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Let $V=[C[0,1], \mathbb{R}]$, the vector space of all continuous real valued functions equipped with an inner product $\langle f, g\rangle:=\int^{1}_0 f.g \ dx$. Let $W \subset V$ be the subspace: $W=\{f(x) \in V \mid \int^{1}_0 f(x) dx =0$}, which is the subspace of all functions that integrate to $0$.

What is $W$'s orthogonal complement?

I have a feeling it is $\{0\}$, but I can't show it to be true. I have a feeling a rigorous treatment of this question requires the concept of denseness of functions in a space, but I have not learnt about functions in a space being dense in another, so I would appreciate if there was an elementary proof. If my intuition is wrong, what is the orthogonal complement then? I can't seem to find the answer to this question online.

I have also read about a proof here of why $(U^{\bot})^{\bot}\neq U$, and that is because of $U^{\bot}=\{0\}$. I have yet to find a counterexample, but I wish to ask this question:

Is it true that if $(U^{\bot})^{\bot}\neq U$, one has $U^{\bot}=\{0\}$?

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  • $\begingroup$ You have at least constant functions in $W^{\perp}$. $\endgroup$
    – Raoul
    Mar 20 '20 at 15:34
  • $\begingroup$ Oh right, that's interesting, but are there any more? $\endgroup$ Mar 20 '20 at 15:47
  • $\begingroup$ I'm quite sure that there are no others. You would like to show that if $\int_0^1 f(x) g(x) \: dx = 0$ for all $g \in W$, then $f$ is constant. WLOG, you can assume that $f(0) = 0$ and then show that $f$ is equal to 0. Probably a proof by contradiction by building a nice $g$ would work, but I am not exactly sure of how to do that. $\endgroup$
    – Raoul
    Mar 20 '20 at 16:11
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I follow up on my earlier comment, and claim that $W^{\perp}$ is the set of constant functions. So take $f \in W^{\perp}$. This means that for all $g \in W$, $$ \int_0^1 f(x) g(x) \: \mathrm{d}x = 0. $$ Now, take two points $a < b$ in $(0,1)$. Consider the function $g_n$ that is made of piecewise linear, with

  • $g_n(x) = 0$ for $x \in [0,a-1/n]$,
  • $g_n(a) = n$,
  • $g_n(x) = 0$ for $x \in [a+1/n,b-1/n]$,
  • $g_n(b) = - n$,
  • $g_n(x) = 0$ for $x \in [b+1/b,0]$.

Then $g_n \in W$, so $\int_0^1 f(x) g_n(x) \: \mathrm{d}x = 0$, which means $$ \int_{a-1/n}^{a+1/n} f(x) g_n(x) \: \mathrm{d}x = \int_{b-1/n}^{b+1/n} f(x) g_n(x) \: \mathrm{d}x. $$ Each "spike" of $g_n$ is an approximation of identity, so the LHS converges to $f(a)$ as $n \to + \infty$, and the RHS converges to $f(b)$. If you have never seen this, you can show it using the continuity of $f$, standard $\epsilon / \delta$ arguments, and the fact that the integral of each "spike" is 1.

Finally, you get that $f(a) = f(b)$ for all $a,b \in (0,1)$, and you conclude that by continuity that $f$ is constant on $[0,1]$.

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An alternative solution, which might be appealing if you are familiar with Fourier series and a little $L^2$ theory, or it might pique your interest even if you haven't.

In particular we will use that $V\subseteq L^2[0,1]$, (this is an isometric embedding in fact, meaning, with the norm induced by the inner product you have mentioned, distance is preserved in the inclusion).

Let $g\in W^{\perp}$ and Fourier expand: $$ g(x)=\sum_{n\in \mathbb{Z}} c_ne^{2\pi i n x} $$ in the sense that $$ \int_0^1\left|g-\sum_{n=-N}^N c_ne^{2\pi i n x}\right|^2\to 0 $$ as $N\to \infty$ and where $c_n=\int_0^1 g(x)e^{-2\pi i n x}\mathrm dx $.

But noting that $\int_0^{1}e^{-2\pi i n x}\mathrm dx=0$ for $n\ne 0$, then by assumption $c_n=0$ for $n\ne 0$.

So, as $L^2$ functions, $g(x)=c_0$, which implies $g=c_0$ almost everywhere. By continuity of $g$, this gives equality everywhere.

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    $\begingroup$ That's definitely more elegant. $\endgroup$
    – Raoul
    Mar 21 '20 at 21:28
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    $\begingroup$ Thank you @Raoul, it requires some machinery to be fair $\endgroup$ Mar 22 '20 at 3:00

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