Let $A$ be the set of all continuous functions $f:[0,1] \to [0, \infty)$ satisfying $\displaystyle\int_0^x f(t)dt\geq f(x), \ \ \forall x \in [0,1]$. Prove that $|A|=1$.
Proof:
$f(x)=0, \forall x\in[0,1]$ satisfies the condition. We show that this is the only function possible.
Suppose the contrary. So, $\exists \ c \in [0,1]$ such that $f(c)=k > 0$. By neighbourhood property of continuous functions, we can find a $\delta>0$ such that $f(t)>0, \ \ \forall t \in [c-\delta,c]\subset [0,1]$. Let $\inf_{t\in[c-\delta,c]} f(t) =m_c >0 $ [since it has to attain its infimum in $[c-\delta,c]$ so it must be greater than $0$].
So, $\displaystyle\int_0^{c-\delta}f(t)dt+\int_{c-\delta}^cf(t)dt\geq\int_{c-\delta}^cf(t)dt\geq m_c\delta \geq f(c)$ i.e.,
a contradiction to our assumption since $f(c)\geq m_c$ and $0<\delta<1$, so $f(c)>m_c\delta$. [equality cannot hold]
Is this correct? Please verify.