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I have this question:

Note: $f$ is assumed to be continuous. screenshot of question

Suppose $\int_{a}^{b} f(x) v(x) d x=0$ for all $v \in C_{0}^{1}(a, b) .$ Show that $f(x)=0$ for all $x \in[a, b]$. What about if $\int_{a}^{b} f(x) v(x) d x=0$ for all $v \in C_{0}^{2}(a, b)$, or for all $v \in C_{0}^{3}(0,1) ?$

I have been trying to figure out an angle of attack but I am completely stumped. First I am a little thrown off by the subscript in $C^2_0$ someone said it means the derivative vanishes at the endpoints but I am not sure.

One thing I wanted to say is, if this applies for all $v$ then it applies for $v$ constant, but all that really tells me is that $F(x) = \int f(x) dx$ is equal at the endpoints.

I don't really want the answer as I want to try to solve this myself, but a hint as to a useful theorem to read, an example or a little trick would be awesome.

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  • $\begingroup$ No it is not necessarily continous $\endgroup$
    – Makogan
    Commented Feb 4, 2022 at 6:26
  • $\begingroup$ Sorry I went back and checked, yes f is continuous, my bad, let me edit. $\endgroup$
    – Makogan
    Commented Feb 4, 2022 at 6:49
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    $\begingroup$ Hint: suppose $f(x_0) \neq 0$. Continuity allows you to find an interval $I$ around $x_0$ so that the function varies around it by at most $\epsilon$. Consider a cutoff function $v$ supported in that interval... $\endgroup$
    – user752785
    Commented Feb 4, 2022 at 6:52
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    $\begingroup$ Do note that while the result of this question may be important in the calculus of variations, it's not actually a problem about the calculus of variations but rather about a separating set of functionals on $C(a,b).$ $\endgroup$ Commented Feb 4, 2022 at 8:00
  • $\begingroup$ @TheHype excuse the silly question, what is a cut off function? $\endgroup$
    – Makogan
    Commented Feb 4, 2022 at 8:29

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