Strichartz, page 123

On page 123, Strichartz makes the claim that it is simple to see the equivalence between two somewhat vaguely worded statements. The equivalence is essentially true, and while it is simple in one direction, does not appear to be simple in the other. It seems a mistake to claim it was simple, but worse still, the direction in which it seems non-obvious has very little importance and only serves as a major distraction to the reader.

The important direction is that the Lipschitz condition implies uniform continuity and this was shown in class and is included here as Fact 1.

Fact 1 Suppose that $f:\Re \to \Re$ and there exists a constant M such that for all x,x₀, it follows that $\vert f(x)-f(x_0)\vert \leq M\vert x-x_0\vert.$ Then for all 1/m we can say that if $\vert x-x_0\vert < \frac{1}{mM}$ then |f(x)-f(x₀)| < 1/m. Hence f is uniformly continuous.

Proof Suppose the assertion is true. Let 1/m be given and suppose $\vert x-x_0\vert < \frac{1}{Mm}$ then

$$\vert f(x)-f(x_0)\vert \leq M\vert x-x_0\vert < \frac{M}{Mm} = 1/m$$

as claimed. That f is uniformly continuous now follows easily. Let 1/m be given and pick $1/n = \frac{1}{Mm}.$ If $\vert x-x_0\vert < 1/n = \frac{1}{Mm}$ then it follows from the above that |f(x)-f(x₀)| < 1/m. This establishes uniform continuity.     QED

Note that in proving uniform continuity, it is not always the case that 1/n is a constant multiple of 1/m. A simple example where 1/n=1/m² is shown in Fact 2.

Fact 2 Let $f(x) = \sqrt{x}.$ Then f is uniformly continuous on $[0,\infty).$

Proof Let 1/m be given. Pick $1/n= \frac{1}{m^2}$ Suppose $x,x_0 \geq 0$ and $\vert x-x_0\vert < 1/n= \frac{1}{m^2}.$ There are two cases. In case 1, if $x< \frac{1}{m^2}$ and $x_0 < \frac{1}{m^2}$ then

$$\vert\sqrt{x} - \sqrt{x_0}\vert \leq max(\sqrt{x},\sqrt{x_0}) < \sqrt{1/m^2} = 1/m$$

In case 2, if $x \geq \frac{1}{m^2}$ or $x_0 \geq \frac{1}{m^2}$ then $\sqrt{x} + \sqrt{x_0} \geq max(\sqrt{x},\sqrt{x_0}) \geq \sqrt{\frac{1}{m^2}} = 1/m.$ Hence

$$\vert\sqrt{x} - \sqrt{x_0}\vert = \frac{\vert x-x_0\vert}{\s... ...0}} \leq \frac{\vert x-x_0\vert}{1/m} < \frac{1/m^2}{1/m} = 1/m$$

The two cases cover all possibilities and thus if x,x₀>0 and $\vert x-x_0\vert < \frac{1}{n}$then $\vert\sqrt{x} - \sqrt{x_0}\vert < 1/m.$ Hence f is uniformly continuous.    QED

It turns out that if the statements are formalized properly, the reverse direction of the equivalence claim made by Strichartz is in fact true, but due to the fact that 1/m cannot be made large for positive integers m, it is more difficult to show. As mentioned earlier, it is of little importance, but for completeness, the formal statement and the proof is included below.

Fact 3 Suppose that $f:\Re \to \Re$ and that there exists a positive integer M such that for all 1/m we can say that if $\vert y-y_0\vert < \frac{1}{mM}$ then |f(y)-f(y₀)| < 1/m. Then $\vert f(x)-f(x_0)\vert \leq M\vert x-x_0\vert$ for all $x,x_0 \in \Re.$

Proof Suppose the hypothesis holds for a positive integer M. Let $x,x_0\in \Re$ be given. For all positive integers m there exists a positive integer N such that

$$\frac{N-1}{mM} \leq \vert x-x_0\vert < \frac{N}{mM}.$$

We will use these inequalities in the following forms:

$$\frac{\vert x-x_0\vert}{N} < \frac{1}{mM} ~~~{\rm and}~~~ \frac{N-1}{m} \leq M\vert x-x_0\vert.$$

Now, for $i=1,2,\dots,N$ let x_i = x₀ + i(x-x₀)/N. Thus x_N=x and $\vert x_i - x_{i-1}\vert = \vert x-x_0\vert/N < \frac{1}{mM}$ for $i=1,\dots,N$ and so by the hypothesis $\vert f(x_i)- f(x_{i-1})\vert < \frac{1}{m}.$ Therefore,

$$\begin{eqnarray*} \vert f(x)-f(x_0\vert & = & \vert f(x_N)-f(x_0)\vert \\ & = &... ...frac{N-1}{m} + \frac{1}{m} \leq M\vert x-x_0\vert + \frac{1}{m}. \end{eqnarray*}$$

Thus we have shown $\vert f(x)-f(x_0)\vert < M\vert x-x_0\vert + \frac{1}{m}$ for all positive integers m and it follows that
$\vert f(x)-f(x_0)\vert \leq M\vert x-x_0\vert.$     QED