Problems that can be shown by ”an epsilon of room”.

Question

This might not be a question suitable here so apologies from posting it if so.

In Terence Tao’s blog he refers to something called ”An epsilon of room” and this seems to be a helpful way to prove for example inequalities in analysis. He goes on to say that

If one has to show that $X \le Y$, try proving that $X \le Y+ \varepsilon$ for any $\varepsilon >0$.

I tried to find examples online from where I could gather a collection of problems that can be shown by his proposed method, but there isn’t much to find. Or perhaps most of the problems are ”masked” in a way that I don’t see this particular trick being used. If anyone knows some of these kind of problems I would appreciate if they could be posted here.

Answer 1

Here are some statements that come to mind:

1. The Lindelöf's theorem and its variants
2. The theorem that every probability measure defined on a Polish space is inner regular
3. The theorem that the continuous extension of a Lipschitz-continuous function that had previously only been defined on a dense subset is again Lipschitz-continuous with the same Lipschitz constant

Answer 2

This happens frequently in the theory of elliptic and parabolic partial differential equations.

Here is a very classical example.

A $C^{2}$ function $u$ in an open set $\Omega$ is harmonic if its Laplacian vanishes everywhere, i.e. $-\Delta u(x) = 0$ for all $x \in \Omega$.

Maximum principle: If $u : \overline{\Omega} \to \mathbb{R}$ is continuous in $\overline{\Omega}$ and twice continuously differentiable in $\Omega$, then $\max\left\{u(x) \, \mid \, x \in \Omega\right\} = \max\left\{u(y) \, \mid \, y \in \partial \Omega\right\}$.

Proof: It's convenient to argue by contradiction. Suppose, then, that the conclusion is false, i.e. $\max\{u(z) \, \mid \, z \in \partial \Omega\} < \max\{u(x) \, \mid \, x \in \Omega\}$.

I will take you up on that offer of $\epsilon$ of room. Consider the function $v_{\epsilon}(x) = \max\{u(x) \, \mid \, x \in \Omega\} - \epsilon \|x\|^{2}$. $u - v_{\epsilon}$ has a maximum $x_{\epsilon}$ somewhere in $\overline{\Omega}$. For $\epsilon$ small enough, $x_{\epsilon} \in \Omega$ --- otherwise, $u$ would attain its maximum somewhere on the boundary. (We have "room" to work with.) Thus, since $x_{\epsilon}$ is an local maximum (in the interior), I can apply the the second derivative test to find \begin{equation*} -\Delta u(x_{\epsilon}) + \Delta v_{\epsilon}(x_{\epsilon}) = -\Delta (u - v_{\epsilon})(x_{\epsilon}) \geq 0. \end{equation*} so $-\Delta u(x_{\epsilon}) \geq -\Delta v_{\epsilon}(x_{\epsilon}) = 2 \epsilon d$. This is impossible since $u$ is harmonic.

With experience, one realizes there is a geometric interpretation of this proof that is perfectly suited to the intuition that "we have room," but it would be cumbersome to explain here.