设|f(x)|在[0,1]上最大值为|f(a)|,0≤a≤1
则|f(a)|=|∫[0->a]f'(t)dt|≤p∫[0->a]|f(t)|dt
≤p∫[0->a]|f(a)|dt=ap|f(a)|
∴|f(a)|(1-ap)≤0,而0≤ap≤p0,∴|f(a)|≤0,即|f(a)|=0
∴而x∈[0,1]时,|f(x)|≤|f(a)|=0
∴|f(x)|=0,即f(x)=0,x∈[0,1]