### 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.
Did you exchange
A walk on part in the war
For a lead role in a cage? (Roger Waters)

gugo82
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### Re: An Integral Inequality Involving Composite Functions

gugo82 ha scritto: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.
Did you exchange
A walk on part in the war
For a lead role in a cage? (Roger Waters)

gugo82
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Messaggio: 21867 di 24015
Iscritto il: 12/10/2007, 23:58
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