lim_{\left(h_{1},h_{2}\cdots,h_{n}\right)\rightarrow\left(0,0,\cdots,0\right)}\left\{ \frac{f\left(a_{1}+h_{1},a_{2}+h_{2},\cdots,a_{n}+h_{n}\right)-f\left(a_{1},a_{2},\cdots,a_{n}\right)-\left(A_{1}h_{1}+A_{2}h_{2}+\cdots+A_{n}h_{n}\right)}{\sqrt{h_{1}^{2}+h_{2}^{2}+\cdots+h_{n}^{2}}}\cdot\sqrt{h_{1}^{2}+h_{2}^{2}+\cdots+h_{n}^{2}}\right\} \\ & =0 \end{align*} となるので連続となる。

逆は一般的に成り立たない。

反例で示す。
1変数関数\(f\left(x\right)=\left|x\right|\)は連続であるが\(x=0\)で全微分可能でない。
従って逆は一般的に成り立たない。