R^n as a Normed Space; Vector-Valued Functions: Examples and Counterexamples

Refuted claim: if $f:[a,b]\to {\mathbb{R}}^m$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there is $\xi \in (a,b)$ with

$f(b)-f(a)=f^{\mathrm{\prime }}(\xi )(b-a)\mathrm{.}$

That is the equality form of the mean value theorem (The mean value theorem, as the case $g(x)=x$ of Cauchy's: for $f$ continuous on $[a,b]$ with $a<b$ and differentiable on $(a,b)$ there is $c\in (a,b)$ with $f(b)-f(a)=f^{\mathrm{\prime }}(c)(b-a)$), which is true for $m=1$ and false for $m\ge 2$. What survives is the inequality $\parallel f(b)-f(a){\parallel }_2\le M(b-a)$ of The mean value inequality: if $f:[a,b]\to {\mathbb{R}}^m$ is continuous and differentiable on $(a,b)$ with $\parallel f^{\mathrm{\prime }}{\parallel }_2\le M$, then $\parallel f(b)-f(a){\parallel }_2\le M(b-a)$, and this item is the witness showing that the inequality cannot be upgraded.

The witness. Take $m=2$, $[a,b]=[0,1]$ and $f:[0,1]\to {\mathbb{R}}^2$ with components $f_0(t)=t^2$ and $f_1(t)=t^3$ (Integer powers $a^m$, Vector-valued functions $f:A\to {\mathbb{R}}^m$, their limits and continuity, with the dictionary to the metric notions).

Why this curve and not the classical one. The crispest classical witness is $t\mapsto (\cos t,\sin t)$ on $[0,2\pi ]$, whose derivative has constant norm $1$ while the endpoints coincide. The trigonometric functions are introduced later in the reading order than this page, so they may not be used here; the polynomial curve above carries the same refutation with the material available. This substitution is recorded here, in the item itself, so that a reader who knows the classical example is told why it is absent rather than left to suppose that this library does not know it.

• What exactly fails. The scalar mean value theorem is applied in the proof of The mean value inequality: if $f:[a,b]\to {\mathbb{R}}^m$ is continuous and differentiable on $(a,b)$ with $\parallel f^{\mathrm{\prime }}{\parallel }_2\le M$, then $\parallel f(b)-f(a){\parallel }_2\le M(b-a)$ to the auxiliary function $t\mapsto ⟨u,f(t)⟩$ for the fixed vector $u=f(b)-f(a)$, and it does produce a point $\xi$. That $\xi$ is a mean value point of that real function, and it depends on $u$; there is no reason for the two coordinates to be served by one and the same point, and on this curve they are not: the first coordinate demands $\xi =1\mathrm{/}\iota (2)$ and the second demands $\iota (3){\xi }^2=1$.

• The curve is not pathological. Both components are polynomials, so $f$ is differentiable at every real (For a natural $n\ge 1$ the function $x\mapsto x^n$ is differentiable everywhere with derivative $\iota (n)x^{n-1}$; for $n=0$ it is the constant $1$, with derivative $0$; for a natural $n\ge 1$ the function $x\mapsto x^{-n}$ is differentiable at every $x\mathrm{\ne }0$ with derivative $-\iota (n)x^{-n-1}$; consequently every polynomial function is differentiable at every real, with the derivative computed term by term) and $f^{\mathrm{\prime }}$ is continuous. No smoothness hypothesis would rescue the equality form.

• The bound of The mean value inequality: if $f:[a,b]\to {\mathbb{R}}^m$ is continuous and differentiable on $(a,b)$ with $\parallel f^{\mathrm{\prime }}{\parallel }_2\le M$, then $\parallel f(b)-f(a){\parallel }_2\le M(b-a)$ is not merely true but sharp, and the companion item A curve for which the mean value inequality is an equality, showing the constant cannot be improved exhibits a curve on which it is an equality. The two items together show that the failure of the equality form is not a failure of the bound.

