[EX] Two Integral Inequalities

[gugo82]

Exercise:

1. Let \(f,g:[a,b]\to \mathbb{R}\) be smooth functions (say, continuous functions in \([a,b]\)), let \(f\) be nonincreasing and let \(f(x)\geq 0\) and \(0\leq g(x)\leq 1\) in \([a,b]\).
Prove that:
\[\tag{S}
\int_{b-\gamma}^b f(x)\ \text{d} x \leq \int_a^b f(x)\ g(x)\ \text{d} x\leq \int_a^{a+\gamma} f(x)\ \text{d} x
\]
where:
\[
\gamma :=\int_a^b g(x)\ \text{d} x\; .
\]

Hints:

For the upper inequality, consider the auxiliary function:
\[
\Delta (x) := \int_a^{a+\int_a^x g(t)\ \text{d} t} f(t)\ \text{d} t - \int_a^x f(t)\ g(t)\ \text{d} t
\]
and try to prove that \(\Delta (x)\geq 0\) in \([a,b]\).
For the lower inequality, write down the upper inequality with \(1-g(x)\) in place of \(g(x)\) and use some algebra.

2. Can (S) be modified in order to make it work also in the case \(\sup_{[a,b]} g>1\)?

Hint:

Consider the normalized function \(\hat{g}(x):= \frac{1}{\sup_{[a,b]} g}\ g(x)\) and...

Another question for those who know a little Lebesgue Integration Theory:

3. Is it possible to generalize inequalities (S) for \(f\in L^1(a,b)\) and \(g\in L^\infty(a,b)\)?

Hint:

Use some approximation theorem from Integration Theory.

[gugo82]

@ Paolo90:

1. Ok.
Here's an alternative proof of the leftmost inequality.

Let $G(x):=1-g(x)$. Obviously, $0\leq G(x)\leq 1$ therefore the rightmost inequality applies to $f$ and $G$ and it yields:
\[
\tag{1}
\int_a^b f(x)\ G(x)\ \text{d} x \leq \int_a^{a+\Gamma} f(x)\ \text{d} x
\]
with $\Gamma = \int_a^b G(x)\ \text{d} x$.
Using some algebra we find:
\[
\begin{split}
a+\Gamma &= a+\int_a^b (1-g(x))\ \text{d} x \\
&= a+(b-a)-\int_a^b g(x)\ \text{d} x \\
&= b-\int_a^b g(x)\ \text{d} x \\
&= b-\gamma\; ,
\end{split}
\]
with $\gamma = \int_a^b g(x)\ \text{d} x$, and:
\[
\begin{split}
\int_a^b f(x)\ G(x)\ \text{d} x &= \int_a^b f(x)\ (1-g(x))\ \text{d} x \\
&= \int_a^b f(x)\ \text{d} x - \int_a^b f(x)\ g(x)\text{d} x\; ;
\end{split}
\]
hence (1) rewrites:
\[
\int_a^b f(x)\ \text{d} x - \int_a^b f(x)\ g(x)\text{d} x \leq \int_a^{b-\gamma} f(x)\ \text{d} x
\]
or finally:
\[
\int_a^b f(x)\ g(x)\text{d} x \geq \int_{b-\gamma}^b f(x)\ \text{d} x
\]
which is the claim.

2. Correct.

3. Your idea is the right one. Follow it.

Neverthless, one can give an alternative proof (which does not make use of approximation) on the following lines:

Let $f\in L^1(a,b)$ and $g\in L^\infty(a,b)$ satisfy the assumptions; w.l.o.g., we can assume $f$ and $g$ be defined and finite everywhere in $]a,b[$.
Then (in what follows $\text{d} x$ stands for the Lebesgue measure):
\[
\begin{split}
\int_a^{a+\gamma} f(x)\ g(x)\ \text{d} x - \int_a^b f(x)\ g(x)\ \text{d} x &= \int_a^{a+\gamma} f(x)\ g(x)\ \text{d} x - \int_a^{a+\gamma} f(x)\ g(x)\ \text{d} x \\
&\phantom{=}- \int_{a+\gamma}^b f(x)\ g(x)\ \text{d} x\\
&= \int_a^{a+\gamma} f(x)\ (1-g(x))\ \text{d} x - \int_{a+\gamma}^b f(x)\ g(x)\ \text{d} x\\
&\geq f(a+\gamma)\ \int_a^{a+\gamma} (1-g(x))\ \text{d} x - \int_{a+\gamma}^b f(x)\ g(x)\ \text{d} x\\
&= f(a+\gamma)\ \left( \gamma - \int_a^{a+\gamma} g(x)\ \text{d} x\right) \\
&\phantom{=}- \int_{a+\gamma}^b f(x)\ g(x)\ \text{d} x\\
&= f(a+\gamma)\ \left( \int_a^b g(x)\ \text{d} x - \int_a^{a+\gamma} g(x)\ \text{d} x\right) \\
&\phantom{=}- \int_{a+\gamma}^b f(x)\ g(x)\ \text{d} x\\
&= f(a+\gamma)\ \int_{a+\gamma}^b g(x)\ \text{d} x - \int_{a+\gamma}^b f(x)\ g(x)\ \text{d} x\\
&= \int_{a+\gamma}^b (f(a+\gamma) -f(x))\ g(x)\ \text{d} x\\
&\geq 0\; ,
\end{split}
\]
which is the leftmost inequality to be proved.
The righmost inequality may be proved using the argument above.

