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We have a space $V=C ([-1,1]; \mathbb{R})$ with inner product:

$$\langle f,g\rangle=\int_{-1}^1 f(x)g(x) dx.$$

Are the even function closed in $V$ in norm induced by inner product?

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    $\begingroup$ You perhaps meant $\int_{-1}^1$ $\endgroup$ Apr 9 '18 at 13:41
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There is a continuous map: $T:V\to V$ defined as $(Tf)(x)=f(x)-f(-x).$ (Left for you to prove: $T$ is continuous.)

Now the space of even functions is $T^{-1}(\{\mathbf 0\})$ where $\mathbf 0$ is the zero function, and $\{\mathbf 0\}$ is closed in $V.$ (Why?)

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  • $\begingroup$ Thanks! Now, all is clear for me. It was realy helpful. $\endgroup$
    – Kornel
    Apr 9 '18 at 16:06
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Assume $(f_n)_n$ is a sequence of even functions in $C[-1,1]$ such that $f_n\xrightarrow{L^2} f$. We have:

$$\|x\mapsto f(x) - f(-x)\|_2 = \lim_{n\to\infty} \underbrace{\|x \mapsto f_n(x) - f_n(-x)\|_2}_{=0} = 0$$

so $f(x) = f(-x), \forall x \in [-1,1]$.

For an alternative proof, note that the space of even functions and the space of odd functions are orthogonal complements of each other (see here for example). Then in particular, both subspaces are closed.

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We note that $$\int_{-1}^1 (f(x) - f_n(x))^2 dt = \int_{0}^1 (f(x) - f_n(x))^2 dt + \int_{0}^1 (f(-x) - f_n(x))^2 dt \rightarrow 0,$$ thus as each integral is positive $f$ is even as required.

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