# Is the subspace of odd functions of $L^2(-1,1)$ convex?

I am trying to show that the Hilbert space $H =L^2(-1,1)$ can be decomposed into the direct orthogonal sum $O\oplus E$ where $O$ is the subspace of odd functions, and $E$ is the subspace of even functions.

I know that the subspace of odd functions of $H$ is closed, and I want to now use the closest point property of Hilbert spaces, which states that if a set $A$ is a non-empty closed convex subset of $H$, then for any $f \in H$ there is a unique point of $O$ which is closer to $f$ than any other point of $O$.

I am not familiar with using convexity when the sets are spaces of functions.

So I have that $O$ is closed, but I'm not sure if it is convex? If I can show that, the rest of the proof is straightforward.

• Surely any vector subspace of a real (or complex) vector space is convex? – Lord Shark the Unknown Aug 31 '17 at 12:40
• Can't you just use the definition of convexity? Take two functions, compute the convex combination and show it is an odd function. – user8469759 Aug 31 '17 at 12:40
• Note that for any $f \in L^2(-1, 1)$ we have $f(x) = \frac{f(x) + f(-x)}{2}+\frac{f(x)-f(-x)}{2}$, where $x \mapsto \frac{f(x) + f(-x)}{2}$ is an even function, and $x \mapsto \frac{f(x) - f(-x)}{2}$ is an odd function. Furthermore, if $g$ is even, and $h$ is odd, then $\langle g, h \rangle = \int_{-1}^{1} f\overline{g} = 0$, as it is an integral of an odd function on a symmetric interval. – mechanodroid Aug 31 '17 at 12:45
• @mechanodroid How does that show the subspace of odd functions is closed though? ...we need to show that it contains all its limit points..i.e. that any sequence of odd functions in $L^2(-1,1)$ converges to an odd function. – eurocoder Sep 1 '17 at 6:07
• I'm with Lord Shark here. It's a subspace, so it's convex, as all subspaces are. – Michael Grant Sep 2 '17 at 2:31

There are several ways to show $O\oplus E = H$.

The first is a direct proof I have provided in the comments:

We have shown that every $f \in H$ permits a decomposition $f = g + h$, where $g \in O$, $h \in E$. This showes that $O + V = H$. Secondly, we showed $O \perp E$, which means that the sum $O \oplus E$ is orthogonal. It is not necessary to show that any of the subspaces is closed.

The second proof is the one you attempted:

We will show $O^\perp = E$ and $E^\perp = O$. From this follows that $O$ (and $E$) is a closed subspace because orthogonal of a set is always a closed subspace.

Let's prove that $O^\perp = E$, as $E^\perp = O$ is similar. From $O \perp E$ we have $E \subseteq O^\perp$. To prove $O^\perp \subseteq E$, take $g \in O^\perp.$ For every $f \in O$ we have: $\newcommand\diff{\mathop{}\!\mathrm{d}}$ $\newcommand\inner[2]{\langle #1, #2 \rangle}$ \begin{align}0 &= \inner{f}{g} \\ &= \int_{-1}^1 f(t)\overline{g(t)} \diff{t} \\ &= \int_{-1}^0 f(t)\overline{g(t)} \diff{t} + \int_{0}^1 f(t)\overline{g(t)} \diff{t} \\ &= -\int_{1}^0 f(-t)\overline{g(-t)} \diff{t} + \int_{0}^1 f(t)\overline{g(t)} \diff{t} \\ &= -\int_{0}^1 f(t)\overline{g(-t)} \diff{t} + \int_{0}^1 f(t)\overline{g(t)} \diff{t} \\ &= \int_{0}^1 f(t)\overline{\bigl( g(t) - g(-t)\bigr)} \diff{t} \end{align}

Since $t \mapsto g(t)-g(-t)$ is an odd function, by setting this for $f$ we have: $$0 = \int_{0}^1 \bigl( g(t) - g(-t)\bigr)\overline{\bigl( g(t) - g(-t)\bigr)} = \int_{0}^1 \left|g(t) - g(-t) \right|^2 \diff{t}$$

If we define $h : (0, 1) \to \mathbb{C}$ as $h(t) = g(t) - g(-t), \forall t \in (0, 1)$, the previous equality states that $\lVert h \rVert_{L^2(0,1)}^2 = 0$ where $\lVert \cdot \rVert_{L^2(0,1)}$ denotes the norm on the space $L^2(0,1)$. This implies $h = 0$, that is $g(t) = g(-t), \forall t \in (0, 1)$, which implies that $g$ is an odd function. Thus, $O^\perp \subseteq E$, which completes the proof of $O^\perp = E$.

Now, back to your original question, $O$ is convex because it is a vector subspace of $H$:

Take $\lambda \in [0, 1]$ and $f, g \in O$: $$\lambda f + (1-\lambda)g \in O$$

since $O$ is closed under linear combinations.

Now you can use the closest point property for $O$ and complete the proof.

Remark: Using the closest point property for a closed subspace of a Hilbert space to prove this kind of thing is essentially reproducing a proof of a result usually known as Riesz projection theorem:

Let $H$ be a Hilbert space, and $M$ a closed subspace of $H$. Then $H = M \oplus M^\perp$ holds.

• Excellent answer thanks! – eurocoder Sep 2 '17 at 10:03