Let $a\_1,\ldots,a\_n$ be real numbers such that $\sum\_i a\_i^2 =1$ and let $X\_1,\ldots,X\_n$ be independent, uniformly distributed, Bernoulli $\pm 1$ random variables. Define the random variable


$S:= \sum\_i X\_i a\_i $


Are there absolute constants $\epsilon >0$ and $\delta<1$ such that for every $a\_1,\ldots,a\_n$


${\mathbb P} [|S| \leq \epsilon ] \leq \delta$ ?



Anti-concentration of Bernoulli sums
The answer to your amended question is yes. In fact, for any $\epsilon\in[0,1)$ we have
$$
\mathbb{P}(\vert S\vert > \epsilon)\ge (1-\epsilon^2)^2/3.
$$
So, we can take $\delta = 1-(1-\epsilon^2)^2/3$. This is the $L^0$ version of the [Khintchine inequality](http://en.wikipedia.org/w/index.php?title=Khintchine_inequality&oldid=295124090).


To prove it, you can use $\mathbb{E}[X\_iX\_j^3]=0$ for $i\not=j$ and $X\_i^4=X\_i^2X\_j^2=1$ to get
$$
\begin{align}
\mathbb{E}[S^4]&=\sum\_ia\_i^4+3\sum\_{i\not=j}a\_i^2a\_j^2=3\left(\sum\_ia\_i^2\right)^2-2\sum\_ia\_i^4\\\\
&\le 3.
\end{align}
$$
The [Paley-Zygmund](http://en.wikipedia.org/w/index.php?title=Paley%25E2%2580%2593Zygmund_inequality&oldid=366994687) inequality gives
$$
\begin{align}
\mathbb{P}(\vert S\vert >\epsilon)&\ge(1-\epsilon^2)^2\frac{\mathbb{E}[S^2]^2}{\mathbb{E}[S^4]}\\\\
&\ge(1-\epsilon^2)^2/3.
\end{align}
$$


This bound gives $\delta=2/3$ for $\epsilon=0$. By considering the example with $a\_1=a\_2=1/\sqrt{2}$ and $a\_i=0$ for $i > 2$, which satisfies $\mathbb{P}(S=0)=1/2$ we see that it is necessary that $\delta\ge1/2$. In fact, a simple argument noting that the distribution is symmetric under a sign change for $X\_1$ (as mentioned by Luca in the comments) shows that $\mathbb{P}(S=0)\le1/2$.


See also the paper [On Khintchine inequalities with a weight](http://arxiv.org/abs/0909.2586v1), where they prove the same bound as I just did above. Also, using the optimal constants for the $L^p$ Khintchine inequality, as in Lemma 3 of that paper, gives an improved bound for $\mathbb{P}(\vert S\vert\le\epsilon)$ tending to $1-2e^{-2+\gamma}\approx0.517$ as $\epsilon$ goes to zero, which is close to optimal.


