# 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 $f:[0,1] \rightarrow \mathbb{R}$ with the properties: $f(0) = 0$, $f(1) = 1$ and $f''\leq 0$.

Prove: If $f \in A$ and if

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

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

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 $f:[0,1] \rightarrow \mathbb{R}$ with the properties: $f(0) = 0$, $f(1) = 1$ and $f''\leq 0$.

Prove: If $f \in A$ and if

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

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

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

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

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

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