An Integral Inequality Involving Composite Functions

Problem:

Let $f,g:[0,1] -> [0,1]$ be continuous functions and $f$ be increasing.
Prove that:
\[
\int_0^1 f(g(x))\ \text{d} x \leq \int_0^1 f(x)\ \text{d} x + \int_0^1 g(x)\ \text{d} x\;.
\]

Hints:

Testo nascosto, fai click qui per vederlo

Consider the function $f(u) - u$, with $u in [0,1]$, and prove that $f(u) - u \leq \int_0^1 f(x) "d" x$; then use Mean Value Theorem to conclude.

Problem:

Let $f,g:[0,1] -> [0,1]$ be continuous functions and $f$ be increasing.
Prove that:
\[
\int_0^1 f(g(x))\ \text{d} x \leq \int_0^1 f(x)\ \text{d} x + \int_0^1 g(x)\ \text{d} x\;.
\]

Hints:

Testo nascosto, fai click qui per vederlo

Consider the function $f(u) - u$, with $u in [0,1]$, and prove that $f(u) - u \leq \int_0^1 f(x) "d" x$; then use Mean Value Theorem to conclude.

Since $0<= f(u) <= 1$, we also have $0<= f(u)*u<= u <=1$ for each $u in [0,1]$, hence $f(u) - u <= f(u) - f(u)*u = f(u)*(1-u) = int_u^1 f(u) " d"x$.
Since $f$ is nonnegative and increasing in $[0,1]$ we can bound from above the latter integral with $int_u^1 f(x) " d"x <= int_0^1 f(x) " d" x$, therefore:
\[
f(u) - u \leq \int_0^1 f(x)\ \text{d} x
\]
for each $u in [0,1]$.

In order to finally prove our inequality, let us observe that Mean Value Theorem and previous inequality (with $u = g(xi)$) yield $int_0^1 ( f(g(x)) - g(x)) " d"x = f(g(xi)) - g(xi) <= int_0^1 f(x) " d"x$, which implies the claim.