Refuted claim: the inequality of The mean value inequality: if $f:[a,b]\to {\mathbb{R}}^m$ is continuous and differentiable on $(a,b)$ with $\parallel f^{\mathrm{\prime }}{\parallel }_2\le M$, then $\parallel f(b)-f(a){\parallel }_2\le M(b-a)$ can be improved: there is a real $c<1$ such that for every $m\ge 1$, every $a<b$ and every $f:[a,b]\to {\mathbb{R}}^m$ continuous on $[a,b]$ and differentiable on $(a,b)$ with $\parallel f^{\mathrm{\prime }}{\parallel }_2\le M$ there,

$\parallel f(b)-f(a){\parallel }_2\le cM(b-a)\mathrm{.}$

The witness. Take $m=2$, $[a,b]=[0,1]$ and $f:[0,1]\to {\mathbb{R}}^2$ with $f_0(t)=t$ and $f_1(t)=0$. Then $\parallel f^{\mathrm{\prime }}(t){\parallel }_2=1$ for every $t\in (0,1)$, so $M=1$ is admissible, and

$\parallel f(1)-f(0){\parallel }_2=1=M(1-0)\mathrm{.}$

The inequality of The mean value inequality: if $f:[a,b]\to {\mathbb{R}}^m$ is continuous and differentiable on $(a,b)$ with $\parallel f^{\mathrm{\prime }}{\parallel }_2\le M$, then $\parallel f(b)-f(a){\parallel }_2\le M(b-a)$ is therefore an equality on this curve, and no constant smaller than $1$ can stand in front of $M(b-a)$.

• The two witnesses on this page say opposite-looking things and are consistent. $f(t)=(t^2,t^3)$ on $[0,1]$: no $\xi$ satisfies $f(1)-f(0)=f^{\mathrm{\prime }}(\xi )$ shows that the equality form of the mean value theorem fails for $m\ge 2$: there need be no $\xi$ with $f(b)-f(a)=f^{\mathrm{\prime }}(\xi )(b-a)$. The present item shows that the inequality of The mean value inequality: if $f:[a,b]\to {\mathbb{R}}^m$ is continuous and differentiable on $(a,b)$ with $\parallel f^{\mathrm{\prime }}{\parallel }_2\le M$, then $\parallel f(b)-f(a){\parallel }_2\le M(b-a)$ is nevertheless sharp. Together they say that the correct vector-valued statement is an inequality, and that it is the best inequality of its shape.

• Why the equality is attained here and not there. On the curve above the derivative is constant, so it points in one direction and the displacement accumulates with no cancellation. On $t\mapsto (t^2,t^3)$ the direction of $f^{\mathrm{\prime }}$ turns as $t$ increases, and the displacement is strictly shorter than the length the bound allows: there $\parallel f(1)-f(0){\parallel }_2=\sqrt{2}$ while the bound is $\sqrt{\iota (13)}$.

• The curve is as simple as it can be. Its image is a segment of the first coordinate axis, and the second component is present only so that the codomain is ${\mathbb{R}}^2$ rather than ${\mathbb{R}}^1$; the same computation in ${\mathbb{R}}^m$ for any $m\ge 1$ gives the same equality.

On ${\mathbb{R}}^2$ with the norms of The $p$-norms $\parallel x{\parallel }_p$ for rational $p\ge 1$, and $\parallel x{\parallel }_{\mathrm{\infty }}$, the comparison chain of The finite and reverse triangle inequalities for a norm; and for $n\ge 1$ every norm $N$ on ${\mathbb{R}}^n$ satisfies $N(x)\le C\parallel x{\parallel }_1$ and is Lipschitz, hence continuous, for $d_2$ clause 3 reads

$\parallel x{\parallel }_{\mathrm{\infty }}\le \parallel x{\parallel }_2\le \parallel x{\parallel }_1\le \iota (2)\parallel x{\parallel }_{\mathrm{\infty }},\parallel x{\parallel }_1\le \sqrt{\iota (2)}\parallel x{\parallel }_2\mathrm{.}$

