Thread: Set of twice differentiable functions

1. Set of twice differentiable functions

Hi all,

Need some help with this simple? proof.

Let A be the set of all twice differentiable functions $\displaystyle f:[0,1] \rightarrow \mathbb{R}$ with the properties: $\displaystyle f(0) = 0$, $\displaystyle f(1) = 1$ and $\displaystyle f''\leq 0$.

Prove: If $\displaystyle f \in A$ and if

$\displaystyle \int_0^1 f(x) dx \leq \int_0^1 g(x) dx$

for all $\displaystyle g \in A$, then $\displaystyle f(x) = x$ with all $\displaystyle x \in [0,1]$.

Thanks.

2. Originally Posted by Irrational
Hi all,

Need some help with this simple? proof.

Let A be the set of all twice differentiable functions $\displaystyle f:[0,1] \rightarrow \mathbb{R}$ with the properties: $\displaystyle f(0) = 0$, $\displaystyle f(1) = 1$ and $\displaystyle f''\leq 0$.

Prove: If $\displaystyle f \in A$ and if

$\displaystyle \int_0^1 f(x) dx \leq \int_0^1 g(x) dx$

for all $\displaystyle g \in A$, then $\displaystyle f(x) = x$ with all $\displaystyle x \in [0,1]$.

Thanks.
suppose first that there exists $\displaystyle 0 < a < 1$ such that $\displaystyle f(a) \leq a.$ then by the mean value theorem $\displaystyle f(a)=af'(b),$ for some $\displaystyle 0 < b < a.$ thus $\displaystyle f'(b) \leq 1.$ again, by the mean value theorem

$\displaystyle 1-f(a)=(1-a)f'(c),$ for some $\displaystyle a < c < 1.$ but since $\displaystyle f'$ is decreasing we have $\displaystyle f'(c) \leq f'(b) \leq 1$ and hence $\displaystyle 1-f(a) \leq 1-a,$ i.e. $\displaystyle f(a) \geq a.$ thus $\displaystyle f(a)=a.$ so $\displaystyle f(x) \geq x, \ \forall x \in [0,1].$

now suppose $\displaystyle f(t) > t,$ for some $\displaystyle 0 < t < 1.$ then, since $\displaystyle f$ is continuous, there exists $\displaystyle \delta > 0$ such that $\displaystyle f(x) > x,$ for all $\displaystyle 0 \leq t-\delta < x < t + \delta \leq 1.$ but then we'll have:

$\displaystyle \int_0^1 (f(x) - x) \ dx \geq \int_{t-\delta}^{t + \delta} (f(x) -x) \ dx > 0.$ thus $\displaystyle \int_0^1 (f(x) - x) \ dx > 0$ and hence $\displaystyle \int_0^1 f(x) \ dx > \int_0^1 x \ dx.$ contradiction! so there's no $\displaystyle 0 < t < 1$ with $\displaystyle f(t) > t$ and we're done.