(1) Proof: If the condition is-1 ≤ x ≤ 1, │f(x)│≤ 1, take x=0 to get │c│=│f(0)│≤ 1, that is │ c │≤. 0, g(x)=ax+b is the increasing function on [- 1, 1], ∴ g (- 1) ≤ g (x) ≤ g (1), ∫ γ f (. ∴ g (1) = a+b = f (1)-c ≤ │ f (1) │+│ c ≤ 2, g (-1) =-a+b =-f. ∴ g (- 1) ≥ g (x) ≥ g ( 1)，∫│f(x)∴g(- 1)=-a+b =-f(- 1)+c≤│f(- 1)│+│c≤2，g (65438∴ c = f (0) =- 1。 Because when-1 ≤ x ≤ 1, f(x) ≥- 1, that is, f(x)≥f(0), according to the properties of quadratic function, the straight line x.