Proving f=0 on [a,b] with Analysis

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SUMMARY

The discussion centers on proving that a continuous function \( f \) on the interval \([a,b]\) is identically zero given that \( f_n(x) = \int_a^x f_{n-1}(t) dt \) results in \( f_n(x) = 0 \) for some \( n \) dependent on \( x \). The proof utilizes the properties of integrals and the fundamental theorem of calculus, establishing that if \( f_n \) is constant and zero, then all preceding functions \( f_{n-1}, f_{n-2}, \ldots \) must also be zero. Thus, it concludes that \( f \equiv 0 \) on \([a,b]\).

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Homework Statement


[tex]f[/tex] is continuous on [a,b]
[tex]f_{1}(x)=\int^x_a f(t)dt[/tex]
[tex]f_{2}(x)=\int^x_a f_{1}(t)dt[/tex]
...
[tex]\forall x\in[a,b],\exists n[/tex] depends on x , such that [tex]f_{n}(x)=0.[/tex]
prove that [tex]f\equiv0.[/tex]

Homework Equations


mathematical analysis

The Attempt at a Solution


copy the taylor theorem 's proof?
but I get nothing.
 
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If, for some n, fn is identically 0, then it is a constant and so its derivative is identically equal to 0. But, by the fundamental theorem, the derivative of [itex]f_n(x)= \int_a^x f_{n-1}(t)dt[/itex] is fn-1(x). Therefore, if fn(x) is identically 0 on [a, b], so is fn-1(x).
 

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