Proof of Basic Properties of Differentiability on an Open Subset of a Real Inner Product Space

Throughout, $p$ denotes $Du(x)$ where relevant, and we use repeatedly that $d(x,x+z)=|z|$ for $x,z\in E$: indeed $x-(x+z)=-z$ in the vector space $E$, so $d(x,x+z)=|-z|=|z|$ by Real Inner Product Spaces (iii) and Elementary Identities in a Real Inner Product Space (iv).

Claim 1. Let $\delta$ be the radius supplied by Fréchet Differentiability and the Gradient on an Open Subset of a Real Inner Product Space (i) for $\varepsilon$. Every $z\in E$ with $|z|<\delta$ satisfies $x+z\in U$ and, by claims 2 and 5 of Properties of the Absolute Value in an Ordered Field and The Cauchy–Schwarz Inequality in a Real Inner Product Space,

$$\bigl|u(x+z)-u(x)\bigr|\le\bigl|u(x+z)-u(x)-\langle p,z\rangle\bigr|+\bigl|\langle p,z\rangle\bigr|\le\varepsilon\,|z|+|p|\,|z|=\bigl(|p|+\varepsilon\bigr)|z| .$$

Claim 2. Let $\delta_{1}$ be the radius supplied by claim 1 for the positive number $1$, and put $M=|p|+1$. Since $0\le|p|$ by Real Inner Product Spaces (ii) and $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field, claim 3 of Elementary Order Arithmetic in an Ordered Field gives $0<M$.

Let $\varepsilon\in\mathbb{R}$ be positive and let $\delta$ be the lesser of $\delta_{1}$ and $\varepsilon M^{-1}$ (claim 9 of Elementary Order Arithmetic in an Ordered Field), which is positive because it is one of them. Let $y\in U$ satisfy $d(x,y)<\delta$ and put $z=y-x$, so that $x+z=y$ and $|z|=d(x,y)<\delta\le\delta_{1}$. By claim 1,

$$\bigl|u(y)-u(x)\bigr|\le M\,|z|<M\,\delta\le\varepsilon ,$$

the strict step by claim 10 of Elementary Order Arithmetic in an Ordered Field and the last by claim 5 of Elementary Arithmetic in an Ordered Field applied to $\delta\le\varepsilon M^{-1}$ with the nonnegative multiplier $M$. Hence $d_{\mathbb{R}}(u(y),u(x))<\varepsilon$, and $u$ is continuous at $x$ relative to $U$ by Continuous Map Between Metric Spaces.

If $u$ is differentiable on $U$ it is differentiable at every point of $U$, hence continuous at every point of $U$ relative to $U$, that is continuous on $U$. Members of $C^{1}(U)$ and of $C^{2}(U)$ are differentiable on $U$ by The Classes $C^1$ and $C^2$ on an Open Subset of a Real Inner Product Space (i) and The Classes $C^1$ and $C^2$ on an Open Subset of a Real Inner Product Space (ii).

Claim 3. Suppose first that $u$ has a local maximum at $x$ relative to $U$, and let $\delta_{0}$ be a positive real number such that every $y\in U$ with $d(x,y)<\delta_{0}$ satisfies $u(y)\le u(x)$ (Local Maximum of a Function Relative to a Subset of a Metric Space). Let $\varepsilon\in\mathbb{R}$ be positive, let $\delta_{1}$ be the radius supplied by Fréchet Differentiability and the Gradient on an Open Subset of a Real Inner Product Space (i) for $\varepsilon$, and let $\delta$ be the lesser of $\delta_{0}$ and $\delta_{1}$ (claim 9 of Elementary Order Arithmetic in an Ordered Field), which is positive because it is one of them.

Let $z\in E$ satisfy $|z|<\delta$ and abbreviate $A=u(x+z)-u(x)$ and $R=A-\langle p,z\rangle$, so that $|R|\le\varepsilon|z|$. Since $x+z\in U$ and $d(x,x+z)=|z|<\delta_{0}$, we have $A\le 0$. Now $\langle p,z\rangle=A+(-R)$, and $-R\le|-R|=|R|\le\varepsilon|z|$ by claims 3 and 2 of Properties of the Absolute Value in an Ordered Field; adding the inequalities $A\le 0$ and $-R\le\varepsilon|z|$, which is permitted by the compatibility of the order with addition (an axiom of Ordered Field) together with transitivity, gives

$$\langle p,z\rangle\le\varepsilon\,|z| .$$

If instead $u$ has a local minimum at $x$ relative to $U$, then $0\le A$ and the same computation, using $R\le|R|\le\varepsilon|z|$ and $-\langle p,z\rangle=R+(-A)$, gives $-\langle p,z\rangle\le\varepsilon|z|$.

