Proof of Segment Derivatives, the Second-Order Taylor Expansion, and the Second-Order Condition at a Local Extremum

Throughout we use that $d(x,x+w)=|w|$ for $x,w\in E$, by Real Inner Product Spaces (iii) and Elementary Identities in a Real Inner Product Space (iv), and that by An Open Interval is an Interval All of Whose Points Are Interior the open interval $(-2,2)$ is an interval every point of which is interior, so that differentiability at a point of it in the sense of Derivative of a Real Function on an Interval is meaningful. We also use that products and inverses of positive real numbers are positive (claims 5 and 7 of Elementary Order Arithmetic in an Ordered Field).

Claim 1. Let $t\in(-2,2)$. Then $|t|<2$ by claim 9 of Properties of the Absolute Value in an Ordered Field, so claim 5 of Elementary Arithmetic in an Ordered Field with the nonnegative multiplier $|z|$ gives $|t|\,|z|\le 2\,|z|$, and $2|z|<r$ by hypothesis, so $|t|\,|z|<r$ by claim 2 of Elementary Order Arithmetic in an Ordered Field. Since $d(x,x+tz)=|tz|=|t|\,|z|$ by Elementary Identities in a Real Inner Product Space (iv), we get $x+tz\in B_{d}(x,r)\subseteq U$, and $g$ is defined.

Fix $t\in(-2,2)$ and put $y=x+tz$ and $p=Du(y)$, which exists because $y\in B_{d}(x,r)$.

If $z=0_{E}$, then $g$ takes the constant value $u(x)$ and $\langle p,z\rangle=0$ by Elementary Identities in a Real Inner Product Space (ii), so for every positive $\varepsilon$ any positive $\delta$ witnesses the condition of Derivative of a Real Function on an Interval, the difference quotient being $0$; thus $g'(t)=0=\langle p,z\rangle$.

Suppose $z\ne 0_{E}$, so that $0<|z|$ by Elementary Identities in a Real Inner Product Space (iii). Let $\varepsilon\in\mathbb{R}$ be positive and let $\delta_{0}$ be the radius supplied by Fréchet Differentiability and the Gradient on an Open Subset of a Real Inner Product Space (i) at $y$ for the positive number $\varepsilon\cdot 2^{-1}\,|z|^{-1}$, so that

$$\bigl|u(y+w)-u(y)-\langle p,w\rangle\bigr|\le\varepsilon\cdot 2^{-1}\,|z|^{-1}\,|w|\qquad\text{whenever }|w|<\delta_{0}.$$

Put $\delta=\delta_{0}\,|z|^{-1}$, a positive real number, and let $s\in\mathbb{R}$ satisfy $0<|s|<\delta$ and $t+s\in(-2,2)$. Then $|sz|=|s|\,|z|<\delta\,|z|=\delta_{0}$ by claim 10 of Elementary Order Arithmetic in an Ordered Field, so the displayed bound applies with $w=sz$ and gives

$$\bigl|u(y+sz)-u(y)-s\,\langle p,z\rangle\bigr|\le\varepsilon\cdot 2^{-1}\,|z|^{-1}\,|s|\,|z|=\varepsilon\cdot 2^{-1}\,|s| ,$$

where $\langle p,sz\rangle=s\,\langle p,z\rangle$ by Elementary Identities in a Real Inner Product Space (i). Since $y+sz=x+(t+s)z$, the left-hand side equals $|g(t+s)-g(t)-s\langle p,z\rangle|$, and dividing by the positive number $|s|$ (claim 5 of Elementary Arithmetic in an Ordered Field, with claims 1 and 4 of Properties of the Absolute Value in an Ordered Field to move the factor inside the absolute value) gives

$$\Bigl|\frac{g(t+s)-g(t)}{s}-\langle p,z\rangle\Bigr|\le\varepsilon\cdot 2^{-1}<\varepsilon$$

by claim 8 of Elementary Order Arithmetic in an Ordered Field, so that the left-hand side is strictly smaller than $\varepsilon$ by claim 2 there. Hence $g$ is differentiable at $t$ with $g'(t)=\langle p,z\rangle=\langle Du(x+tz),z\rangle$.