***

@ Rigel:

4. The alternative proof given above implies that the assumption $f(x)\geq 0$ is not necessary to our inequalities to hold.

***

Exercise:

5. One can easily see that both inequalities (S) become equalities simultaneously when $g(x)=0$ or $g(x)=1$ for any fixed $f$.
Find a function $\hat{f}$ such that the converse is not true.
In other words, find $\hat{f}$ in such a way that there exists a function $\hat{g}$, which is neither identically zero nor identically one, such that both inequalities (S) become equalities.

Hint:

What happens if $\hat{f}$ is symmetric with respect to $\frac{a+b}{2}$?

6 (Open question, at least for me). In general, is it possible to characterize the simultaneous equality case in both (S)?
I.e., is it possible to find all the functions $g$ such that (S) becomes a chain of equalities for any $f$?

[Paolo902]

"gugo82":Exercise:

1. Let \(f,g:[a,b]\to \mathbb{R}\) be smooth functions (say, continuous functions in \([a,b]\)), let \(f\) be nonincreasing and let \(f(x)\geq 0\) and \(0\leq g(x)\leq 1\) in \([a,b]\).
Prove that:
\[\tag{S}
\int_{b-\gamma}^b f(x)\ \text{d} x \leq \int_a^b f(x)\ g(x)\ \text{d} x\leq \int_a^{a+\gamma} f(x)\ \text{d} x
\]
where:
\[
\gamma :=\int_a^b g(x)\ \text{d} x\; .
\]

Let
\[
\omega(x) := \int_a^{a+\int_a^x g(s)ds} f(t)dt - \int_a^x f(t)\ g(t)dt.
\]
Clearly we have that \( \omega \in C^{1}([a,b])\) and $\omega(a)=0$. Some routine calculations (in particular, using the chain rule) show that
\[
\omega'(x) = f\left(a+\int_a^x g(s)ds\right)g(x)-f(x)g(x) = \left[ f\left(a+\int_a^x g(s)ds\right)-f(x)\right] g(x).
\]
We claim that $\omega'(x)\ge 0$ for every $x \in [a,b]$. Indeed, $g(x)\ge 0$ (1) for every $x \in [a,b]$; on the other side, we have
\[
a + \int_a^x g(s)ds \le a +\int_a^x 1ds \le a+(x-a)=x
\]
since \( \sup g \le 1\); hence, using the fact that $f$ is nonincreasing, \(f\left(a+\int_a^x g(s)ds\right) \ge f(x)\) (2). Now from (1) and (2) we get $\omega'(x) \ge 0$: in particular, $\omega'(b)\ge 0$ which is the first inequality we wanted to prove.

Now, in order to prove the second inequality we consider the function
\[
\alpha(x) := \int_a^x f(s)g(s)ds - \int_{x-\int_a^xg(s)ds}^xf(t)dt
\]
Note that, exactly as $\omega$, we have $\alpha(a)=0$ and we want to prove $\alpha(b)\ge 0$. Calculations are quite similar: we get
\[
\alpha'(x)=\left[f\left(x-\int_a^xg(s)ds \right) - f(x) \right](1-g(x)) \ge 0
\]
for every $x \in [a,b]$.

"gugo82":2. Can (S) be modified in order to make it work also in the case \(\sup_{[a,b]} g>1\)?

Suppose \( \max_{[a,b]} g =M \) (it's a max by Weierstrass). Then we can apply the above argument to the normalized function
\[
\tilde{g}(x):= \frac{g(x)}{M}
\]
and we obtain
\[
\int_{b-\tilde{\gamma}}^b f(x)dx \leq \int_a^b f(x)\tilde{g}(x)dx \leq \int_a^{a+\tilde{\gamma}} f(x)dx
\]
where [tex]\tilde{\gamma} = \int_a^b \tilde{g}dx[/tex]. Since \(\tilde{\gamma} = \frac{1}{M}\gamma \) we obtain the following version of inequality (S):
\[
\int_{b-\frac{1}{M}\gamma}^b f(x)dx \leq \frac{1}{M}\int_a^b f(x)g(x)dx \leq \int_a^{a+\frac{1}{M}\gamma} f(x)dx.
\]

"gugo82":3. Is it possible to generalize inequalities (S) for \(f\in L^1(a,b)\) and \(g\in L^\infty(a,b)\)?

The first idea that springs to my mind is the following: we can find a sequence of continuous functions $f_n$ s.t. $f_n \to f$ in $L^1$ (by density). We can argue the same thing concerning $g$: it's true that continuous functions are not dense in $L^\infty$, but (since $\Omega = (a,b)$ has finite measure) we have $1 \in L^1(\Omega)$ and hence, by Holder, $1 g = g \in L^1$. So we have a sequence of continuous functions $g_n \to g$ in $L^1$, i.e. $\gamma_n \to \gamma$ (where $\gamma_n:=\int_a^b g_n d\mu$ and $\gamma := \int_a^b g d\mu$ as above).
But now I've run out of ideas. How can we conclude?

Thanks for the nice exercise and for your help.

[gugo82]

@ Rigel: Nice!

[Rigel1]

4. Show that the assumption \(f\geq 0\) in not necessary.

Hint:

Apply (S) to the non-negative function \(f(x) - f(b)\).