18.3 The Mean Value Theorem

Definition 18.3.1 (Right-Differentiable). Let $-\infty < a < b < \infty$, $E$ be a separated topological vector space, $f: [a, b] \to E$, and $x \in [a, b)$, then $f$ is right-differentiable at $x$ if

\[D^{+}f(x) = \lim_{t \downto 0}\frac{f(x + t) - f(x)}{t}\]

exists.

Lemma 18.3.2. Let $-\infty < a < b < \infty$, $f, g \in C([a,b]; \real)$, and $N \subset [a, b]$ be at most countable such that:

$f, g$ are right-differentiable on $[a, b] \setminus N$.
For every $x \in [a, b] \setminus N$, $D^{+}f(x) \le D^{+}g(x)$.

then for any $x \in [a, b]$, $f(x) - f(a) \le g(x) - g(a)$.

Proof. First assume that for every $x \in [a, b] \setminus N$, $D^{+}f(x) < D^{+}g(x)$.

Let $\seq{x_n}$ be an enumeration of $N$. For each $x \in [a, b]$, let $N(x) = \bracs{n \in \natp|x_n \in [a, x)}$. Let $\eps > 0$ and define

\[S = \bracs{x \in [a, b] \bigg | f(y) - f(a) \le g(y) - g(a) + \eps\sum_{n \in N(x)}2^{-n} \quad \forall y \in [a, x]}\]

then by continuity of $f$ and $g$, $S$ is closed.

Let $x \in S$ and suppose that $s < b$. If $x \in [a, b] \setminus N$, then since

\[\lim_{t \downto 0}\frac{f(x + t) - f(x)}{t}< \lim_{t \downto 0}\frac{g(x + t) - f(x)}{t}\]

there exists $\delta > 0$ such that $f(x + t) - f(x) < g(x + t) < g(x)$ for all $t \in [0, \delta)$. In which case, $[x, x + \delta) \subset S$.

If $x = x_{n} \in N$, then by continuity of $f$ and $g$, there exists $\delta > 0$ such that $f(x + t) - f(x) \le g(x + t) - g(x) + 2^{-n}$ for all $t \in (0, \delta)$. Hence $[x, x + \delta) \subset S$ as well.

Therefore $S = [a, b]$. As this holds for all $\eps > 0$, $f(b) - f(a) \le g(b) - g(a)$.

Finally, suppose that $D^{+}f(x) \le D^{+}g(x)$ for all $x \in [a, b] \setminus N$. Let $\eps > 0$ and $h(x) = g(x) + \eps(x - a)$. By the preceding case, $f(b) - f(a) \le g(b) - g(a) + \eps(b - a)$. As this holds for all $\eps > 0$, $f(x) - f(a) \le g(x) - g(a)$.$\square$

Theorem 18.3.3. Let $-\infty < a < b < \infty$, $E$ be a separated locally convex space, $f \in C([a, b]; E)$, $g \in C([a, b]; \real)$, $N \subset [a, b]$ be at most countable, and $B \subset E$ be a closed convex set such that:

$f, g$ are right-differentiable on $[a, b] \setminus N$.
For each $x \in [a, b] \setminus N$, $D^{+}f(x) \in D^{+}g(x)B$.
$g$ is non-decreasing.

then

\[f(b) - f(a) \in [g(b) - g(a)]B\]

Proof. Let $\phi \in E^{*}$ and $x \in [a, b] \setminus N$, then since $g$ is non-decreasing,

\begin{align*}\lim_{t \downto 0}\frac{\phi(f(x + t) - f(x))}{t}&\le \lim_{t \downto 0}\sup_{z \in B}\frac{\phi(g(x + t) - g(x)z)}{t}\\ D^{+}(\phi \circ f)(x)&\le \sup_{z \in B}\phi(z) \cdot \lim_{t \downto 0}\frac{\phi(g(x + t) - g(x))}{t}\\ D^{+}(\phi \circ f)(x)&\le D^{+}g(x) \cdot \sup_{z \in B}\phi(z)\end{align*}