Each of these four constants is attained, so none can be improved:

• $e_0=(1,0)$ has $\parallel e_0{\parallel }_{\mathrm{\infty }}=\parallel e_0{\parallel }_2=\parallel e_0{\parallel }_1=1$, so the first and second inequalities are equalities there;
• $(1,1)$ has $\parallel (1,1){\parallel }_{\mathrm{\infty }}=1$, $\parallel (1,1){\parallel }_2=\sqrt{\iota (2)}$ and $\parallel (1,1){\parallel }_1=\iota (2)$, so the third and fourth inequalities are equalities there.

The general theorem For $n\ge 1$ all norms on ${\mathbb{R}}^n$ are equivalent supplies constants but no attaining vectors; that is what this computation adds.

Unit balls. Writing $B_p:=\{x\in {\mathbb{R}}^2:\parallel x{\parallel }_p\le 1\}$ for $p\in \{1,2,\mathrm{\infty }\}$, the chain gives $B_1\subseteq B_2\subseteq B_{\mathrm{\infty }}$, and both inclusions are strict: $(1,1)$ lies in $B_{\mathrm{\infty }}$ and not in $B_2$, and $(3\mathrm{/}\iota (5))(1,1)$ lies in $B_2$ and not in $B_1$. The scalar has to be chosen strictly between $1\mathrm{/}\iota (2)$ and $1\mathrm{/}\sqrt{\iota (2)}$: at the endpoint $1\mathrm{/}\iota (2)$ the vector $(1,1)\mathrm{/}\iota (2)$ has $\parallel \cdot {\parallel }_1=1$ and so still lies in $B_1$.

• Sharpness is not the same as equivalence. For $n\ge 1$ all norms on ${\mathbb{R}}^n$ are equivalent asserts that constants exist and produces some; nothing in it says which are smallest. The computation above supplies attaining vectors, and those are what make the constants of the chain best possible on ${\mathbb{R}}^2$.

• Both attaining vectors are extreme in the expected way. A vector with a single nonzero coordinate makes all three norms agree; a vector whose two coordinates have equal absolute value spreads the mass as evenly as possible and is where $\parallel \cdot {\parallel }_1$ is largest relative to the other two. On ${\mathbb{R}}^n$ the same two vectors give equality with $\iota (n)$ and $\sqrt{\iota (n)}$ in place of $\iota (2)$ and $\sqrt{\iota (2)}$; only the case $n=2$ is verified here.

• The strictness computation in step 2.3 is arithmetic, not geometry. The scalar $3\mathrm{/}\iota (5)$ was chosen to lie strictly between $1\mathrm{/}\iota (2)$ and $1\mathrm{/}\sqrt{\iota (2)}$; any scalar in that open interval would serve, and the interval is nonempty exactly because $\sqrt{\iota (2)}<\iota (2)$.

Refuted claim: every norm on ${\mathbb{R}}^2$ arises from a symmetric bilinear form, that is, for every norm $N$ there is a function $B:{\mathbb{R}}^2\times {\mathbb{R}}^2\to \mathbb{R}$ that is symmetric and additive and homogeneous in each argument, with $N(x)=\sqrt{B(x,x)}$ for every $x$.

The witness is $N:=\parallel \cdot {\parallel }_1$ on ${\mathbb{R}}^2$ (The $p$-norms $\parallel x{\parallel }_p$ for rational $p\ge 1$, and $\parallel x{\parallel }_{\mathrm{\infty }}$), and the obstruction is the parallelogram law, which every such $N$ satisfies (Cauchy-Schwarz $\mid ⟨x,y⟩\mid \le \parallel x{\parallel }_2\parallel y{\parallel }_2$ with its equality case, the triangle inequality for $\parallel \cdot {\parallel }_2$, the parallelogram law and polarisation clause 3 is the instance for the Euclidean form, and the general computation is two lines of bilinearity, done below) and which $\parallel \cdot {\parallel }_1$ fails at $x=e_0$, $y=e_1$.

