- #1

psie

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- TL;DR Summary
- I have a question about a proof on Lipschitz continuity of a vector-valued function.

I'm reading Ordinary Differential Equations by Andersson and Böiers, although this is more related to multivariable calculus. There is a Lemma regarding Lipschitz continuity which I have a question about. Below ##\pmb{f}:\mathbf{R}^{n+1}\to \mathbf{R}^n ## is a vector-valued function defined by ##\pmb{f}=\pmb{f}(t,\pmb{x})##, where ##\pmb x## is a vector in ##\mathbf{R}^n## that depends on ##t##.

Here's my understanding of the notation. In ##(2)##, ##\frac{\partial \pmb{f}}{\partial x_i}(t,\pmb{y}+s(\pmb{x}-\pmb{y}))## is a vector evaluated at ##(t,\pmb{y}+s(\pmb{x}-\pmb{y}))##. ##\frac{\partial}{\partial x_i}## denotes the partial derivative of ##\pmb{f}(t,x_1,\ldots,x_n)## with respect ##x_i## (a better notation in ##(2)## would probably be ##\frac{\partial}{\partial z_i}##, as ##\pmb{x}## is already being used to denote a vector). More precisely, if ##f_1,\ldots,f_n## are the components of ##\pmb{f}##, then $$\frac{\partial \pmb{f}}{\partial x_i}(t,\pmb{y}+s(\pmb{x}-\pmb{y}))=\begin{pmatrix}

\frac{\partial f_1}{\partial x_i}(t,\pmb{y}+s(\pmb{x}-\pmb{y})) \\ \vdots \\ \frac{\partial f_n}{\partial x_i}(t,\pmb{y}+s(\pmb{x}-\pmb{y}))

\end{pmatrix}.\tag4$$

My question is that I do not really understand how ##(3)## is obtained. For the sake of brevity, I will write ##\frac{\partial \pmb{f}}{\partial x_i}## when I mean ##\frac{\partial \pmb{f}}{\partial x_i}(t,\pmb{y}+s(\pmb{x}-\pmb{y}))##. Suppose ##n=2##. Taking the norm (##\ell^2## norm to be exact) of ##(1)## and using ##\left\lVert\int_a^b \pmb{f}\right\rVert\le\int_a^b\left\lVert\pmb{f}\right\rVert##, we get

\begin{align}

\lVert\pmb{f}(t,\pmb{x})-\pmb{f}(t,\pmb{y})\rVert_2&=\left\lVert\int_0^1\sum_{i=1}^2\frac{\partial \pmb{f}}{\partial x_i}(x_i-y_i)\mathrm{d}s\right\rVert_2 \tag5\\

&\leq \int_0^1\left\lVert\sum_{i=1}^2\frac{\partial \pmb{f}}{\partial x_i}(x_i-y_i) \right\rVert_2 \mathrm{d}s \tag6

\end{align}

Next, using the triangle inequality, the norm inside the integral gets bounded by

\begin{align}

\left\lVert\sum_{i=1}^2\frac{\partial \pmb{f}}{\partial x_i}(x_i-y_i) \right\rVert_2 &\leq\left\lVert \frac{\partial \pmb{f}}{\partial x_1}(x_1-y_1)\right\rVert_2+\left\lVert \frac{\partial \pmb{f}}{\partial x_2}(x_2-y_2)\right\rVert_2. \tag7

\end{align}

The above terms on the right-hand side equal

\begin{align}

\sqrt{\left(\frac{\partial f_1}{\partial x_1}\right)^2+\left(\frac{\partial f_2}{\partial x_1}\right)^2}|x_1-y_1|+\sqrt{\left(\frac{\partial f_1}{\partial x_2}\right)^2+\left(\frac{\partial f_2}{\partial x_2}\right)^2}|x_2-y_2|.\tag8

\end{align}

Letting ##K=\max\limits_{i,j\in\{1,2\}}\sup\limits_{\Omega}\left|\frac{\partial f_i}{\partial x_j}\right|##, ##(8)## gets bounded by

\begin{align}

\sqrt{2}K|x_1-y_1|+\sqrt{2}K|x_2-y_2|=\sqrt{2}K(|x_1-y_1|+|x_2-y_2|) \tag9

\end{align}

which is clearly not equal to ##nK|\pmb{x}-\pmb{y}|##. I suspect I am doing something wrong, but I'm unable to spot it.

Lemma.Assume that ##\Omega\subseteq \mathbf{R}\times\mathbf{R}^n## is a convex and bounded set, and that the function ##\pmb{f}## is continuously differentiable in a neighborhood of ##\overline{\Omega}##. Then ##\pmb{f}## is Lipschitz continuous in ##\Omega##.

