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

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.

• Desired counterexamples are given in the answers by Holo and Przenysław Scherwentke. The former, along with David C. Ulrich's answer, highlight my mistake, that is that, while 1. and 2. are correct, they only imply the maximum for positive functions, and $F$ need not be positive. So (a) is proved, since $G$ is positive and satisfies 1. and 2., but (b) is not rendered impossible, because $F$ is in general not positive, // – MickG Mar 30 '18 at 17:03
• // and so the maximum in $[-M,M]$ is in fact not necessarily a global maximum. I accept Holo's answer because he both provided a counterexample and pointed out what I was getting wrong. [Continued from a deleted answer by OP.] – quid Mar 30 '18 at 21:20

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

• Did you mean "You should say that" or "You showed that", or something else? – MickG Mar 30 '18 at 17:05
• @MickG "you showed", yes. My apologies, and thanks – ℋolo Mar 30 '18 at 17:31

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 $F(x)\ne0$.

A desired counterexample.

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