In either case the resulting inequality holds for every $z\in E$ with $|z|<\delta$. Applying it to $-z$, which also satisfies $|-z|=|z|<\delta$ by Elementary Identities in a Real Inner Product Space (iv), and using $\langle p,-z\rangle=-\langle p,z\rangle$ from Elementary Identities in a Real Inner Product Space (i), we obtain both $\langle p,z\rangle\le\varepsilon|z|$ and $-\langle p,z\rangle\le\varepsilon|z|$; the second gives $-\varepsilon|z|\le\langle p,z\rangle$ by claim 4 of Elementary Order Arithmetic in an Ordered Field, so claim 6 of Properties of the Absolute Value in an Ordered Field yields

$$\bigl|\langle p,z\rangle\bigr|\le\varepsilon\,|z| .$$

Since $\varepsilon$ was an arbitrary positive real number, Vanishing of Uniformly Small Linear, Quadratic and Bilinear Terms in a Real Inner Product Space (i) gives $p=0_{E}$.

Claim 4. Since $U'$ is open in $(E,d)$ and $x\in U'$, Open Subset of a Metric Space provides a positive $\rho$ with $B_{d}(x,\rho)\subseteq U'$; as $d(x,x+z)=|z|$, every $z\in E$ with $|z|<\rho$ satisfies $x+z\in U'$.

Suppose $u$ is differentiable at $x$ with gradient $p$, let $\varepsilon$ be positive, let $\delta_{1}$ be the radius supplied by Fréchet Differentiability and the Gradient on an Open Subset of a Real Inner Product Space (i), and let $\delta$ be the lesser of $\delta_{1}$ and $\rho$, which is positive because it is one of them. Every $z$ with $|z|<\delta$ satisfies $x+z\in U'$ and, because $u|_{U'}$ agrees with $u$ at $x$ and at $x+z$, the inequality of that clause with $u|_{U'}$ in place of $u$. Hence $u|_{U'}$ is differentiable at $x$ with gradient $p$, and $D(u|_{U'})(x)=Du(x)$ by the uniqueness in Fréchet Differentiability and the Gradient on an Open Subset of a Real Inner Product Space (ii). This argument used nothing about $x$ beyond its lying in the open set $U'$ and $u$ being differentiable there, so it applies verbatim at any point of $U'$ at which $u$ is differentiable: for every subset $S\subseteq U'$ such that $u$ is differentiable at every point of $S$, the restriction $u|_{U'}$ is differentiable at every point of $S$ with $D(u|_{U'})(y)=Du(y)$ for $y\in S$. In particular, if $u$ is differentiable on $U$ then $u|_{U'}$ is differentiable on $U'$ with $D(u|_{U'})(y)=Du(y)$ for every $y\in U'$.

Now let $b\in\mathrm{Sym}(E)$ be a second derivative of $u$ at $x$, and let $\rho_{0}$ be a radius as in The Second Derivative and the Hessian on an Open Subset of a Real Inner Product Space (i): thus $B_{d}(x,\rho_{0})\subseteq U$ and $u$ is differentiable at every point of $B_{d}(x,\rho_{0})$. Let $\rho'$ be the lesser of $\rho_{0}$ and $\rho$, which is positive. Then $B_{d}(x,\rho')\subseteq U'$, and by the previous paragraph applied with $S=B_{d}(x,\rho')$ the restriction $u|_{U'}$ is differentiable at every point of $B_{d}(x,\rho')$ with the same gradients as $u$. Given a positive $\varepsilon$, let $\delta$ be a radius witnessing the condition of that clause for $u$, $b$ and $\rho_{0}$, and let $\delta'$ be the lesser of $\delta$ and $\rho'$; then $\delta'$ is positive, $\delta'\le\rho'$, and the displayed inequality of that clause holds for all $w,y\in E$ with $|w|<\delta'$, with $u|_{U'}$ in place of $u$. Hence $b$ is a second derivative of $u|_{U'}$ at $x$, and $D^{2}(u|_{U'})(x)=D^{2}u(x)$ by the uniqueness in The Second Derivative and the Hessian on an Open Subset of a Real Inner Product Space (ii).

Finally, suppose $u\in C^{1}(U)$. Then $u|_{U'}$ is differentiable on $U'$ and its gradient map is the restriction of $Du$ to $U'$, which is continuous on $U'$ by claim 1 of Restriction Stability of Continuity and of the Derivative; hence $u|_{U'}\in C^{1}(U')$ by The Classes $C^1$ and $C^2$ on an Open Subset of a Real Inner Product Space (i). If moreover $u\in C^{2}(U)$, then by the previous paragraph $u|_{U'}$ has at every $y\in U'$ the second derivative $D^{2}u(y)$, so its Hessian map is the restriction of $D^{2}u$ to $U'$ and is continuous on $U'$ by the same claim; hence $u|_{U'}\in C^{2}(U')$ by The Classes $C^1$ and $C^2$ on an Open Subset of a Real Inner Product Space (ii), with $D^{2}(u|_{U'})(y)=D^{2}u(y)$ for every $y\in U'$.