Proof.The line segment between two points ##(t,\pmb{x})## and ##(t,\pmb{y})## in ##\Omega## is contained in ##\Omega## by the assumption of convexity. Hence \begin{align} \pmb{f}(t,\pmb{x})-\pmb{f}(t,\pmb{y})&=\int_0^1\frac{\mathrm{d}}{\mathrm{d}s}\pmb{f}(t,\pmb{y}+s(\pmb{x}-\pmb{y}))\mathrm{d}s \tag1 \\ &=\int_0^1\sum_{i=1}^n\frac{\partial \pmb{f}}{\partial x_i}(t,\pmb{y}+s(\pmb{x}-\pmb{y}))(x_i-y_i)\mathrm{d}s,\tag2\end{align} by the chain rule. Set ##K=\max_i\sup_{\Omega}\left|\frac{\partial f}{\partial x_i}\right|## (which exists by assumption). Then $$|\pmb{f}(t,\pmb{x})-\pmb{f}(t,\pmb{y})|\leq\int_0^1 nK|\pmb{x}-\pmb{y}|\mathrm{d}s=nK|\pmb{x}-\pmb{y}|.\tag3$$

Here's my understanding of the notation. In ##(2)##, ##\frac{\partial \pmb{f}}{\partial x_i}(t,\pmb{y}+s(\pmb{x}-\pmb{y}))## is a vector evaluated at ##(t,\pmb{y}+s(\pmb{x}-\pmb{y}))##. ##\frac{\partial}{\partial x_i}## denotes the partial derivative of ##\pmb{f}(t,x_1,\ldots,x_n)## with respect ##x_i## (a better notation in ##(2)## would probably be ##\frac{\partial}{\partial z_i}##, as ##\pmb{x}## is already being used to denote a vector). More precisely, if ##f_1,\ldots,f_n## are the components of ##\pmb{f}##, then $$\frac{\partial \pmb{f}}{\partial x_i}(t,\pmb{y}+s(\pmb{x}-\pmb{y}))=\begin{pmatrix}

\frac{\partial f_1}{\partial x_i}(t,\pmb{y}+s(\pmb{x}-\pmb{y})) \\ \vdots \\ \frac{\partial f_n}{\partial x_i}(t,\pmb{y}+s(\pmb{x}-\pmb{y}))

\end{pmatrix}.\tag4$$

My question is that I do not really understand how ##(3)## is obtained. For the sake of brevity, I will write ##\frac{\partial \pmb{f}}{\partial x_i}## when I mean ##\frac{\partial \pmb{f}}{\partial x_i}(t,\pmb{y}+s(\pmb{x}-\pmb{y}))##. Suppose ##n=2##. Taking the norm (##\ell^2## norm to be exact) of ##(1)## and using ##\left\lVert\int_a^b \pmb{f}\right\rVert\le\int_a^b\left\lVert\pmb{f}\right\rVert##, we get

\begin{align}

\lVert\pmb{f}(t,\pmb{x})-\pmb{f}(t,\pmb{y})\rVert_2&=\left\lVert\int_0^1\sum_{i=1}^2\frac{\partial \pmb{f}}{\partial x_i}(x_i-y_i)\mathrm{d}s\right\rVert_2 \tag5\\

&\leq \int_0^1\left\lVert\sum_{i=1}^2\frac{\partial \pmb{f}}{\partial x_i}(x_i-y_i) \right\rVert_2 \mathrm{d}s \tag6

\end{align}

Next, using the triangle inequality, the norm inside the integral gets bounded by

\begin{align}

\left\lVert\sum_{i=1}^2\frac{\partial \pmb{f}}{\partial x_i}(x_i-y_i) \right\rVert_2 &\leq\left\lVert \frac{\partial \pmb{f}}{\partial x_1}(x_1-y_1)\right\rVert_2+\left\lVert \frac{\partial \pmb{f}}{\partial x_2}(x_2-y_2)\right\rVert_2. \tag7

\end{align}

The above terms on the right-hand side equal

\begin{align}

\sqrt{\left(\frac{\partial f_1}{\partial x_1}\right)^2+\left(\frac{\partial f_2}{\partial x_1}\right)^2}|x_1-y_1|+\sqrt{\left(\frac{\partial f_1}{\partial x_2}\right)^2+\left(\frac{\partial f_2}{\partial x_2}\right)^2}|x_2-y_2|.\tag8

\end{align}

Letting ##K=\max\limits_{i,j\in\{1,2\}}\sup\limits_{\Omega}\left|\frac{\partial f_i}{\partial x_j}\right|##, ##(8)## gets bounded by

\begin{align}

\sqrt{2}K|x_1-y_1|+\sqrt{2}K|x_2-y_2|=\sqrt{2}K(|x_1-y_1|+|x_2-y_2|) \tag9

\end{align}

which is clearly not equal to ##nK|\pmb{x}-\pmb{y}|##. I suspect I am doing something wrong, but I'm unable to spot it.