(2) For every boolean function , ,

(2′) For boolean function of full degree , .

Proof of (2′)=>(2): Suppose a Boolean function has degree , take a max monomial in , say define then has full degree , it follows form (2′) that

As a reminder for the rest of the argument , consider also the following statement that Huang proved in his recent paper:

(1) For a unbalanced partition of , one of the induced subgraph has maximum degree at least .

What we need here is (1)=>(2′)=>(2). We saw (2″) ==> (2). The implication (1)=>(2′) follows from multiplying the Boolean function with the parity function.

]]>I tried to read the paper by Gotsman and Linial but was puzzled by their proof the equivalence theorem. In particular I don’t see how 2 was equivalent to 2’ in their proof. It’s tricial that 2 implies 2’ but 2’ seems to be only about the special case where the degree of f is n, the number of its variables. Without the reverse direction the polynomial equivalence is degree and sensitivity seems to break down. Could you help me make sense of the subtlety behind it? Thanks.

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