Claim 2. Put $p=Du(x)$ and $b=D^{2}u(x)$, and let $\rho$ be a positive real number with $B_{d}(x,\rho)\subseteq U$ and $u$ differentiable at every point of $B_{d}(x,\rho)$, as in The Second Derivative and the Hessian on an Open Subset of a Real Inner Product Space (i). Let $\varepsilon\in\mathbb{R}$ be positive and let $\delta_{0}$ be the radius that clause supplies for $\varepsilon$, so that $\delta_{0}\le\rho$ and

$$\bigl|\langle Du(x+w)-p,y\rangle-b(w,y)\bigr|\le\varepsilon\,|w|\,|y|\qquad\text{for all }w,y\in E\text{ with }|w|<\delta_{0}.$$

Put $\delta=\delta_{0}\cdot 2^{-1}$, which is positive by claim 8 of Elementary Order Arithmetic in an Ordered Field, and let $z\in E$ satisfy $|z|<\delta$. Then $2\,|z|<\delta_{0}\le\rho$ by claim 10 of Elementary Order Arithmetic in an Ordered Field, and in particular $|z|<\rho$, so $x+z\in B_{d}(x,\rho)\subseteq U$.

Since $u$ is differentiable at every point of $B_{d}(x,\rho)$ and $2|z|<\rho$, claim 1 applies with $r=\rho$: the function $g:(-2,2)\to\mathbb{R}$ with $g(\tau)=u(x+\tau z)$ is defined and differentiable at every $\tau\in(-2,2)$, with $g'(\tau)=\langle Du(x+\tau z),z\rangle$.

Write $\beta=\langle p,z\rangle$ and $\gamma=b(z,z)$, and let $\psi:(-2,2)\to\mathbb{R}$ be given by $\psi(\tau)=u(x)+\tau\,\beta+\tau^{2}\,\gamma\cdot 2^{-1}$. We check that $\psi$ is differentiable at every $\tau\in(-2,2)$ with $\psi'(\tau)=\beta+\tau\,\gamma$. Indeed, for $s\in\mathbb{R}$ with $s\ne 0$ and $\tau+s\in(-2,2)$, the field identity $(\tau+s)^{2}-\tau^{2}=2\tau s+s^{2}$ gives

$$\psi(\tau+s)-\psi(\tau)=s\,\beta+(2\tau s+s^{2})\,\gamma\cdot 2^{-1}=s\,(\beta+\tau\gamma)+s^{2}\,\gamma\cdot 2^{-1},$$

so that, dividing by $s$,

$$\Bigl|\frac{\psi(\tau+s)-\psi(\tau)}{s}-(\beta+\tau\gamma)\Bigr|=\bigl|\gamma\cdot 2^{-1}\bigr|\,|s|$$

by claims 1 and 4 of Properties of the Absolute Value in an Ordered Field. Given a positive $\varepsilon'$, put $L=|\gamma\cdot 2^{-1}|+1$, which is positive because $0\le|\gamma\cdot 2^{-1}|$ by claim 1 of Properties of the Absolute Value in an Ordered Field and $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field, so that $0<L$ by claim 3 there, and put $\delta'=\varepsilon'\,L^{-1}$. For $0<|s|<\delta'$, claim 5 of Elementary Arithmetic in an Ordered Field and claim 10 of Elementary Order Arithmetic in an Ordered Field give $|\gamma\cdot 2^{-1}|\,|s|\le L\,|s|<L\,\delta'=\varepsilon'$, so that the displayed difference quotient is strictly smaller than $\varepsilon'$ in absolute value by claim 2 of Elementary Order Arithmetic in an Ordered Field, as required by Derivative of a Real Function on an Interval.

By Derivative of a Sum and of a Difference the function $\phi=g-\psi$ on $(-2,2)$ is differentiable at every $\tau\in(-2,2)$, with

$$\phi'(\tau)=\langle Du(x+\tau z),z\rangle-\beta-\tau\gamma=\langle Du(x+\tau z)-p,z\rangle-b(\tau z,z),$$

using Elementary Identities in a Real Inner Product Space (i) and the homogeneity of $b$ in its first argument (Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (i)). If $\tau\in(-2,2)$ satisfies $|\tau|\le 1$, then $|\tau z|=|\tau|\,|z|\le|z|<\delta\le\delta_{0}$ by claim 5 of Elementary Arithmetic in an Ordered Field, so the second displayed bound of this claim applies with $w=\tau z$ and $y=z$ and gives, again by claim 5 of Elementary Arithmetic in an Ordered Field,

$$\bigl|\phi'(\tau)\bigr|\le\varepsilon\,|\tau z|\,|z|\le\varepsilon\,|z|^{2} .$$

Now $0$ and $1$ lie in $(-2,2)$ and $0<1$, so Mean Value Theorem on an Open Interval provides $c\in(0,1)$ with $\phi(1)-\phi(0)=\phi'(c)\,(1-0)=\phi'(c)$; since $0<c<1$ we have $|c|=c\le 1$, so the bound above applies at $\tau=c$. Finally $\phi(0)=u(x)-u(x)=0$ and

$$\phi(1)=u(x+z)-u(x)-\langle p,z\rangle-\tfrac{1}{2}\,b(z,z),$$

whence $|u(x+z)-u(x)-\langle Du(x),z\rangle-\tfrac{1}{2}D^{2}u(x)(z,z)|=|\phi'(c)|\le\varepsilon\,|z|^{2}$, as asserted.

