Problem: Let $L^p(\mu)$ be the space of equivalence classes of functions $f$ defined on the unit circle and measurable with respect to the Lebesgue measure $\mu$, such that
$$ \|f\|_p = \left(\int_{\mathbb{T}} \quad |f|^p \, d\mu \right)^{1/p} < + \infty. $$
Decide for which values of $p \in [1, +\infty]$ the following statement holds:
"If $E$ is a non-empty, closed, and convex subset of $L^p(\mu)$, then there exists a unique $f_0 \in E$ such that $\|f_0\|_p \leq \|f\|_p$ for all $f \in E$."
Proof:
Consider the different cases:
For $p \in (1, \infty)$.
The real function $ t \longmapsto |t|^p $ is strictly convex, meaning that for any $ t_0, t_1 \in \mathbb{R} $ with $ t_0 \neq t_1 $, we have
$$ |\lambda t_0 + (1 - \lambda) t_1|^p < \lambda |t_0|^p + (1 - \lambda) |t_1|^p \quad \forall \lambda \in (0, 1). $$
Let $ E $ be a closed, convex subset of $ L^p(\mu) $ and let $ m = \inf_{f \in E} \|f\|_p $. This is well-defined since $E$ is non-empty.
a) Existence of the minimum:
By the definition of infimum, we can construct a sequence $ \{f_n\}_{n = 1}^{\infty} \subset E $ such that $$ \|f_n\|_p < m + \frac{1}{n} \quad \forall n \in \mathbb{N}. $$
Since $ \{f_n\}_{n = 1}^{\infty} $ is bounded $( \|f_n\|_p \leq 2 \quad \forall n \in \mathbb{N} )$ and $ E $ is closed, it follows from the Bolzano-Weierstrass theorem that there exists a convergent subsequence $ \{f_{n_k}\}_{k = 1}^{\infty} $. For simplicity, rename this subsequence $ f_n $ and let $ f_0 = \lim_{n \to \infty} f_n $. By Fatou's Lemma,
$$ \|f_0\|_p \leq \liminf_{n \to \infty} \|f_n\|_p = m, $$
thus $ \|f_n\|_p = m $ and $ f_0 \in E $ since $ E $ is closed.
b) Uniqueness:
Assume $ \exists f, g \in E $ such that $ \|f\|_p = \|g\|_p = m $ and $ f \neq g $.
Since $ E $ is convex, $ (\lambda f + (1 - \lambda) g) \in E \quad \forall \lambda \in (0, 1) $. Using the convexity of $ t \longmapsto |t|^p $, we get
\begin{align*} m^p &= \|\lambda f + (1 - \lambda) g\|_p^p = \int_{\mathbb{T}} \quad |\lambda f + (1 - \lambda) g|^p \, d\mu < \int_{\mathbb{T}} \left( \lambda |f|^p + (1 - \lambda) |g|^p \right) \, d\mu \\ &= \lambda \|f\|_p^p + (1 - \lambda) \|g\|_p^p = \lambda m^p + (1 - \lambda) m^p = m^p, \end{align*}
which is a contradiction.
- For $ p = 1 $.
Let $ E = \{f \in L^1(\mu) : \|f\|_1 = 1 \text{ and } f(t) \geq 0 \text{ a.e. } \mathbb{T}\} $.
$ E $ is convex since if $f, g \in E $, it is clear that $ \lambda f + (1 - \lambda) g \geq 0 $ a.e. and
\begin{align*} \|\lambda f + (1 - \lambda) g\|_1 &= \int_{\mathbb{T}} \quad |\lambda f + (1 - \lambda) g | \, d\mu \\ &= \int_{\mathbb{T}} \left( \lambda f + (1 - \lambda) g \right) \, d\mu \\ &= \lambda \|f\|_1 + (1 - \lambda) \|g\|_1 = \lambda + (1 - \lambda) = 1. \end{align*}
Let $ f(t) = \frac{1}{2\pi} $ and $ g(t) = \frac{1}{\pi} \mathcal{X}_{[0, \pi]}(t) $. Clearly, $ \|f\|_1 = \|g\|_1 = 1 $ and $ 1 = \inf_{f \in E} \|f\| $, but $ f \neq g $.
- For $ p = \infty $.
Consider
$$ E = \{f \in L^{\infty}(\mu) : f(t) = 1 \text{ a.e. } t \in [0, \pi) \text{ and } f(t) \in [-1, 1] \text{ a.e. } t \in [\pi, 2 \pi)\}. $$
Clearly, $ \|f\|_{\infty} = 1 \quad \forall f \in E $. For any $ \lambda \in (0, 1) $, we have $ \lambda f(t) + (1 - \lambda) g(t) = 1 $ a.e. $ t \in [0, \pi) $ and $ (\lambda f(t) + (1 - \lambda) g(t)) \in [-1, 1] $ a.e. $ t \in [\pi, 2\pi) $, so $ E $ is convex. However, $ f(t) = 1 $ and $ g(t) = \mathcal{X}_{[0, \pi)}(t) $ satisfy
$$ \|f\|_{\infty} = \|g\|_{\infty} = \inf_{h \in E} \|h\|_{\infty} = 1, $$
but $f \neq g $.
Therefore, the statement is only true for $ p \in (1, \infty) $.
