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Let $f:[-\pi,\pi] \rightarrow R $ be continuous. Pick out cases which imply $f\equiv0$ $$(a) \int_{-\pi}^{\pi} x^n f(x) dx=0 ,\quad\forall n\geq0 \\(b)\int_{-\pi}^{\pi} f(x) cos(nx) dx=0 ,\quad\forall n\geq0\\(c)\int_{-\pi}^{\pi}f(x) sin(nx) dx=0 ,\quad\forall n\geq1.$$ The question is asked in NBHM(India) examination. My doubt here is that if we put $f(x)=1$ in (b) and (c) then the integrals vanish. And (a) seems possible choice but answer key says that all options are correct. Please correct where I am making mistake or the answers are wrong. Thank you for help.

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    For $f=1$, $(b)$ doesn't vanish for $n=0$. But it's wrong anyway: just choose $f$ to be an odd function. – 2017-01-11
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    @ open ball : yes, u r right but what about other options then? – 2017-01-11

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You are right. (b) and (c) do not imply that $f\equiv0$.

Suppose that (a) holds. Then we have $\int_{-\pi}^{\pi} p(x) f(x) dx=0$ for all polynomials $p$.

Since f is continuous, by the approximation theorem of Weierstraß there is a sequence $(p_n)$ of polynomials such that $(p_n)$ converges uniformly t0 $f$ on $[- \pi, \pi]$. Hence $(p_nf)$ converges uniformly t0 $f^2$ on $[- \pi, \pi]$. Therefore

$\int_{-\pi}^{\pi} p_n(x) f(x) dx \to \int_{-\pi}^{\pi} f^2(x) dx$.

Thus we get $\int_{-\pi}^{\pi} f^2(x) dx=0$. Since $f$ is continuous, we derive $f\equiv0$

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    Take the function $f(x)=x$ – 2017-01-11
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    What's an example where we don't have $f \equiv 0$ if $f$ isn't continuous? I think that $f^2 \ge 0$ is sufficient – 2017-01-11
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    @BCLC: The standard non-continuous example is $f(x)=0$ unless $x=0$ when $f(0)=1$. A recurring theme in elementary real-analysis courses is that disturbing a function at a single point does not change the values of any integrals. – 2017-01-11
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    @JyrkiLahtonen Oh okay thanks but without continuity can we conclude $f = 0 \ \mu-\text{a.e.}$ for some (any?) $(S, \Sigma, \mu)$ ? – 2017-01-11
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    @BCLC I guess it depends on whether the functions you test $f$ against (here the monomials) span a dense enough subspace. Here things are also simplified by the fact that we are working inside a Hilbert space, $L^2([-\pi,\pi])$. I don't know about the possible generalizations. I'm not an expert on this. – 2017-01-11
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    @BCLC : Can you please tell how (b), (c) are wrong? – 2017-01-11
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    @SubhashChandBhoria for $(b)$, take $f(x) = x$, and for $(c)$ take $f(x) = 1$. – 2017-01-11
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    @OpenBall Thank you. I verified. – 2017-01-11
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    @BCLC $\int x^n f = 0$ implies, by Weierstrass, that for any $g \in C([-\pi,\pi])$, $\int fg = 0$. Assuming that $f\in L^2$, by the $L^2$ density of continuous functions, there exists $\phi_n \in C([-\pi,\pi]) \to_{L^2} f$, and we have: $0 \leftarrow \|f - \phi_n\|_{L^2}^2 = \|f\|^2_{L^2} + \|\phi_n\|^2_{L^2} - 2 \int f\phi_n \to 2\|f\|_{L^2}^2$. Hence $\|f\|_{L^2} = 0$ and $f = 0$ a.e. – 2017-01-11
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    @JyrkiLahtonen thanks! ^-^ – 2017-01-11
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    @OpenBall what? My question is without continuity – 2017-01-11
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    @BCLC where do you see a continuity assumption in my work? – 2017-01-12
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    @OpenBall '$L^2$ density of continuous functions'? – 2017-01-12
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    @BCLC continuous functions are dense in $L^2$ with respect to the norm $\|\cdot\|_{L^2}$. In fact, test functions (i.e. infinitely differentiable, compactly supported functions) are dense in $L^p$ for $\|\cdot\|_{L^p}$ for each $1\le p < \infty$. – 2017-01-12
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    @OpenBall Is that something in functional analysis? I haven't taken that. But I'm guessing it's some kind of approximation? Like f may not be continuous but since continuous functions are dense* in L2, we still have so and so leading to f=0 a.e.? *This is some kind of functional analysis equivalent of rational numbers' being dense in R (with respect I guess to some norm)? – 2017-01-14
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    @BCLC yes for your first, second and third questions. I'm not sure what kind of analogy you mean in your 4th question, but yes one uses the density of a certain class of functions in the $L^p$ space, where "density" is with respect to the $L^p$ norm. – 2017-01-14
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    @OpenBall thanks! – 2017-01-14