Let $h\colon[0,1]\to \mathbb{R}^+$ be any bounded measurable non-negative function with a unique maximum at $a$ and $h$ is continuous at $a$. For $\lambda>0$ define $h_\lambda(x)=C_\lambda h(x)^\lambda$ where $C_\lambda$ normalizes such that $\displaystyle\int_0^1 h_\lambda(x)\,dx=1$. $f$ is any continuous function on $[0,1]$ and $\epsilon>0$. Are the following assertions true?

1)$\displaystyle\lim\limits_{\lambda\to\infty}\int_{h(x)\le h(a)−\epsilon}h_λ(x)f(x)\,dx=0$ and $\displaystyle\lim\limits_{\lambda\to\infty}\int_0^1 h_λ(x)f(x)\,dx=f(a)$. In words, the limit of $h_\lambda$ is the Dirac delta function $\delta(\cdot-a)$.

2) If $h$ is continuous, $h_\lambda(x)$ converges uniformly to $0$ in $\{x: h(x) \le h(a)-\epsilon\}$.

It seems correct since the exponentiation suppresses the part somewhere below $h(a)$ while heightens the part of $h$ near $a$. The difficulty seems to be giving a partitions of $[0,1]$ where I can utilize the exponentiation to either suppress or heighten $h$ by the exponentiation of $\lambda$ after $h$ is normalized by $C_\lambda$.

This question is a generalization of my more specific case, and was suggested by one commentator whuber. However, he did not supply a proof and I am stumped by the difficulty described the last paragraph.

  • $\begingroup$ I have now edited my question and provided more background, my attempt at its solution and my difficulty. Could the down-voter and closing proponent revise their votes? Or, explicate their reason if they maintain their position? $\endgroup$
    – Hans
    May 9 '15 at 23:55

In this answer I will assume that $h$ is defined on $\mathbb R$ instead of $[0,1]$ and with $\int h <\infty$ (so that I do not need to write $\cap [0,1]$ all the time). Also I will make an extra assumption on $h$ (You will see).

For any $d>0$ we have

\begin{equation} \begin{split} \bigg|\int_{\mathbb R} &h_\lambda(x) f(x) dx -f(a) \bigg| =\bigg|\int_{\mathbb R} h_\lambda(x) (f(x) -f(a)) dx \bigg| \\ &\le \bigg|\int_{|x-a|\le d} h_\lambda (x) (f(x)-f(a)) dx \bigg|+ \bigg|\int_{|x-a|> d} h_\lambda (x) (f(x)-f(a)) dx\bigg|\\ &\le \int_{|x-a|\le d} h_\lambda (x) \big|f(x)-f(a)\big| dx + \max \big|f(x)-f(a)\big|\bigg|\int_{|x-a|> d} h_\lambda (x) dx\bigg|\\ \end{split} \end{equation}

Let now $\epsilon>0$. Then as $f$ is continuous there is $c>0$ so that $|f(x) - f(a)|<\epsilon/2$ whenever $|x-a|<c$. So as long as $c>d$ we have

$$\bigg|\int_{\mathbb R} h_\lambda(x) f(x) dx -f(a) \bigg| \le \frac{\epsilon}{2} + \max\big|f(x)-f(a)\big|\bigg|\int_{|x-a|>d} h_\lambda (x) dx\bigg|$$

Now the more difficult task is to show that

$$(*)\ \ \ \ \bigg|\int_{|x-a|> d} h_\lambda (x) dx\bigg| \to 0 \ \ \text{as } \lambda \to \infty.$$

For this one I will use that $$\|h\|_{\infty, d} < \|h\|_\infty$$

for all $d>0$, where $\|h\|_{\infty, d}$ is the $L^\infty$ norm of $h$ when restricted to $\{|x-a|>d\}$ (Note the strict inequality. That is not implied by your assumption).

Now note that for any nonnegative integrable functions $h$ we have

$$\bigg(\int_\mathbb{R} h^\lambda\bigg)^\frac{1}{\lambda} \to \|h\|_\infty\ \ \text{as }\lambda \to \infty$$

So as by definition,

$$C_\lambda = \frac{1}{\int_\mathbb{R} h^\lambda} \Rightarrow , C_\lambda^\frac{1}{\lambda} \to \frac{1}{\|h\|_\infty}$$

this implies that

$$\bigg(\int_{|x-a|>d} h_\lambda\bigg)^\frac{1}{\lambda} = C_\lambda^\frac{1}{\lambda} \bigg(\int_{|x-a|>d} h^\lambda\bigg)^\frac{1}{\lambda}\to \frac{\|h\|_{\infty, d}}{\|h\|_\infty} <1$$

by the extra assumption I made. Thus there is $\lambda_0$ so that

$$\bigg(\int_{|x-a|>d} h_\lambda\bigg)^\frac{1}{\lambda} <B <1$$

for all $\lambda >\lambda_0$. In particular,

$$\int_{|x-a|>d} h_\lambda < B^\lambda$$

for all $\lambda >\lambda_0$. As $B<1$, $B^\lambda \to 0$. This implies (*) and we are done.

(Note that continuity of $h$ at $a$ is not needed, only some sort of "continuity in measure at $a$" is used).

  • $\begingroup$ Excellent! Questions: by $\|h\|_{\{|x-a|>d\}}$ you mean $\|hI(|x-a|>d)\|_\infty$, where $I$ is the characteristic function, right? In the case $h$ is continuous, we can conclude $\|hI(|x-a|>d)\|_\infty< \|h\|_\infty$, right? If so, what weaker conditions would you suggest to maintain that result? $\endgroup$
    – Hans
    May 10 '15 at 18:56
  • $\begingroup$ @Hans: Yes, you are right. The answer is edited. If your $h$ is continuous, then the extra assumption is satisfied. $\endgroup$
    – user99914
    May 10 '15 at 19:14
  • $\begingroup$ Great. Would you suggest a few example conditions of "continuity in measure at $a$"? Would those imply $\|h\|_{\infty,d}<\|h\|_\infty$? $\endgroup$
    – Hans
    May 10 '15 at 19:19
  • $\begingroup$ @Hans: I am using that term quite loosely. I was think of a function which has unique max at a, but is not continuous there. However it has the property that $||hI||_\infty = f(a)$, where $I = \{|x-a|<c\}$ for all $c$. $\endgroup$
    – user99914
    May 10 '15 at 19:38
  • $\begingroup$ I understand you were using the term loosely and I am very comfortable with the notion. I am just wondering if you can name examples of some commonly used conditions that can also guarantee the strict inequality on the maximum norm. $\endgroup$
    – Hans
    May 10 '15 at 19:43

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