Let $0<C<1$, show $\exists \delta_n \to 0$ s.t if $A_k\subseteq[0,1], 1\leq k\leq n$ and $\lambda(A_k)=C$ then $\lambda(A_i\cap A_j)\geq(1-\delta_n)C^2$ for some $i,j$
I make use of the hint. By Holders $\|F\|_1\leq\|F\|_2$. We pick $\delta_n=\frac{1}{C(n-1)}\to 0$. $\|F\|_1^2=n^2C^2$ and $$\|F\|_2^2=nC +\sum_{i\not=j}^{n}\sum_{i}^{n}\lambda(A_i\cap A_j)$$
Now assume by contradiction that the conclusion does not hold. Then $$\|F\|_2^2=nC +\sum_{i\not=j}^{n}\sum_{i}^{n}\lambda(A_i\cap A_j)\leq nC+n(n-1)(1-\frac{1}{C(n-1)})C^2$$
However considering $n^2C^2 \leq nC+n(n-1)(1-\frac{1}{C(n-1)})C^2$ we find a contradiction. Thus $\delta_n$ must works!
Is this correct? Perhaps you could comment on the choice of $\delta_n$. How I found it is: I let $\delta_n$ be arbitrary and assumed that towards contradiction that $\lambda(A_i\cap A_j)<(1-\delta_n)C^2$ and just played around with the inequality I got from Holder to see what $delta_n$ would not only give a contradiction but also converge to 0. Is there a more efficient way of doing this?