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Consider all polynomials $\mathbb R[x]$ and the subspace of polynomials of degree $0$, which we will refer to by the letter $U$. Is this subspace closed with respect to the inner product: $$\langle f,g \rangle = \int_0^1 f(x)g(x)dx$$

My teacher says that they must not be closed, since if they were, then $(U^{\perp})^{\perp} = U$, but $(U^\perp)^{\perp} = \mathbb R[x]$. I'm having a hard time believing this without seeing an actual example of a limit point of $U$ that is not in $U$.

So if/since it is not closed, could someone please provide an example of one of its limit points not in the space itself?

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I don't see how $(U^\perp)^\perp=\mathbb R[x]$. $U^\perp$ contains exactly the polynomials that average to $0$ over $[0,1]$, and these polynomials are certainly not orthogonal to themselves for a start, so their orthogonal complement can't be all of $\mathbb R[x]$. –  joriki Dec 15 '12 at 22:36

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I think your teacher is wrong, and that the subspace is closed indeed.

Fix a $p(x) \in \mathbb{R}[x]$, and assume it is a limitpoint of the subspace of constant polynomials. Consider $$g(c) = \int_0^1 (p(x)-c)^2 dx = \langle p(x)-c, p(x)-c \rangle$$ So the infimum of $g(c)$ should be 0.

We try to find the infimum of $g(c)$. Actually, as $c \to \pm \infty$, $g(c) \to \infty$, we see that $g$ should have a minimum somewhere. Let's try to compute it.

$$g'(c) = \int_0^1 -2(p(x)-c)dx = -2 \left(\int_0^1 p(x)dx - c \right)$$ So the critical point is $c = \int_0^1 p(x)dx$. At the critical point, $$g(c) = \int_0^1 p(x)^2 dx - 2c \int_0^1 p(x)dx + c^2 = \int_0^1 p(x)^2 dx - \left(\int_0^1 p(x) \right)^2 \ge 0$$ by Cauchy Schwarz inequality, with equality holds only when $p(x)$ is a constant. When $p(x)$ s not a constant, then the last inequality is strict, meaning that $\langle p(x)-c, p(x)-c \rangle = g(c) > 0$ for all $c$. So the subspace is really closed.

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Or you could argue that $g(c)$ is bounded away from $0$ for large $|c|$, so the infimum is the infimum over a finite closed interval; $g(c)$ is continuous and thus attains it minimum on a compact set; so if the infimum is $0$, the minimum is $0$; but this only happens if $p(x)$ is constant. –  joriki Dec 16 '12 at 8:03

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