determine if the subset $S = \{ f \in \mathcal{C}(\mathbb{R}): \int_0^1f(x)\;dx= 0 \}$ is a subring of $\mathcal{C}(\mathbb{R})$.

I know to prove this, I must show that:

$S$ is nonempty, $S$ is closed under subtraction, and $S$ is closed under multiplication.

Clearly, $S$ is nonempty because it contains $0$, but I am not sure how to show closure under multiplication and subtraction.

  • 1
    $\begingroup$ Subtraction can be done by splitting the intergral. I don't think it's closed under multiplication, and a counterexample is quite simple: Compress one period of the sin curve into $[0,1]$ by scaling, so that it's integral is zero. However integral of $\sin^2$ function is not zero. $\endgroup$ – астон вілла тереса лисбон Apr 15 '16 at 5:42
  • $\begingroup$ Im not sure what you mean by scaling. The integral of sin would be .459, not 0. $\endgroup$ – p.l Apr 15 '16 at 5:51
  • $\begingroup$ No. See, $\sin x$ is $2\pi$-periodic. By multiplying $x$ by $2\pi$, we get the function $\sin(2\pi x)$.Now the integral of this function from $0$ to $1$ would be zero. But the integral of $\sin^2 (2\pi x)$ is not zero, it is one-half. $\endgroup$ – астон вілла тереса лисбон Apr 15 '16 at 5:57

Consider $f(x)=\sin (2\pi x)$. Then $f \in S$. But $f^2 \not \in S$ because $\int_0^1 \sin^2 (2\pi x)\, dx \neq 0$. Hence not closed under multiplication.

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