On The Local Convexity of $L^p$ Spaces for $0<p<1$

Let $(X,\mathcal{M},\mu)$ be a measure space.

In this post, I wish to rigorously prove the following theorem.

Theorem: If $(X,\mathcal{M},\mu)$ is such that $L^1(X,\mathcal{M},\mu)$ is infinite dimensional, then $L^p(X,\mathcal{M},\mu)$ with $0<p<1$ is not locally convex.

Before that, we first prove the following important lemma.

Lemma: If $\mu$ takes only finitely many values then $L^1(X,\mathcal{M},\mu)$ is finite dimensional

Proof of lemma: Let $a_1$ be the smallest non-zero, finite value that $\mu$ takes on all measurable sets. Let $A_1\in\mathcal{M}$ be any set such that $\mu(A_1)=a_1.$ If no such $a_1$ exists, we skip directly to $(\star).$

Assume we have constructed sets $A_1,A_2,\dots,A_k$ for any $k\in\mathbb{N}$ such that $\mu(A_j)=a_j$ for all $j\le k,j\in\mathbb{N}.$ Let $a_{k+1}$ be the smallest non-zero, finite value that $\mu$ takes on all measurable subsets of $X\setminus \cup_{j=1}^k A_j.$ Let $A_{k+1}\in\mathcal{M}$ be any set such that $A_{k+1}\subset X\setminus \cup_{j=1}^k A_j$ and $\mu(A_{k+1})=a_{k+1}.$

Claim 1: The process terminates after a finite number of iterations.

Proof of claim: Assume, for a contradiction, that it does not.

We note that $(a_n)_{n=1}^\infty$ is a non-decreasing sequence: if $a_{n+1}<a_n$ for some $n\in\mathbb{N}$ then $0<\mu(A_{n+1})<\mu(A_{n})<\infty$ and $A_{n+1}\subset X\setminus \cup_{j=1}^n A_j\subset X\setminus \cup_{j=1}^{n-1} A_j$ which contradicticts the definition of $a_n.$

Since the process does not terminate and the sequence $(a_n)_{n=1}^\infty$ is non-decreasing, there must exist some $k\in\mathbb{N}$ such that $a_j=a_k$ for all $j>k,j\in\mathbb{N}$ since $\mu$ can only take on a finite number of values. Now, for any $n>k,n\in\mathbb{N},$ we have $\mu(\cup_{j=k+1}^nA_j)=\sum_{j=k+1}^n\mu(A_j)=\sum_{j=k+1}^na_k=(n-k-1)a_k$ by disjointness. Since this holds for all $n>k$ and $0<a_n<\infty$, we see that $\mu$ takes on infinitely many values, which is a contradiction. Thus, our process terminates.

Say, it terminates after $N\in\mathbb{N}\cup\{0\}$ steps to give sets $A_1$ through $A_N.$ ($\star$) Note that the case where no such sequence can be formed at all (i.e. $ N=0$) is handled vacuously from here onwards (considering an empty collection of such sets).

Claim 2: All subsets of $A_n$ are either null or have the same measure as $A_n$ for all $n\in\mathbb{N}.$

Proof of claim: If, for any $n\in\mathbb{N},$ there exists a $B_n\subset A_n$ such that $0<\mu(B_n)<\mu(A_n)=a_n$ then it contradicts the definition of $a_n$ as $B_n\subset A_n\subset X\setminus \cup_{j=1}^{n-1}A_j.$ So our claim is proved.

Continuation of the proof of lemma: Let $A_0=X\setminus \cup_{n=1}^N A_n.$ We have $X=\sqcup_{n=0}^NA_n.$ Since the process terminated after $N$ steps, we must have $\mu(B)=0$ or $\mu(B)=\infty$ for all $B\subset A_0,B\in\mathcal{M}.$

So for any $f\in L^{1}(\mu),$ we have $\int_{A_0}|f|d\mu\le \int |f|d\mu<\infty.$ Let $E_n=\{x\in X:|f(x)|>1/n\}$ for all $n\in\mathbb{N}.$ We have $$\mu(A_0\cap E_n)=n\int_{A_0\cap E_n}1/nd\mu\le n\int_{A_0\cap E_n}|f|d\mu\le n\int_{A_0}|f|d\mu<\infty$$

By the dichotomy established before, we have $\mu(A_0\cap E_n)=0$ for all $n\in\mathbb{N}.$ Since $\{f\neq 0\}=\{|f|>0\}$ and $\{x\in A_0:|f(x)|>0\}=\cup_{n=1}^\infty (A_0\cap E_n)$ with $\mu(\cup_{n=1}^\infty (A_0\cap E_n))\le \sum_{n=1}^\infty \mu(A_0\cap E_n)=0,$ we have $f=0$ a.e in $A_0.$

Claim 3: All integrable simple functions are constant a.e in the sets $A_1,A_2,\dots,A_N.$

Proof of claim: Let $\phi=\sum_{i=1}^n \xi_i\chi_{E_i}$ be any measurable simple function in standard form.

