1.2 Mean Value Theorem

The mean value theorem and its corollary provide the analytical tools needed to verify contraction properties of Newton-like operators.

Definition 1.2.1 Operator norm

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Let \((X, \| \cdot \| _X)\) and \((Y, \| \cdot \| _Y)\) be normed linear spaces and let \(A: X \rightarrow Y\) be a continuous linear map. The operator norm on \(A\) is given by

\begin{equation*} \| A\| := \sup _{x \in X \setminus \{ 0\} } \frac{\| Ax\| _Y}{\| x\| _X} = \sup _{\| x\| _X = 1} \| Ax\| _Y. \end{equation*}

Proposition 1.2.2 Properties of operator norm

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Let \((X, \| \cdot \| )\) be a normed linear space and let \(A, B: X \rightarrow X\) be linear maps. Then:

(1) \(\| Ax\| \leq \| A\| \| x\| \) for all \(x \in X\)

(2) \(\| AB\| \leq \| A\| \| B\| \) (submultiplicativity)

Theorem 1.2.3 Mean Value Theorem

Suppose that \(U \subset \mathbb {R}^n\) is open and that \(f: U \rightarrow \mathbb {R}^n\) is \(C^1\). Consider \(x, y \in U\) such that the line segment \((1-t)x + ty\), \(t \in [0,1]\), is contained in \(U\). Then,

\begin{equation*} f(y) - f(x) = \left(\int _0^1 \text{Df}((1-t)x + ty) \, dt\right)(y-x), \end{equation*}

where \(\text{Df}\) denotes the Jacobian of \(f\).

Proof ▶

For each \(i \in \{ 1,\ldots ,n\} \), define \(g_i: [0,1] \rightarrow \mathbb {R}\) by \(g_i(t) = f_i((1-t)x + ty)\). By the fundamental theorem of calculus and chain rule:

\begin{equation*} f_i(y) - f_i(x) = g_i(1) - g_i(0) = \int _0^1 g’_i(t) \, dt = \int _0^1 \text{D}f_i((1-t)x + ty) \cdot (y-x) \, dt. \end{equation*}

Collecting all components yields the result.

Corollary 1.2.4 Mean Value Inequality

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Consider an open set \(U \subset \mathbb {R}^n\). Let \(f: U \rightarrow \mathbb {R}^n\) be a \(C^1\) function. Fix a point \(x_0 \in U\) and assume \(\overline{B}_\rho (x_0) \subset U\) for some \(\rho {\gt} 0\). Then, for all \(x, y \in \overline{B}_\rho (x_0)\),

\begin{equation*} \| f(y) - f(x)\| \leq \left(\sup _{z \in \overline{B}_\rho (x_0)} \| \text{Df}(z)\| \right) \| y-x\| , \end{equation*}

where \(\| \text{Df}(\cdot )\| \) denotes the operator norm.

Proof ▶

The line segment from \(x\) to \(y\) lies in \(\overline{B}_\rho (x_0)\) by convexity. By the Mean Value Theorem:

\begin{equation*} \| f(y) - f(x)\| = \left\| \int _0^1 \text{Df}((1-t)x + ty)(y-x) \, dt \right\| \leq \int _0^1 \| \text{Df}((1-t)x + ty)\| \| y-x\| \, dt. \end{equation*}

Since \((1-t)x + ty \in \overline{B}_\rho (x_0)\) for all \(t \in [0,1]\), we have \(\| \text{Df}((1-t)x + ty)\| \leq \sup _{z \in \overline{B}_\rho (x_0)} \| \text{Df}(z)\| \), yielding the result.