proof : ∫_0 ^1 f^2 (t)dt=0 ⇒ f=0


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Question Number 203807 by aba last updated on 28/Jan/24

$proof : \int_{0}^{1} f^{2} (t) dt = 0 \Rightarrow f = 0$
	Answered by JDamian last updated on 28/Jan/24

	$f^{2} (t) ⩾ 0 \forall t$
	Answered by witcher3 last updated on 29/Jan/24
	$f is continus we can f(x)= { ((1 is x∈IQ)),((0 if x∉IQ)) :} ∫_0 ^1 f^2 (x)dx=∫_(x∈IQ ∩[0,1]) 1 dx=μ[[0,1]∩Q]=0 or f≠0 if f is continus in[0,1] suppse f≠0 ⇒∃a∈]0,1[ ∣f(a)≠0 ⇒∃ ε>0 such ∀x∈“I=[a−ε,a+ε] f(x)≠0 ⇒∀x∈I f^2 (x)>0 since f is continus is bounded in compact interval ⇒y∈I such f^2 (x)≥f^2 (y)>0 0≥∫_0 ^1 f^2 (x)dx≥∫_I f^2 (x)dx≥∫_I f^2 (y)=f^2 (y).(2ε) ⇒f^2 (y).2ε=0⇒f(y)=0 absurd ⇒f=0$
	$f is continus$ $we can f (x) = {\begin{cases} 1 is x \in IQ \\ 0 if x \notin IQ \end{cases}$ $\int_{0}^{1} f^{2} (x) dx = \int_{x \in IQ \cap [0, 1]} 1 dx = μ [[0, 1] \cap Q] = 0$ $or f \neq 0$ $if f is continus in [0, 1]$ $suppse f \neq 0 \Rightarrow \exists a \in] 0, 1 [∣ f (a) \neq 0$ $\Rightarrow \exists ϵ > 0 such \forall x \in ‘ ‘ I = [a - ϵ, a + ϵ] f (x) \neq 0$ $\Rightarrow \forall x \in I f^{2} (x) > 0 since f is continus is bounded in compact$ $interval \Rightarrow y \in I such f^{2} (x) ⩾ f^{2} (y) > 0$ $0 ⩾ \int_{0}^{1} f^{2} (x) dx ⩾ \int_{I} f^{2} (x) dx ⩾ \int_{I} f^{2} (y) = f^{2} (y) . (2 ϵ)$ $\Rightarrow f^{2} (y) . 2 ϵ = 0 \Rightarrow f (y) = 0 absurd$ $\Rightarrow f = 0$
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