If $u\in C^{2}(U)$ and $x\in U$, then $u$ is differentiable on $U$ and has a second derivative at $x$ by The Classes $C^1$ and $C^2$ on an Open Subset of a Real Inner Product Space (ii), so the above applies.

Claim 3. By The Second Derivative and the Hessian on an Open Subset of a Real Inner Product Space (i) the function $u$ is differentiable at every point of a ball $B_{d}(x,\rho)\subseteq U$, in particular at $x$, so Basic Properties of Differentiability on an Open Subset of a Real Inner Product Space (iii) gives $Du(x)=0_{E}$ in both cases. Write $b=D^{2}u(x)$.

Suppose $u$ 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). By Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (iii) and $0_{\mathrm{Sym}}(w,w)=0$, it suffices to show that $b(w,w)\le 0$ for every $w\in E$.

Let $w\in E$. If $w=0_{E}$, then $0_{E}=0\cdot 0_{E}$ by claim 3 of Elementary Identities in a Vector Space and homogeneity in the first argument (Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (i)) gives $b(w,w)=0\cdot b(0_{E},0_{E})=0$. So assume $w\ne 0_{E}$, whence $0<|w|$ and $0<|w|^{2}$.

Let $\eta\in\mathbb{R}$ be positive and put $\varepsilon=\eta\,\bigl(2\,|w|^{2}\bigr)^{-1}$, a positive real number. By claim 2 there is a positive $\delta_{1}$ such that every $z\in E$ with $|z|<\delta_{1}$ satisfies $x+z\in U$ and, since $\langle Du(x),z\rangle=\langle 0_{E},z\rangle=0$ by Elementary Identities in a Real Inner Product Space (ii),

$$\bigl|u(x+z)-u(x)-\tfrac{1}{2}\,b(z,z)\bigr|\le\varepsilon\,|z|^{2} .$$

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, put $t=(\delta\cdot 2^{-1})\,|w|^{-1}$, a positive real number, and put $z=t\,w$; then $|z|=t\,|w|=\delta\cdot 2^{-1}<\delta$ by Elementary Identities in a Real Inner Product Space (iv) and claim 8 of Elementary Order Arithmetic in an Ordered Field. Since $d(x,x+z)=|z|<\delta_{0}$ and $x+z\in U$, we have $u(x+z)-u(x)\le 0$; abbreviating $A=u(x+z)-u(x)$ and using claim 3 of Properties of the Absolute Value in an Ordered Field, the displayed bound gives $\tfrac{1}{2}b(z,z)-A\le\varepsilon|z|^{2}$ and hence

$$\tfrac{1}{2}\,b(z,z)\le A+\varepsilon\,|z|^{2}\le\varepsilon\,|z|^{2} .$$

By the homogeneity and symmetry of Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (i), $b(z,z)=t^{2}\,b(w,w)$, and $|z|^{2}=t^{2}|w|^{2}$; multiplying by the nonnegative number $(t^{2})^{-1}$ (claim 5 of Elementary Arithmetic in an Ordered Field) and then by $2$ gives

$$b(w,w)\le 2\,\varepsilon\,|w|^{2}=\eta .$$

As $\eta$ was an arbitrary positive real number, claim 1 of Comparison of Real Numbers with Arbitrary Positive Slack gives $b(w,w)\le 0$. Hence $b\preceq 0_{\mathrm{Sym}}$.

If instead $u$ has a local minimum at $x$ relative to $U$, then $0\le A$ in the argument above, and claim 3 of Properties of the Absolute Value in an Ordered Field applied to $A-\tfrac{1}{2}b(z,z)$ gives $-\tfrac{1}{2}b(z,z)\le\varepsilon|z|^{2}-A\le\varepsilon|z|^{2}$; the same rescaling yields $-b(w,w)\le\eta$ for every positive $\eta$, hence $-b(w,w)\le 0$ by claim 1 of Comparison of Real Numbers with Arbitrary Positive Slack and, by claim 4 of Elementary Order Arithmetic in an Ordered Field, $0\le b(w,w)$ for every $w\in E$. Therefore $0_{\mathrm{Sym}}\preceq b$.