For any $j\le N,j\in\mathbb{N},$ $\phi\chi_{A_j}=\sum_{i=1}^n \xi_i\chi_{E_i\cap A_j}.$ If $\mu(E_k\cap A_j)=0$ for all $k\le n,k\in\mathbb{N}$ then it is easy to show that $\phi=0$ a.e in $A_j.$

Otherwise, by claim 2, there exists some $k\le n,k\in\mathbb{N}$ such that $\mu(E_k\cap A_j)=a_j.$ If $\mu(E_{k'}\cap A_j)=a_j$ for some $k'\neq k,k'\le n,k'\in\mathbb{N},$ then $a_j=\mu(A_j)\ge \mu((E_{k}\cup E_{k'})\cap A_j)=\mu(E_k\cap A_j)+\mu(E_{k'}\cap A_j)=2a_j$ by disjointness of $\{E_i\}_{i=1}^n.$ This produces a contradiction as $0<a_j<\infty.$ So there is only one $k$ such that $\mu(E_k\cap A_j)>0.$

Thus, $\phi=\xi_k$ a.e in $A_j$ since $\{x\in A_j:\phi(x)\neq \xi_k\}=\cup_{i\neq k,i\le n}(E_i\cap A_j)$ with $\mu(\cup_{i\neq k,i\le n}(E_i\cap A_j))=\sum_{i\neq k,i\le n}\mu(E_i\cap A_j)=0.$

Since $j$ is arbitrary, we conclude that $\phi$ is constant a.e in the sets $A_1$ through $A_N$. Since we already know that $\phi=0$ a.e in $A_0$ and $X=\sqcup_{j=0}^N A_j,$ we have $\phi=\sum_{j=1}^N\alpha_j\chi_{A_j}$ a.e for some $\{\alpha_1,\alpha_2,\dots,\alpha_N\}\subset \mathbb{C}.$

Claim 4: All functions in $L^{1}(\mu)$ are constant a.e in the sets $A_1,A_2,\dots,A_N.$

Proof of claim: Let $f\in L^{1}(\mu)$ and $\{\phi_n\}_{n=1}^\infty$ a sequence of integrable simple functions defined as in theorem 2.10 (b) (Folland). Using claim 3, $\phi_n=\sum_{j=1}^N\alpha_{nj}\chi_{A_j}$ a.e for some $\{\alpha_{n0},\alpha_{n1},\dots,\alpha_{nN}\}\subset \mathbb{C}$ for all $n\in\mathbb{N}.$ Say, equality holds in $\mathcal{S}_n^c$ for some $\mathcal{S}_n\in\mathcal{M}$ with $\mu(\mathcal{S}_n)=0.$

Since $\phi_n\to f$ a.e as $n\to\infty,$ there exists some set $\mathcal{S}_0\in\mathcal{M},$ such that $\lim_{n\to\infty}\phi_n(x)=f(x)$ for all $x\in \mathcal{S}_0^c$ and $\mu(\mathcal{S}_0)=0.$

Consider any $x\in \cap_{n=0}^\infty\mathcal{S}^c_n.$ If $x\in A_j$ for some $j\in\mathbb{N},j\le N$ then we have $$f(x)=\lim_{n\to\infty}\phi_n(x)=\lim_{n\to\infty}\left(\sum_{i=1}^N\alpha_{ni}\chi_{A_i}(x)\right)=\lim_{n\to\infty}\alpha_{nj}$$

Hence, $f$ is constant in $A_j\cap \left(\cap_{n=0}^\infty\mathcal{S}^c_n\right).$ We call this constant $\alpha_j.$

On the other hand, if $x\in A_0,$ then the same arguments as above show $f(x)=0$. So $f=\sum_{j=1}^N \alpha_j \chi_{A_j}$ for all $x\in \cap_{n=0}^\infty\mathcal{S}^c_n.$ Since $\mu(\cup_{n=0}^\infty\mathcal{S}_n)\le \sum_{n=0}^\infty\mu(\mathcal{S}_n)=0,$ we conclude that $f=\sum_{j=0}^N \alpha_j \chi_{A_j}$ a.e.

Final steps of the proof of lemma: Using claim 4, every function in $L^{1}(\mu)$ can be expressed as a linear combination of functions $\{\chi_{A_1},\chi_{A_2},\dots,\chi_{A_N}\}$ (modulo a.e). Thus, $L^1(\mu)$ is finite dimensional. Hence, proved.

Now, we can prove the original theorem without much extra effort.

