
 Let $X$ and $Y$ be finitedimensional inner product spaces and let $ℒ(X, Y)$ denote the space of linear maps from $X$ to $Y$. Let $f: U → Y$ be a function defined on an open set $U$ of $X$. We say that $f$ has property C at a point $a ∈ U$ if there exists a function $M: U → ℒ(X, Y)$, written $x ↦ M_x ∈ ℒ(X, Y)$, such that $M$ is continuous at $a$ and
$$
f(x)=f(a)+M_x(xa) \text { for all } x ∈ U .
$$
 Show that if $f$ has property C at $a ∈ U$ then $f$ is differentiable at $a$ with $D f_a=M_a$.
 Show conversely, that if $f$ is differentiable at $a$, then $f$ has property C at $a$.
[Hint: Consider the linear functional on $U$ given by $y ↦\left\xa\right\^{1}⟨xa, y⟩$ where $⟨u, v⟩$ denotes the inner product of vectors $u, v ∈ X$. ]
 Let $f: U → Y$ and $a ∈ U$ be as in part (a), and suppose in addition that $V$ is an open subset of $Y$ containing $f(U)$ and $g: V → Z$ is a function on $V$ taking values in a finitedimensional inner product space $Z$. Show, using part (a), that if $f$ is differentiable at $a$ and $g$ is differentiable at $b=f(a)$, then $g ∘ f: U → Z$ is differentiable at $a$ with derivative $D g_b ∘ D f_a$.
[Standard properties of continuous functions may be used without proof.]

 Let $W$ be a subset of $ℝ^n$, and let $p ∈ W$. Define the tangent space $T_p W$ to $W$ at $p$.
 Let $f: ℝ^3 → ℝ$ be the function $f(x, y, z)=x^2+y^2z^21$, and let
\begin{aligned}
S & =\left\{(x, y, z) ∈ ℝ^3: f(x, y, z)=0\right\} \\
R & =\left\{\left(x, y, z, x', y', z'\right) ∈ ℝ^3 × ℝ^3:(x, y, z) ∈ S,\left(x', y', z'\right) ∈ T_{(x, y, z)} S\right\} .
\end{aligned}
Show that $S$ is a submanifold of $ℝ^3$, deduce that $R$ is a submanifold of $ℝ^6=ℝ^3 × ℝ^3$. [You may use any standard results about tangent spaces provided you state them clearly.]
 Suppose that $f$ has property C, and let $M: U → ℒ(X, Y)$ be a function satisfying the required properties. We claim that $f$ is differentiable at $a$ and $D f_a=M_a=\lim _{x → a} M_x$. To see this, it suffices to show that
$$
\frac{\left\f(x)f(a)M_a(xa)\right\}{\xa\} → 0
$$
as $x → a$. But by definition we have $f(x)f(a)=M_x(xa)$, and
$$
\frac{\left\f(x)f(a)M_a(xa)\right\}{\xa\}=\frac{\left\M_x(xa)M_a(xa)\right\}{\xa\}=\left\\left(M_xM_a\right)\left(xa\over\xa\\right)\right\
$$
and the final expression is by definition bounded by $\left\M_xM_a\right\_{∞}$ which tends to zero as $x → a$ by the continuity of $M$.
 Now suppose $f$ is differentiable at $a$. Replacing $f$ with $f(x)f(a)D f_a$, we may assume $f(a)=0$ and $D f_a=0$, so that $f(x)$ is $o(xa)$. We must find a function $M: U → ℒ(X, Y)$ such that $f(x)=M_x(xa)$, that is, we must write $f(x)$ in the form $M_x(xa)$.
The function $M$ is not unique, but one natural possibility is to set $M_a=0$ and, for $x ≠ a$,
$$
M_x(y)=\frac{⟨ xa, y⟩}{\xa\^2} ⋅ f(x)
$$
It is immediate that $M_x(xa)=f(x)$, and, using CauchySchwarz, we see that
$$
\left\M_x(y)\right\ ⩽ \frac{\y\}{\xa\} ⋅\left\f(x)\right\ ⟹\left\M_x\right\_{∞} ⩽ \frac{\f(x)\}{\xa\}
$$
and since $f(x)$ is $o(xa)$ this immediately implies that $M_x$ is continuous at $xa$ as required.
 [Similar but easier, assuming (a), than the standard proof of the Chain Rule]
Using a) we may write $f(x)=f(a)+M_x^f(xa)$ and similarly $g(y)=g(b)+M_y^g(yb)$, where $M^f: U → ℒ(X, Y), M^g: V → ℒ(Y, Z)$. But then $g ∘ f(x)=g(f(a))+M_{f(x)}^g ∘ M_x^f(xa)$. Now $x ↦ M_{f(x)}^g ∘ M_x^f(xa)$ is function from $U$ to $ℒ(X, Z)$, and since the composition of continuous functions is continuous, it is continuous at $a$. Hence, again by a), $g ∘ f$ is differentiable at $a$ as required.  If $C$ is a subset of $ℝ^n$ and $x ∈ C$, let
\begin{gathered}
𝒫(C, x)=\left\{γ:(r, r) → ℝ^n: γ \text { continuously differentiable on }(r, r),\right. \\
γ(0)=x, \text { and } γ((r, r)) ⊆ C\}
\end{gathered}
(where $r>0$ can vary with $γ$). Then $T_x C=\left\{γ'(0): γ ∈ 𝒫(C, x)\right\}$
 First note that $D f_{(x, y, z)}=(2 x, 2 y,2 z)$, and hence has maximal rank (i.e. rank 1) provided $(x, y, z) ≠ 0$. If $(x, y, z)$ lies in $S$, so that $x^2+y^2z^2=1$, then certainly $(x, y, z) ≠(0,0,0)$, so that $S$ is a 2submanifold of $ℝ^3$. Since $S$ is a submanifold, its tangent space at a point $p=(x, y, z) ∈ S$ can be described as:
$$
\bbox[5px, border: 2px solid red]{T_p S=\ker(D f_p)}=\left\{\left(x', y', z'\right) ∈ ℝ^3: 2 x x'+2 y y'2 zz'=0\right\}
$$
Thus we see that
$$
R=\left\{\left(x, y, z, x', y', z'\right) ∈ ℝ^6: x^2+y^2z^2=1,2 xx'+2 yy'2 zz'=0\right\}
$$
That is, $R=\left\{(p, q) ∈ ℝ^3 × ℝ^3: f(p)=0, D f_p(q)=0\right\}$, and hence we can view $R$ as the zeroset of the function $F: ℝ^6 → ℝ^2$ : given by $F(p, q)=\left(f(p), D f_p(q)\right)$. To see that $R$ is a submanifold, it thus suffices to show that $D F$ has full rank at $(p, q) ∈ R$. But if $(p, q)=\left(x, y, z, x', y', z'\right)$ we have
$$
D F_{(p, q)}=\begin{pmatrix}
2 x & 2 y & 2 z & 0 & 0 & 0 \\
2 x' & 2 y' & 2 z' & 2 x & 2 y & 2 z
\end{pmatrix}
$$
Clearly the first and second rows are linearly independent if $(x, y, z) ≠(0,0,0)$ (since then they are both nonzero, and the final three entries of the second row are nonzero). If follows $R$ is a 4submanifold of $ℝ^6$ as required. (Note that there is no condition required of $q$ in order to ensure $D F_{(p, q)}$ has maximal rank.)