Decide for which values of $p \in [1, +\infty]$ the following statement holds.

Question

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:

1. 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.

2. 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 $.

3. 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) $.

Answer 1

There is a crucial mistake in part a) of existence. You claim that since $f_n$ is bounded in $L^p$ it admits a norm convergent subsequence. Unfortunately this is false (in fact, it holds only in finite dimensional spaces). However, $L^p$ for $1<p<\infty$ is a reflexive space and thus we can find a weakly convergent subsequence $(f_{k_n})$ of $(f_n)$. Denoting by $f$ the weak limit of $ f_{k_n}$ and using the lower semicontinuity of the norm with respect to the weak topology you get that $$ \|f\|_p \le \liminf_n \|f_{k_n}\|_p \le m $$ and so $\|f\|_p = m$. You can see more details in my previous answer.

Uniqueness in b) is almost correct. You need to replace

\begin{align*} m^p &= \|\lambda f + (1 - \lambda) g\|_p^p = \dots \end{align*}

by

\begin{align*} m^p & \le \|\lambda f + (1 - \lambda) g\|_p^p = \dots \end{align*}

You also need to verify that the set $E$ is your counterexamples is closed, but this is rather easy.

Answer 2

Problem Statement: We want to determine for which values of $ p \in [1, \infty] $ the following statement holds for $ L^p(\mu) $:

If $ E \subset L^p(\mu) $ is a non-empty, closed, and convex subset, then there exists a unique $ f_0 \in E $ such that $ \|f_0\|_p \leq \|f\|_p $ for all $ f \in E $.

Solution: The correct solution hinges on two key points:

1. Existence of an element of minimal norm in $ E $ for all $ p $ in the range $ 1 < p < \infty $, which relies on reflexivity.
2. Uniqueness of the element of minimal norm in $ E $, which requires both reflexivity and strict convexity.

Corrected Solution:

A. For $ p \in (1, \infty) $

In this case, $ L^p(\mu) $ is reflexive and strictly convex. We proceed with two parts: existence and uniqueness.

a) Existence of a Minimizer

Since $ E \subset L^p(\mu) $ is closed, convex, and non-empty, the infimum $ m = \inf_{f \in E} \|f\|_p $ is well-defined and finite. We aim to show that there exists an $ f_0 \in E $ such that $ \|f_0\|_p = m $.

1. Let $ \{f_n\}_{n=1}^\infty \subset E $ be a sequence such that $ \|f_n\|_p \to m $ as $ n \to \infty $. This sequence is bounded since $ \|f_n\|_p \leq m + 1 $ for large $ n $.
2. By reflexivity of $ L^p $ for $ 1 < p < \infty $, there exists a weakly convergent subsequence $ \{f_{n_k}\} \subset \{f_n\} $ with $ f_{n_k} \rightharpoonup f_0 $ weakly in $ L^p $ for some $ f_0 \in L^p $.
3. Since $ E $ is convex and closed, it is also weakly closed (by Mazur's theorem). Hence, $ f_0 \in E $.
4. By the weak lower semicontinuity of the $L^p $ norm, we have $$ \|f_0\|_p \leq \liminf_{k \to \infty} \|f_{n_k}\|_p = m. $$ Thus, $ \|f_0\|_p = m $, showing that $ f_0 $ is a minimizer in $ E $.

b) Uniqueness of the Minimizer

To establish uniqueness, assume there exist two distinct minimizers $ f_1, f_2 \in E $ such that $ \|f_1\|_p = \|f_2\|_p = m $.

1. Since $ E $ is convex, any convex combination $ \lambda f_1 + (1 - \lambda) f_2 \in E $ for $ \lambda \in (0,1) $.
2. Using the strict convexity of the $ L^p $ norm for $ 1 < p < \infty $, we get $$ \|\lambda f_1 + (1 - \lambda) f_2\|_p < \lambda \|f_1\|_p + (1 - \lambda) \|f_2\|_p = m. $$ This contradicts the minimality of $ m $, so no such distinct minimizers can exist. Thus, the minimizer is unique.

Therefore, for $ p \in (1, \infty) $, a unique minimizer exists in $ E $.

B. For $ p = 1 $

For $ L^1(\mu) $, reflexivity fails, which means weak convergence arguments cannot be applied to ensure the existence of a norm-minimizing element in a general closed convex subset $ E \subset L^1(\mu) $. We provide a counterexample to show that the statement does not hold in general for $ p = 1 $:

1. Let $ E = \{f \in L^1(\mu) : \|f\|_1 = 1, f \geq 0 \text{ a.e.}\} $.
2. Consider two distinct functions $ f(t) = \frac{1}{2\pi} $ and $ g(t) = \frac{1}{\pi} \chi_{[0, \pi]}(t) $, both of which belong to $ E $ and satisfy $ \|f\|_1 = \|g\|_1 = 1 $.
3. Here, $ \inf_{f \in E} \|f\|_1 = 1 $, but both $ f $ and $ g $ achieve this infimum, showing that the minimizer is not unique.

Thus, the statement does not hold for $ p = 1 $.

C. For $ p = \infty $

For $ L^\infty(\mu) $, strict convexity also fails, preventing uniqueness of minimizers.

1. Consider the set $ E = \{f \in L^\infty(\mu) : f(t) = 1 \text{ for } t \in [0, \pi), f(t) \in [-1, 1] \text{ for } t \in [\pi, 2\pi)\} $.
2. For any $ f \in E $, we have $ \|f\|_\infty = 1 $, and different choices of $ f $ in $ E $ give distinct functions with the same $ L^\infty $-norm.

Thus, uniqueness does not hold for $ p = \infty $ either.

Conclusion: The statement holds if and only if $ p \in (1, \infty) $. This range ensures that $ L^p(\mu) $ is both reflexive and strictly convex, guaranteeing the existence and uniqueness of a norm-minimizing element in any closed, convex subset of $ L^p(\mu) $.