What is and is not claimed. What is refuted is the displayed claim, whose hypothesis is a symmetric bilinear form on ${\mathbb{R}}^2$ written out in full. The general converse — that a norm satisfying the parallelogram law is induced by an inner product, the Jordan-von Neumann theorem — is not proved here and is not used here. Abstract inner-product spaces are developed on a separate published page; the refutation below uses only bilinearity and the explicit one-norm calculation (Conventions of this page, the standing $n\ge 1$ hypothesis, and what is taken up elsewhere in the reading order).

• Equivalence of norms says nothing about inner products. By For $n\ge 1$ all norms on ${\mathbb{R}}^n$ are equivalent the norms $\parallel \cdot {\parallel }_1$ and $\parallel \cdot {\parallel }_2$ on ${\mathbb{R}}^2$ are equivalent: they have the same open sets, the same convergent sequences and the same Cauchy sequences. What the computation above shows is that they are nevertheless different norms, and that one of them cannot be written as $\sqrt{B(\cdot ,\cdot )}$ for any symmetric bilinear $B$. Equivalence is a metric statement; the parallelogram law is not.

• Only one instance of the law is needed. The claim is refuted by a single pair $(e_0,e_1)$, and the arithmetic is $\iota (8)\mathrm{\ne }\iota (4)$. No general theory is required, which is exactly why this item can be stated on a page that has no abstract inner products.

• The converse direction is a different theorem. That a norm satisfying the parallelogram law is induced by an inner product is the Jordan-von Neumann theorem, proved by polarisation; Cauchy-Schwarz $\mid ⟨x,y⟩\mid \le \parallel x{\parallel }_2\parallel y{\parallel }_2$ with its equality case, the triangle inequality for $\parallel \cdot {\parallel }_2$, the parallelogram law and polarisation clause 4 contains the polarisation identity for the Euclidean form, but the general theorem needs the abstract theory and is not asserted anywhere in this library.

Refuted claim: a function $g:{\mathbb{R}}^2\to \mathbb{R}$ that is continuous in each variable separately — that is, for which $t\mapsto g(t,b)$ and $t\mapsto g(a,t)$ are continuous on $\mathbb{R}$ for every fixed $a$ and $b$ (Continuity of $f:A\to \mathbb{R}$ at a point of $A$ and on $A$: the $\epsilon$-$\delta$ condition, its agreement with ${\lim }_{x\to c}f(x)=f(c)$ at a limit point, and continuity at an isolated point) — is continuous as a map $({\mathbb{R}}^2,d_2)\to (\mathbb{R},d_{\mathbb{R}})$ (Vector-valued functions $f:A\to {\mathbb{R}}^m$, their limits and continuity, with the dictionary to the metric notions, Continuity of a map between metric spaces, at a point and globally, in the $\epsilon$-$\delta$ form, ${\mathbb{R}}^n$ as the set of functions $n\to \mathbb{R}$, and $d_1$, $d_2$, $d_{\mathrm{\infty }}$ are metrics on it).

The witness. Define $g:{\mathbb{R}}^2\to \mathbb{R}$ by

$g(p):=\frac{p_0{p}_1}{p_0^2+p_1^2}\text{ for }p\mathrm{\ne }0,g(0):=0,$

writing $p=(p_0,p_1)$ for an element of ${\mathbb{R}}^2$, the set of functions $2\to \mathbb{R}$ (${\mathbb{R}}^n$ as the set of functions $n\to \mathbb{R}$, and $d_1$, $d_2$, $d_{\mathrm{\infty }}$ are metrics on it). The quotient is defined for $p\mathrm{\ne }0$ because $p_0^2+p_1^2=\parallel p{\parallel }_2^2>0$ there (The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).