Given a total variation distance from uniform, how well can we bound the probabilities of sub-intervals? I made the following claim, which I now see that I don't know how to prove. Can anyone prove it?
Claim. Let $f$ be a concave and non-negative function on $[0,1]$ with $$\int_0^1 f = 1,$$
$$\int_0^1 |f-1| > 2\varepsilon.$$
Then
$$\min\left(\int_0^{1/4} f , \int_{3/4}^1 f\right) < \frac14 - \frac\varepsilon8.$$
In principle, this claim can be settled via computer algebra. For any $f$ satisfying the conditions, there is a piecewise-linear function $g$ of 10 pieces which satisfies the same conditions and has the same integrals as $f$ and $|f-1|$ on $[0,\frac14]$, $[\frac14,\frac34]$ and $[\frac34,1]$. The existence of such $g$ is just a matter of the existence of 9 $x$'s, 11 $y$'s, and 1 $\varepsilon$ satisfying some multilinear equalities and inequalities, and therefore is algorithmically solvable. In practice, my simplifications in Mathematica get overwhelmed before they finish.
Does anyone see a proof or a counterexample? I'd be happy to see a proof with $8$ replaced by any other small number.
Update: Distributions like $f(x)=\min(cx,c(1-x),\frac12(c-\sqrt{c^2-4c}))$ may provide extreme examples. The graph below shows $c=24$, with a maximum of $12-\sqrt{120}$, and a bound of $\frac14-\varepsilon/3.66$. For $f$ of this form with high $c$, the bound is $\min(\int_0^{1/4}f,\int_{3/4}^1f)<\frac14-\varepsilon/4$.

 A: I claim the following result which implies what you are asking for.
Claim: Let $g$ be a convex function such that $\int_0^1 g=0$.
Then one of $\int_0^{1/4}g$ and $\int_{3/4}^1 g$ is at least $\|g\|_1/18$.
The result implies what you are looking for (up to replacing 8 by 9) by setting $g=1-f$.
Proof: Since $\int_0^1 g=0$, we have $\int g^+=\int g^-=\|g\|_1/2$.
Let $\{x\colon g(x)\le 0\}=[a,b]$.
First suppose that $b\le\frac 34$.
In this case, since $g$ is increasing on $[b,1]$, $\int_{3/4}^1 g\ge \int_b^{b+\frac 14}g$. By convexity, $\int_b^{b+1/4}g\ge \frac1{32}g'(b^+)$
while $\int g^-=-\int_a^b g\le \frac 12(b-a)^2g'(b^+)\le \frac{9}{32}g'(b^+)$. It follows that $\int_{3/4}^1 g\ge \frac 19\int g^-=\frac 1{18}\|g\|_1$. By symmetry, the same applies if $a\ge \frac 14$.
In the other case, $g$ is negative on $[\frac 14,\frac 34]$. Since
$\int g=0$, either $\int_0^{1/2}g\ge 0$ or $\int_{1/2}^1 g\ge0$. Without loss
of generality, we assume that the first of these holds. Hence
$\int_0^{1/4}g\ge -\int_{1/4}^{1/2}g$. We then apply the following lemma with $f=-g$, which gives $\int_{1/4}^{1/2}(-g)\ge \frac 19\int (-g)^+$, or $-\int_{1/4}^{1/2}g\ge \frac 1{18}\int\|g\|_1$, and the claim then follows.
Lemma: Let $f$ be concave on $[\frac 14,1]$ and positive on $[\frac 14,\frac 34]$ (at least). Then $\int_{1/4}^{1/2}f\ge \frac 19\int f^+$.
Proof: We are attempting to find a lower bound for $\int_{1/4}^{1/2}f\big /
\int_0^1 f^+$. Given any $f$, this quantity is reduced if $f$ is replaced
on $[\frac 14,\frac 12]$ by a linear function joining $(\frac 14,f(\frac 14))$ to $(\frac 12,f(\frac 12))$. It is then further reduced by extending
that linear function to $[\frac 14,1]$. Now it is straightforward to see that
the quantity is minimized if $f(x)=x-\frac 14$ on $[\frac 14,1]$, where the ratio is $\frac 19$ as required.
