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Rudin asked:

For $f\in L^{\infty}(\mathbb{R}^{1})$, define $f_{t}(x)=f(x-t)$. Assume that $$\lim_{t\rightarrow 0}|f_{t}-f|_{\infty}=0$$Prove that under these conditions there is a uniformly continuous function $g$ such that $g=f$ almost everywhere.

The suggested strategy is to find a sequence of functions $h_{n}\rightarrow \delta$, and use $g_{n}=h_{n}*f$. It is easy to show that $g_{n}\rightarrow f$ almost everywhere(for example, a proof by contradiction or by dominated convergence theorem). However to bridge from $g_{n}$ to $g$ I encountered a difficulty; the classical Arzelà–Ascoli only works in compact spaces, and $g_{n}$ is not compact supported in general. Therefore it is not clear if $g_{n}\rightarrow g$ uniformly even if we only looking for a subsequence of the original sequence.

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What's $\delta$ here? Anyway, would it not be possible to select a sequence $h_n$ of compactly supported continuous functions with the desired limit? The convolution would preserve the nice properties you're looking for. –  anonymous Jan 13 '13 at 17:35
    
$R$ is not compact. –  Bombyx mori Jan 13 '13 at 17:39
    
Then why not just select a family $h_n$ of $C^{\infty}$ bump functions which form an approximate identity? Then $g_n$ is certainly a continuous function of compact support, since the convolution of two functions is at least as nice as the nicest. –  anonymous Jan 13 '13 at 17:39
    
$h_{n}$ can be as nice as possible, and $g_{n}$ is uniformly continuous since $|g_{n}(x+t)-g_{n}(x)|\le |f_{t}-f|_{\infty}$. But $g_{n}(x)=\int_{A}f(x+y)h(y)dy$ is not necessarily compact supported. It is well defined for any $x$. –  Bombyx mori Jan 13 '13 at 17:41
    
Is there a more direct proof? The condition above is 'essentially' uniform continuity. –  copper.hat Jan 13 '13 at 22:15

2 Answers 2

up vote 1 down vote accepted

Here is another way: (I thought I had an elementary proof, but I had to rely on a not entirely elementary result.)

Let $\phi(t) = \text{esssup}_x |f(x+t)-f(x)|$. We have $\lim_{t\to 0} \phi(t) = 0$.

Since $f \in L^\infty(\mathbb{R})$, we have $f \in L^1(K)$ for any bounded set $K$. Let $f_n(x) = \frac{n}{2} \int_{-\frac{1}{n}}^\frac{1}{n} f(x+t) dt $ (and note that $f_n(x) = \frac{1}{2} \int_{-1}^1f(x+\frac{t}{n}) dt$). Since $f$ is essentially bounded, it follows that $f_n$ is continuous, and hence $f_n \in C([-M,M])$ for any $M>0$.

Suppose $n,m \geq N$, then we have \begin{eqnarray} |f_n(x)-f_m(x)| &\leq& \frac{1}{2} \int_{-1}^1 |f(x+\frac{t}{n})- f(x+\frac{t}{m})|dt \\ &\leq & \frac{1}{2} \int_{-1}^1 |f(x+\frac{t}{n})-f(x)|+|f(x)- f(x+\frac{t}{m})|dt \\ &\leq& \sup_{|t| \leq \frac{1}{N}}\phi(t) \end{eqnarray} Note that by assumption $\lim_{N \to \infty} \sup_{|t| \leq \frac{1}{N}}\phi(t) = 0$. Choose $M>0$, then the above shows that $f_n$ is Cauchy in $C([-M,M])$, and hence converges to some $\hat{f} \in C([-M,M])$. It is clear that if $M'>M$, then is is clear that the limit coincides on $[-M,M]$, hence this procedure defines a continuous function $\hat{f}$ on all of $\mathbb{R}$.

Now for the not entirely elementary part: Since $f \in L^1(K)$ for any bounded $K$, it follows from the Lebesgue differentiation theorem that almost every $x$ is a Lebesgue point, ie, $\lim_{n \to \infty} f_n(x) = f(x)$ a.e. Since $\lim_{n\to \infty} f_n(x) = \hat{f}(x)$ everywhere, and $\hat{f}$ is continuous, we have the desired result.

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Unfortunately, my approach is not that different after all. If one lets $h_n = \frac{n}{2} 1_{[-\frac{1}{n},\frac{1}{n}]}$, then $f_n = f * h_n$. –  copper.hat Jan 14 '13 at 6:50
    
Indeed this is what Rudin hinted explicitly in his book. –  Bombyx mori Jan 14 '13 at 10:01

You do not need Arzelà–Ascoli here; the fact that $(g_n)$ converge uniformly can be shown directly. The key point is that $h_n$ should have small support when $n$ is large. This makes sure that the convolution integral involves only $f_t$ for small $t$.
$$ \sup |g_n-g_m| = \sup_x \left|\int h_n(t)f_t(x)\,dt- \int h_m(t)f_t(x)\,dt\right| \\ \le \epsilon + \sup_x \left|\int h_n(t)f(x)\,dt- \int h_m(t)f(x)\,dt\right| \\ = \epsilon + \sup_x \left|f(x)\cdot 1 - f(x)\cdot 1 \right| = \epsilon $$ Here the second step is based on $\|f_t-f\|_\infty\le \epsilon$.

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Thanks! I should ask - is there a nice way to show $g_{n}\rightarrow f$ a.e? My proof passing from convergence almost everywhere to convergence in measure (because $h_{n}$ as compact support) and then by contradiction feels over-complicated. The dominated convergence theorem gives pointwise convergence, so I feel I might have made a mistake. –  Bombyx mori Jan 13 '13 at 22:28
    
PavelPM, how did you become 5PM? –  Bombyx mori Jan 14 '13 at 7:42

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