proof-0-1-f-2-t-dt-0-f-0- Tinku Tara January 28, 2024 Algebra 0 Comments FacebookTweetPin Question Number 203807 by aba last updated on 28/Jan/24 proof:∫01f2(t)dt=0⇒f=0 Answered by JDamian last updated on 28/Jan/24 f2(t)⩾0∀t Answered by witcher3 last updated on 29/Jan/24 fiscontinuswecanf(x)={1isx∈IQ0ifx∉IQ∫01f2(x)dx=∫x∈IQ∩[0,1]1dx=μ[[0,1]∩Q]=0orf≠0iffiscontinusin[0,1]suppsef≠0⇒∃a∈]0,1[∣f(a)≠0⇒∃ϵ>0such∀x∈“I=[a−ϵ,a+ϵ]f(x)≠0⇒∀x∈If2(x)>0sincefiscontinusisboundedincompactinterval⇒y∈Isuchf2(x)⩾f2(y)>00⩾∫01f2(x)dx⩾∫If2(x)dx⩾∫If2(y)=f2(y).(2ϵ)⇒f2(y).2ϵ=0⇒f(y)=0absurd⇒f=0 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-203800Next Next post: The-sum-of-all-possible-products-taken-two-at-a-time-out-of-the-numbers-1-2-3-4-is-a-0-b-30-c-30-d-55- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![f is continus we can f(x)= { ((1 is x∈IQ)),((0 if x∉IQ)) :} ∫_0 ^1 f^2 (x)dx=∫_(x∈IQ ∩[0,1]) 1 dx=μ[[0,1]∩Q]=0 or f≠0 if f is continus in[0,1] suppse f≠0 ⇒∃a∈]0,1[ ∣f(a)≠0 ⇒∃ ε>0 such ∀x∈“I=[a−ε,a+ε] f(x)≠0 ⇒∀x∈I f^2 (x)>0 since f is continus is bounded in compact interval ⇒y∈I such f^2 (x)≥f^2 (y)>0 0≥∫_0 ^1 f^2 (x)dx≥∫_I f^2 (x)dx≥∫_I f^2 (y)=f^2 (y).(2ε) ⇒f^2 (y).2ε=0⇒f(y)=0 absurd ⇒f=0](https://www.tinkutara.com/question/Q203820.png)