Orthogonal complement of the set of all real continuous functions that integrate to $0$?

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 $$U^{\bot}=\{0\}$$?

• You have at least constant functions in $W^{\perp}$. Mar 20 '20 at 15:34
• Oh right, that's interesting, but are there any more? Mar 20 '20 at 15:47
• 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. Mar 20 '20 at 16:11

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 $$[0,1]$$.

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.

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