Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've been working on this problem whilst studying for a comprehensive exam, and I've come up with a solution that I don't like. I'd appreciate some critiques. Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a uniformly continuous function with the property that $f\in L^{1}\left(\mathbb{R}\right)$. The claim is that $f$ is necessarily bounded on $\mathbb{R}$. My solution that I don't like is as follows:

Suppose to the contrary that $f$ fails to be bounded on $\mathbb{R}$. Let $M\in\mathbb{R}^+$ be given, and find $x_0\in\mathbb{R}$ satisfying $\left|f\left(x_0\right)\right|>M$. Because $f$ is, by assumption, continuous, there is a $\delta>0$ such that for all $x\in\left(x_0-\delta,x_0+\delta\right)$, we have $\left|f(x)\right|>M$. It follows that $$\int_{\mathbb{R}}{|f|\hspace{3pt}dm}\geq\int_{\left(x_0-\delta,x_0+\delta\right)}{|f|\hspace{3pt}dm}\geq\int_{\left(x_0-\delta,x_0+\delta\right)}{M\hspace{3pt}dm}=M\cdot 2\delta.$$ But $M$ was arbitrary, so we have that, essentially, $$\int_{\mathbb{R}}{|f|\hspace{3pt}dm}\geq M$$ for any $M\in\mathbb{R}^+$, and therefore $$\int_{\mathbb{R}}{|f|\hspace{3pt}dm}=\infty.$$ Hence $f\notin L^{1}(\mathbb{R})$, a contradiction. Therefore it must be the case that our assumption that $f$ was unbounded is false. $\therefore f$ is bounded, as claimed.

Primarily, I don't like the fact that I seemed not to use $uniform$ continuity of $f$ in the proof. I'd appreciate any thoughts, critiques, etc.

share|improve this question
1  
Since there are unbounded integrable continuous functions, a correct proof of the proposition must make use of the uniform continuity. Try to find the mistake in your proof. If you use the uniform continuity there, you can fix it. –  Daniel Fischer Jul 16 at 18:22
    
If $f$ is only continuous, then $\delta$ could depend on $x_0$ and thus on $M$, which would make your proof invalid (why?). But as $f$ is uniformly continuous, you can take $\delta$ independent of $x_0$ (choose $\delta$ first, (how exactly?) before you say anything about $M$). Further exercise: show that $f(x) \to 0$ as $x \to \pm \infty$. –  PhoemueX Jul 16 at 18:28
2  
To elaborate on @DanielFischer comment, imagine a piecewise linear function whose graph is zero for $x < 0$ and for $x \geq 0$ a sequence of triangles, each with area 1, 1/2, 1/4, etc. Then the function is continuous, and integrable (since the integral would be 2), but is evidently unbounded since we can make the height of the triangles arbitrarily large. This function cannot be uniformly continuous. –  user38584 Jul 16 at 18:28
    
In general you may note that if $ x_0$ is a singular point of $f$ then $f$ is uniformly cts. iff $ f→0 $ as $ x→x_0$. So that unbounded near a point $ x_0$ implies that $ f $ not uni –  mwomath Jul 16 at 19:45
1  
I think I've got it: Take $\varepsilon:=1$, and find the corresponding $\delta>0$ in the u.c. condition. $Now$ take $M\in\mathbb{R}^+$, and find a point $x_0$ witnessing the failure of $f$ to be bounded by $M+1$, and then basically repeat the proof as above with the caveat that, since $\delta$ no longer depends upon the choice of $M$, we can send $M\rightarrow\infty$ and contradict integrability. –  anonymous Jul 16 at 21:36

1 Answer 1

up vote 1 down vote accepted

Why not show even more: Actually, $f\in C_0(\mathbb{R})$, that is, $\lim_{x\to-\infty}|f(x)|=\lim_{x\to+\infty}|f(x)|=0$ (we say that $f$ vanishes at $\infty$). This implies that $f$ is bounded.

Let's treat the case $x\to+\infty$. If we didn't have $\lim_{x\to+\infty}|f(x)|=0$, there would exist $\varepsilon>0$ and an increasing sequence $x_n\to\infty$ with $|f(x_n)|>\varepsilon$. Taking subsequences, assume $x_{n+1}>x_n$. Uniform continuity gives us $\delta<1$ such that for every $n$ and for every $x\in (x_n-\delta,x_n+\delta)$, we have $|f(x)|\geq\varepsilon/2$. Then $$\int_\mathbb{R}|f(x)|dx\geq\sum_{n=1}^\infty\int_{x_n-\delta}^{x_n+\delta}|f(x)|dx\geq\sum_{n=1}^\infty \delta\varepsilon=\infty,$$ a contradiction.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.