Let $f\in L^1(\mathbb R)$ and let $F,G:\mathbb R\to\mathbb R$ be the functions defined by: ...

Question

I have applied for a Ph.D. in Trieste and am preparing for the exams. I am having a problem with Problem 8 here. Here is the text.

Let $f\in L^1(\mathbb R)$ and let $F,G:\mathbb R\to\mathbb R$ be the functions defined by:

$$F(x)=\int_x^{x+1}f(t)dt,\qquad\text{and}\qquad G(x)=\left|\int_x^{x+1}f(t)dt\right|.$$

(a) Prove that $G$ has a maximum point on $\mathbb R$.
(b) Give an example of $f\in L^1(\mathbb R)$ such that $F$ has no maximum point on $\mathbb R$.

Now, unless I'm much mistaken (proof at question end), we have:

1. $F,G$ continuous on $\mathbb R$;
2. $F,G$ tend to 0 as $|x|\to\infty$.

With that, by 2., both $F$ and $G$ are less than their sup-norms on $\mathbb R$ whenever $|x|>M$ for $M$ big enough, and by 1. and the compactness of $[-M,M]$ they must have a maximum on $[-M,M]$, which is then a global maximum on $\mathbb R$. So (a) is done, and (b)… is asking me to disprove the maximum of $F$ which I just proved, so it is impossible! Is my reasoning above correct? Are the proofs below correct? Or is there anything I am missing that is wrong in them?

Proofs

$F,G\to0$ as $|x|\to\infty$

I write $|x|\to\infty$ to say $x\to\infty$ or $x\to-\infty$, so let's do $x\to\infty$, and $x\to-\infty$ is proved the same way, more or less. Now:

$$|F(x)|\leq\int_x^{x+1}|f(t)|dt\leq\int_x^{+\infty}|f(t)|dt,$$

which tends to zero for $x\to+\infty$ since $f\in L^1(\mathbb R)$. And of course $G$ is already handled this way.

For $x\to-\infty$:

$$|F(x)|\leq\int_{-\infty}^{x+1}|f(t)|dt.$$

Continuity

We rewrite:

$$F(x)=\int_{-\infty}^{+\infty}f(t)1_{[x,x+1]}(t)dt,$$

$1_A$ being the indicator of $A$. Suppose $x\to x_0$. If $t<x_0$, then $t<x$ eventually, so that $f(t)1_{[x,x+1]}(t)=0$ eventually, hence $f(t)1_{[x,x+1]}\to0$. The same occurs if $t>x_0+1$, whereas of $t\in[x_0,x_0+1]$ then $t\in[x,x+1]$ eventually so that $f(t)1_{[x,x+1]}(t)=f(t)$ eventually. So for $t\neq x_0$ we have $f(t)1_{[x,x+1]}(t)\to f(t)1_{[x_0,x_0+1]}(t)$. Since this leaves out only two points, $x_0$ and $x_0+1$, the convergence is pointwise almost everywhere. All of these functions have absolute values that is at most $|f(t)|$, so by dominated convergence we have:

$$F(x)=\int_x^{x+1}f(t)dt=\int_{-\infty}^{+\infty}f(t)1_{[x,x+1]}(t)dt\to\int_{-\infty}^{+\infty}f(t)1_{[x_0,x_0+1]}(t)dt=\int_{x_0}^{x_0+1}f(t)dt=F(x_0),$$

as $x\to x_0$, proving $F$ is continuous.

$G=|F|$ is the composition of $h(x)=|x|$ and $F$, and since $h,F$ are both continuous we have $G=h\circ F$ is also continuous.

Answer 1

The first example I think of is any negative Gaussian function, in particular $f(x)=-e^{-x^2}$;

The graph:

You can see that for for any $x$ the area between $x$ and $x+1$ is negative.

You showed that for positive functions there is a maximum, but if $f$ is negative the limit to infinity implies that there is minimum, and $0$ is the $\sup$ of $F$, but not the maximum

Answer 2

They're talking about a maximum for $F$, not for $|F|$. If $F<0$ then $F$ has no maximum, since $\sup F(x)=0$ but $F(x)\ne0$.

Answer 3

A desired counterexample.

Consider $f(x)=(2^{-n})$ on $[n,n+1)$, $f(x)=-1$ on $[0,1]$ and zero elsewhere.