Proof of theorem: Using the contrapositive of lemma 1, $\mu$ assumes infinitely many values. So there exists some set $A_1\in\mathcal{M}$ such that $0<\mu(A_1)<\mu(X).$ Let $B_1=X\setminus A_1.$ Note that both $\mu(A_1),\mu(B_1)>0.$ Obviously $\mu$ assumes infinitely many values on subsets of $B_1$ or that on subsets of $A_1.$

Assume, for a contradiction, that it is not true. Let $\mu$ assume values in finite set $K_A$ on measurable subsets of $A_1$ and values in finite set $K_B$ on measurable subsets of $B_1.$ Then, for any $S\in\mathcal{M},$ we have $S=(S\cap A_1)\cup(S\cap B_1).$ So $\mu(S)=\mu(S\cap A_1)+\mu(S\cap B_1)=a+b$ for some $a\in K_A$ and $b\in K_A.$ So $\mu$ assumes only finitely many values (set of possible values is $K_A+K_B$) on $X,$ which is a contradiction. So the claim is proved.

If $\mu$ assumes infinitely many values on measurable subsets of $A_1,$ we set $X_1=A_1$ and $S_1=B_1.$ Otherwise, set $X_1=B_1$ and $S_1=A_1.$

We repeat the process recursively with the new space $(X_1,\mathcal{M}|_{X_1},\mu).$

This process yields a sequence $\{S_n\}_{n=1}^\infty$ of sets such that $\mu(S_n)>0$ for all $n\in\mathbb{N}.$ For any $i>j$ where $i,j\in\mathbb{N}$ we have $S_i\subset X_j$ and $S_j=X_{j-1}\setminus X_j.$ This shows that we have a disjoint sequence of sets.

Assume for a contradiction that $L^p(X,\mathcal{M},\mu)$ is locally convex. So there exists some convex neighborhood $U$ of $\mathbf{0}\in L^p(X,\mathcal{M},\mu)$ such that $U\subset B(\mathbf{0},1).$ Since $U$ is open, we have another open ball $B(\mathbf{0},2\epsilon)\subset U$ for some $\epsilon>0.$

For any $n\in\mathbb{N}$ set $f_n=\left(\frac{\epsilon}{\mu(S_n)}\right)^{1/p}\chi_{S_n}.$ Now, $d(f_n,\mathbf{0})=\int |f_n|^pd\mu=\frac{\epsilon}{\mu(S_n)}\mu(S_n)=\epsilon<2\epsilon$ since $\chi_{S_n}^p=\chi_{S_n}.$ Note that, by our construction, only $S_1$ can have infinite measure, in which case, we just start the sequence from $S_2.$ So the sequences of functions are well-defined.

Thus, $\{f_n\}_{n=1}^\infty\subset B(\mathbf{0},2\epsilon)\subset U$ for all $n\in \mathbb{N}.$ For any $N\in\mathbb{N},$ let $g_N=1/N\sum_{n=1}^N f_n.$ Now, $$\int |g_N|^pd\mu=\int_{X\setminus \cup_{n=1}^N S_n}|g_N|^pd\mu+\sum_{n=1}^N\int_{S_n}|g_N|^pd\mu=\sum_{n=1}^N\int_{S_n}|f_n/N|^pd\mu\text{ (by disjointness of $\{S_n\}_{n=1}^\infty$)}$$

Now, $\int_{S_n}|f_n/N|^pd\mu=\frac{\epsilon/N^p}{\mu(S_n)}\mu(S_n)=\epsilon/N^p.$ Thus, $\int|g_N|^pd\mu=\sum_{n=1}^N\epsilon/N^p=\epsilon N^{1-p}.$

Since $0<p<1,$ we have $1-p>0.$ Thus, there exists some $N_0\in\mathbb{N}$ such that $N_0^{1-p}>1/\epsilon$ since $\lim_{N\to\infty}N^{1-p}=\infty.$ Thus, $d(g_{N_0},\mathbf{0})=\int |g_{N_0}|^pd\mu=\epsilon N_0^{1-p}>1.$ This means $g_{N_0}\notin B(\mathbf{0},1).$ Thus, $g_{N_0}\notin U$ as well since $U\subset B(\mathbf{0},1).$

This contradicticts the convexity of $U$ since $g_{N_0}$ is a finite convex combination of elements in $U$. So our assumption was wrong. Hence, $L^p(X,\mathcal{M},\mu)$ with $0<p<1$ is not locally convex.

Note: The above theorem, hence, implies that $L^p(X,\mathcal{M},\mu)$ is not normable for $p\in(0,1)$ if $L^1(X,\mathcal{M},\mu)$ is infinite dimensional. So, in most “interesting” cases, $L^p$ spaces (for $p\in(0,1)$) are not normable.