I have a question regarding the proof of a proposition. Basically I tried a method, and it seems to work, I don't find a mistake. However I don't use all the conditions in the hypothesis, leading me to believe that there might be a mistake. The proposition goes as follows:
We consider the set $L^p[0,1]$, with $0<p<1$, i.e. the set of functions such that, if $f \in L^p[0,1]$, then:
$$\int_0^1 |f(t)|^p dt < \infty$$
Now we consider for all $f \in L^p[0,1]$ the operator $\Delta$ defined as:
$$\Delta f = \int_0^1 |f(t)|^p dt$$
Then for any $f,g \in L^p[0,1]$, the map $d(f,g) = \Delta (f-g)$ defines a distance that is invariant through translation (already shown).
I want now to show that any subspace of $L^p[0,1]$ that is open and convex, must be either empty or $L^p[0,1]$ itself. So here's my approach:
Let $X$ be an open and convex subspace of $L^p[0,1]$, that is not empty. Let's show that $L^p[0,1] \subset X$ and therefore $X = L^p[0,1]$.
Since $X$ is a vector space, the null function $0 \in X$. And since $X$ is open, there is a real number $r > 0$ such that $0 \in \mathcal{B}(0;r)$, where $\mathcal{B}(0;r)$ is the open ball of center $0$ and radius $r$ with respect to the above defined distance.
Also, for any $f \in L^p[0,1]$ and any $\alpha \in \mathbb{R}$, we have:
$$\Delta (\alpha f) = \int_0^1 |\alpha f(t)|^p dt = | \alpha |^p \int_0^1 |f(t)|^p dt = | \alpha |^p \Delta f$$
With those two properties in mind, we now take any non-zero $f \in L^p[0,1]$ and show that $f \in X$. Let be:
$$g = \alpha f \text{, with } \alpha = (\frac{r}{2\Delta f})^{1/p} \in \mathbb{R}$$
Then we have:
$$d(g,0) = \Delta g = \Delta \alpha f = | \alpha |^p \Delta f = \frac{r}{2} < r$$
Thus $g \in \mathcal{B}(0;r) \subset X$. But as $X$ is a vector space, we also have that $\alpha g \in X$, thus $f \in X$.
This concludes the proof.
Here's my concern: I don't use the fact that $L^p[0,1]$ is convex (nor that it is invariant by translation for that matter). So as I see it, there are either two possibilities:
- There's a mistake somewhere, but I just can't find it.
- Somewhere in the proof I somehow used convexity without actually being aware of it. But I wouldn't know where. The only way I could imagine is by using the definition of $\Delta$ and by $\Delta \alpha f = | \alpha |^p \Delta f$. But in that case, intuitively, it seems to be that for any bounded operator that defines a distance in a vector space, the proposition would be true, as we could just follow the same stream of thought. Intuitively, it seems to me that in that case you would need boundedness to make some statement similar to $\Delta \alpha f = | \alpha |^p \Delta f$ for any operator. But that's conjecture, I don't have a proof for that.
EDIT: So the mistake is that in the original proposition, $X$ is a subset, not a subspace. Convexity is therefore needed. In a subspace, it would be automatically included.
For whoever stumbles upon this and needs the proof, here it is:
Let be $X \neq \emptyset$ a convex open set in $L^p[0,1]$. Let be $f \in L^p[0,1]$. We'll show that $f \in X$, thus $L^p[0,1] \subset X$, thus $X = L^p[0,1]$.
Since the distance is invariant by translation, we can assume that $0 \in X$, and since $X$ is open, there is a real $r > 0$, such that $0 \in \mathcal{B}(0,r)$
$p < 1 \Longrightarrow p-1 < 0$, and thus there is an $n \in \mathbb{N}$ such that $n^{p-1} \Delta f < r$. Now we will sort of split $f$ into $n$ functions defined on smaller segments of $[0,1]$, such that their integral over $[0,1]$ are identical. Basically, there is a set of points $0 = x_0 < x_1 < ... < x_n =1$ such that:
$$\forall i \in \{1,...,n\}: \int_{x_{i-1}}^{x_i} |f(t)|^p = \frac{1}{n} \Delta f$$
So we define for $i \in {1,...,n} $ the functions:
$$g_i(t) = \begin{cases} nf(t) \text{, if t $\in [x_{i-1},x_i]$} \\ 0 \text{, else} \end{cases} $$
We have
$$\Delta g_i = n^{p-1} \Delta f <r$$
thus $g \in \mathcal{B}(0;r) \subset X$.
Finally since $X$ is convex, we have that:
$$f = \frac{1}{n} (g_1 + ... +g_n) \in X$$
That concludes the proof.
