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In this paper, the authors introduce on p.645 ($11^{th}$ page of the paper) the $\sup$ and $\inf$ convolutions for a continuous, bounded function $w:\mathbb{R}^n \times [0, \infty) \to \mathbb{R}$

$$ w^{\varepsilon}(x,t) = \sup_{y,s} \left( w(y,s) - \frac{1}{\varepsilon} \left(|x-y|^2 + (t-s)^2 \right)\right)$$

$$w_\varepsilon(x,t) = \inf_{y,s}\left( w(y,s) + \frac{1}{\varepsilon} \left(|x-y|^2 + (t-s)^2 \right)\right) $$

Amoung the properties they cite for these functions, they list:

  1. $w_\varepsilon \leq w \leq w^\varepsilon$
  2. $||w^\varepsilon||_\infty, ||w_\varepsilon||_\infty \leq ||w||_\infty$
  3. If $w^\varepsilon(x,t) = w(y,s) - \frac{1}{\varepsilon}\left(|x-y|^2 + (t-s)^2 \right)$ then $|x-y|,|t-s| \leq \sqrt{ 2 ||w||_\infty \varepsilon}$
  4. $w^\varepsilon, w_\varepsilon \to w$ uniformly as $\varepsilon \to 0^+$ on compact subsets of $\mathbb{R}^n \times [0, \infty)$

The first three are not so hard to show and build upon one another. For instance, using 2, one can prove 3 by contradiction.

However, 4 is giving me some trouble. The authors indicate that 4 should follow from the uniform continuity of $w$ on compact sets. My attempt was the following:

We want to show that for given $\delta>0$, there is a $\gamma>0$ such that $\gamma > \varepsilon >0$ implies $|w(x,t) - w^\epsilon(x,t)| < \delta$.

For given $\varepsilon >0$, 3 implies $$w^\varepsilon(x,t) = \sup_{|x-y|,|t-s| \leq \sqrt{ 2 ||w||_\infty \varepsilon}} \left( w(y,s) - \frac{1}{\varepsilon} \left(|x-y|^2 + (t-s)^2 \right)\right)$$ So, we can replace the $\sup$ with a max. Thus, for any $\varepsilon$ there are $y_\varepsilon, s_\varepsilon$ such that

$$ w^\epsilon(x,t) = w(y_\varepsilon,s_\varepsilon) - \frac{1}{\varepsilon} \left(|x-y_\varepsilon|^2 + (t-s_\varepsilon)^2 \right)$$

Thus, $|w(x,t) - w^\epsilon(x,t)| \leq |w(x,t) - w(y_\varepsilon, s_\varepsilon)| + \frac{1}{\varepsilon} \left(|x-y_\varepsilon|^2 + (t-s_\varepsilon)^2 \right)$

This is where I ran into trouble. The first term I can bound by uniform continuity. However, I do not see how I can bound the second term by anything that goes to $0$, since we might have $|x-y_\varepsilon| = \sqrt{ 2 ||w|_\infty \epsilon}$ , giving:

$$ \frac{1}{\varepsilon} \left(|x-y_\varepsilon|^2 + (t-s_\varepsilon)^2 \right) \geq 2 ||w||_\infty$$

I must be missing something but I cannot tell what it is. If someone could let me know what it is, I would be very appreciative. How do we make sure that the last term goes to 0? Does anyone have a hint?

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msm
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1 Answers1

2

Since $w^\epsilon\geq w$, we have $|w^\epsilon(x,t)-w(x,t)| = w^\epsilon(x,t) - w(x,t) = w(y_\epsilon,s_\epsilon) - \frac{1}{\epsilon}(|x-y_\epsilon|^2+|t-s_\epsilon|^2) - w(x,t) \leq w(y_\epsilon, s_\epsilon) - w(x,t).$

Elcyc
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