By Lemma 18.3.2,

\[\phi(f(b) - f(a)) \le (g(b) - g(a)) \cdot \sup_{z \in B}\phi(z)\]

Suppose that $f(b) - f(a) \not\in [g(b) - g(a)]B$. Given that $B$ is closed and convex, by the Hahn-Banach theorem, there exists $\phi \in E^{*}$ such that

\[\phi(f(b) - f(a)) > \sup_{x \in B}\phi[(g(b) - g(a))b] = \phi(g(b) - g(a)) \cdot \sup_{x \in B}\phi(x)\]

which is impossible. Therefore $f(b) - f(a) \in [g(b) - g(a)]B$.$\square$

Theorem 18.3.4 (Mean Value Theorem). Let $-\infty < a < b < \infty$, $E$ be a separated locally convex space, $S \subset [a, b]$ be at most countable, and $f \in C([a, b]; E)$ be differentiable on $(a, b) \setminus N$, then

\[f(b) - f(a) \in \overline{\text{Conv}\bracs{Df(x)(b - a)| x \in (a, b) \setminus N}}\]

Proof. By Proposition 18.2.9, $f$ is right-differentiable on $(a, b) \setminus N$ with

\[D^{+}f(x) = Df(x) \in \bracs{Df(y)|y \in (a, b) \setminus N}\]

for all $x \in (a, b)$. Let $g(x) = x$, then by Theorem 18.3.3,

\[f(b) - f(a) \in \overline{(b - a)\text{Conv}\bracs{Df(x)|x \in (a, b) \setminus N}}\]

$\square$

Theorem 18.3.5 (Mean Value Theorem). Let $E$ be a topological vector space, $F$ be a separated locally convex space, $V \subset E$ be open and star shaped at $x \in V$, $f: V \to F$ be Gateau-differentiable on $V$, then for any $y \in V$,

\[f(y) - f(x) \in \overline{\text{Conv}\bracs{Df(z)(y - x)|z \in (x, y)}}\]

where $[x, y] = \bracs{(1 - t)x + ty|y \in [0, 1]}$.

Proof. Let $g: [0, 1] \to F$ be defined by $g(t) = f((1 - t)x + ty)$. Since $f$ is Gateaux-differentiable, $g$ is differentiable by the chain rule Proposition 18.2.7 with $Dg(t) = Df((1 - t)x + ty)(y - x)$, and continuous by Proposition 18.2.9.

By the Mean Value Theorem,

\[f(y) - f(x) = g(1) - g(0) \in \overline{\text{Conv}\bracs{Dg(t)|t \in [0, 1]}}= \overline{\text{Conv}\bracs{Df(z)(y - x)|z \in (x, y)}}\]

$\square$

Proposition 18.3.6. Let $E$ be a topological vector space, $F$ be a separated locally convex space, $V \subset E$ be open and connected, $f: V \to F$ be Gateaux-differentiable on $V$ such that $Df(x) = 0$ for all $x \in V$, then $f$ is constant.

Proof. Let $x \in V$, then for any $U \in \cn(0)$ circled with $U + x \subset V$ and $y \in U + x$,

\[f(y) - f(x) \in \overline{\text{Conv}\bracs{Df(z)(y - x)|z \in (x, y)}}= \bracs{0}\]

by the Mean Value Theorem.

Fix $x \in V$ and let $W$ be the connected component of $\bracs{y \in V|f(x) = f(y)}$ containing $x$. For any $y \in W$, by Proposition 8.1.11, there exists $U \in \cn(0)$ circled such that $U + y \subset V$. Therefore $W \supset W \cup (U + y)$, and $W$ is open by Lemma 4.4.3.

For any $y \in \ol W \cap V$, there exists $U \in \cn(0)$ circled such that $U + y \subset V$. As $y \in \ol W \cap V$, $U \cap W \ne \emptyset$. Thus $f(y) = f(x)$, $y \in W$, and $W$ is relatively closed.

Since $V$ is connected and $W \subset V$ is both open and closed, $W = V$.$\square$

Jerry's Digital Garden

18.3 The Mean Value Theorem

Direct References

Jerry's Digital Garden

Direct References