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For a Boolean function $f:\{0,1{\}}^n\to \{0,1\}$ under the uniform probability measure, some coordinate has influence of a variable at least $c\mathrm{Var}(f)\log (n)/n$, where $c>0$ is an absolute constant and $I_i(f)$ is the probability that flipping coordinate $i$ changes $f$. For variance bounded below, this is order $\log (n)/n$. The tribes function shows that this order cannot be improved for general balanced Boolean functions.

A Boolean function takes each vector of bits to either zero or one. Use independent uniform input bits; the influence of a variable is the probability that flipping that bit changes the value. A quite fair Boolean function has $1/4\le \mathbb{E}f\le 3/4$. All logarithms in the estimates below are natural, except where a base is explicitly written.

Here is a balanced tribes construction with unused coordinates valid for every $n\ge 2$. For each $w\ge 1$ put $m_w=\lfloor (\log 2)2^w\rfloor$, and choose the largest $w$ with $w{m}_w\le n$. Partition $w{m}_w$ of the bits into $m=m_w$ blocks of size $w$. Let $f$ be one when at least one block is all ones, and ignore the other bits. This is a tribes function. With $s=2^{-w}$, its zero probability is $(1-s{)}^m$.

For $w=1$ the function is a single bit and has mean $1/2$. For $w\ge 2$, $ms\le \log 2$ and $ms\ge \log 2-s$. The elementary inequalities $-s/(1-s)\le \log (1-s)\le -s$ giveThus both output probabilities are between one quarter and three quarters.

A used bit is pivotal exactly when the other $w-1$ bits of its block are one and every other block fails. Hence its influence is $2^{1-w}(1-2^{-w}{)}^{m-1}\le 2^{1-w}$; unused bits have zero influence. Maximality of $w$ gives $n<(w+1)m_{w+1}\le (w+1)(\log 2)2^{w+1}$. Also $w\le {\log }_{2}n$ for $n\ge 2$: this is immediate for $w=1$, while for $w\ge 2$ one has $w{m}_w\ge 2^w$. ThereforeThis proves the requested existence with room in the constant. The dimension-one boundary case cannot satisfy the printed logarithmic bound: a quite fair function of one bit is nonconstant and has influence one, whereas $10\log 1=0$. Thus the statement requires the usual nontrivial range $n\ge 2$ (or its intended large-$n$ interpretation).

Use the uniform probability measure on the Boolean hypercube. The influence of a variable is the flip probabilityThe polynomial defining $f$ is interpreted over ${\mathbb{F}}_2$. Flipping the first coordinate adds $x_2+\cdots +x_n$ modulo two. This nonempty parity is balanced for $n\ge 2$, so

An explicit Tribes function works for every $n\ge 2$. Choose the largest integer $w\ge 1$ with $w{2}^w\le n$, let $m=\lfloor 2^w\log 2\rfloor$, and take the OR of $m$ disjoint ANDs of $w$ coordinates. Ignore the unused coordinates. Its zero probability is $(1-2^{-w}{)}^m$, between $1/4$ and $3/4$, and a used coordinate has influenceHere $n<2^{w+1}(w+1)$ by maximality. For the balance claim, $w=1$ gives probability $1/2$; for $w\ge 2$, $m\ge 2^w\log 2-1$ gives zero probability at most $e^{-\log 2+2^{-w}}<3/4$, while $(1-1/q{)}^q\ge 1/4$ for $q=2^w\ge 2$ gives a lower bound $(1/4{)}^{\log 2}>1/4$. Only the example was requested, but these checks fix its parameters unambiguously.

For the lower bound we prove the required consequence of Beckner's inequality with constants. Write the Fourier-Walsh transform as $f={\sum }_S\hat{f}(S){\chi }_S$, where ${\chi }_S(x)=(-1{)}^{{\sum }_{i\in S}{x}_i}$ and $\hat{f}(S)=\mathbb{E}(f{\chi }_S)$. The discrete derivative of a Boolean function is $D_{i}f=(f-f(\cdot +e_i))/2$. Since $f$ takes values zero and one, $D_{i}f$ takes values $0,\pm 1/2$. Consequently Parseval's identity givesThe factor four is required for the printed $0/1$ convention. Also $V:=\mathrm{Var}(f)={\sum }_{S\ne \varnothing }|\hat{f}(S){|}^2\ge 3/16$ under the balance hypothesis.

The true Beckner's inequality states $\Vert T_{\rho }h{\Vert }_q\le \Vert h{\Vert }_p$ for $1<p\le q<\infty$, $0\le \rho \le \sqrt{(p-1)/(q-1)}$, with $T_{\rho }h={\sum }_S{\rho }^{|S|}\hat{h}(S){\chi }_S$. Apply it with $p=3/2$, $q=2$, $\rho =1/\sqrt{2}$ to $h_i={\chi }_{\{i\}}{D}_{i}f$, a function independent of coordinate $i$. Its nonzero values still have magnitude $1/2$. We obtainLet $a={\max }_i{I}_i(f)>0$. For an integer $\ell \ge 1$, the low-degree Fourier weight satisfiesFor the high-degree part the total influence identity givesTogether these imply $4V\le na(2^{\ell -1}{a}^{1/3}+1/(\ell +1))$.

Suppose, seeking a contradiction, that $a<\log n/(100n)$, with natural logarithms. Then $a<1/(100e)$ and $\ell =\lfloor \log (1/a)/(6\log 2)\rfloor \ge 1$. Hence $2^{\ell -1}{a}^{1/3}\le a^{1/6}/2$. Put $y=\log n$. The first term on the right is bounded bythe middle expression is maximized at $y=7$. Moreover $\log (1/a)\ge y/2$, since $y\le 100e^{y/2}$. ThusWe would have $4V<1/10$, contradicting $4V\ge 3/4$. ThereforeThis derives the stated Kahn-Kalai-Linial theorem bound from the actual noise operator on the Boolean hypercube inequality, rather than using the desired conclusion as an input.

The total influence isThis Fourier-Walsh transform identity uses $\{-1,1\}$-valued $f$ and the flip-probability convention for influence of a variable. For $\{0,1\}$-valued $f$, the right-hand side is multiplied by four. Indeed Parseval's identity applied to the discrete derivative of a Boolean function gives $\Vert D_{i}f{\Vert }_2^2={\sum }_{S\ni i}\hat{f}(S{)}^2$; sum over coordinates and apply the chosen normalization.

A tribes function is an OR of AND blocks: partition $mw$ coordinates into $m$ blocks of size $w$, and let $f=1$ when at least one block consists entirely of ones. For independent random variables with each bit uniform, $\mathbb{P}(f=0)=(1-2^{-w}{)}^m$. The influence of a variable is $2^{-(w-1)}(1-2^{-w}{)}^{m-1}$, because every other bit in its block must be one and every other block must fail. Taking $m$ approximately $(\log 2)2^w$ keeps both output probabilities bounded away from zero and gives influences of order $\log (mw)/(mw)$. This shows the order in the Kahn-Kalai-Linial theorem is sharp.