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I'm reading Proposition 0.7. in this lecture note.

Let $C$ be an open convex subset of a normed space $X$ and $f: C \to \mathbb{R}$ convex.

(a) If $f$ is u.s.c., then $f$ is continuous on $C$.

(b) If $X$ is a Banach space and $f$ l.s.c., then $f$ is continuous on $C$.

The proof of (b) uses the completeness of $X$ through the Baire Category Theorem.

Could you provide an example that (b) does not hold in case $X$ is not complete?

Akira
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2 Answers2

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Let $X = c_c$ be the space of finite sequences, equipped with the $\ell^2$-norm. Then, $f \colon c_c \to \mathbb R$, $$ f(x) := \sum_{n = 1}^\infty n x_n^2 $$ is convex, lower semicontinuous (via Fatou), but not continuous.

gerw
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I emailed to the author (professor Libor Vesely of Università degli Studi di Milano) of the notes and luckily got his response containing a counter-example.

$X$ is the normed space and $c_{00}$ the space of sequences that are definitely null equipped with the max-norm, and $f$ is the $\ell_1$-norm on $X$.

Below is my elaboration on the author's ideas.

Let $c_{0}$ be the space of null sequences endowed with the supremum norm $| \cdot |_\infty$.

  1. $(c_0, | \cdot |_\infty)$ is a Banach space.

Indeed, let $(x_n)_{n\in \mathbb N}$ be a Cauchy sequence in $c_0$ with $x_n := (x_{nk})_{k\in \mathbb N}$. For each $\varepsilon>0$, there is $N \in \mathbb N$ such that $$ |x_n-x_m|_\infty = \sup_{k \in \mathbb N} |x_{nk}-x_{mk}| < \varepsilon \quad \forall m,n>N. $$

It follows that $(x_{nk})_{n\in \mathbb N}$ is a Cauchy sequence for each $k \in \mathbb N$. Let $a_k := \lim_n x_{nk} \in \mathbb R$ and $a := (a_k)_{k \in \mathbb N}$.

  • First, we will prove that $x_n \to a$ in $| \cdot |_\infty$. In fact, $$ \begin{align} & |x_{nk}-x_{mk}| < \varepsilon \quad \forall m>N, n>N, k\in \mathbb N \\ \implies & \lim_m |x_{nk}-x_{mk}| = |x_{nk}-\lim_mx_{mk}| = |x_{nk}-a_{k}|\le \varepsilon \quad \forall n>N, k\in \mathbb N \\ \implies & \sup_{k \in \mathbb N}|x_{nk}-a_{k}| = |x_n - a|_\infty\le \varepsilon \quad \forall n>N. \end{align} $$

  • Second, we prove that $a \in c_0$, i.e., $a_k \to 0$ as $k \to \infty$. As above, $|x_{nk}-a_{k}|\le \varepsilon$ for all $n>N, k\in \mathbb N$. This implies $$ \lim_k |x_{nk}-a_{k}| = |\lim_k x_{nk}-\lim_ka_{k}| =|\lim_k a_k| \le \varepsilon \quad \forall n>N. $$ Because $\varepsilon>0$ is arbitrarily chosen, we get $\lim_k a_k=0$.

Let $c_{00}$ be the space of sequences which have only finitely many nonzero elements. Then $c_{00} \subset c_0$.

  1. $c_{00}$ is not closed in $c_{0}$.

Indeed, consider a sequence $(x_n)_{n\in \mathbb N}$ with $x_n := (x_{nk})_{k\in \mathbb N}$ where $x_{nk} := \frac{1}{k}$ if $k \le n$ and $0$ otherwise. It follows that $x_n$ converges to $(1, 1/2, 1/3,1/4, \ldots) \notin c_{00}$. It follows that $(c_{00}, |\cdot|_\infty)$ is an incomplete normed space.

Consider the map $$ f: c_{00} \to \mathbb R, x \mapsto |x|_{\ell_1} $$

Clearly, $f$ is convex.

  1. $f$ is l.s.c.

Let $(x_n) \subset c_{00}$ and $a \in c_{00}$ such that $x_n \to a$ in $|\cdot|_\infty$. We need to prove $$ f(a) \le \liminf_n f(x_n). $$

Define $\bar a \in c_{00}$ by $\bar a_k := |a_k|$. Similarly, we define $\bar x_n \in c_{00}$ by $\bar x_{nk} := |x_{nk} |$. First, $x_n \to a$ in $|\cdot|_\infty$ implies $x_{nk} \xrightarrow{n \to \infty} a_k$ and thus $\bar x_{nk} \xrightarrow{n \to \infty} \bar a_k$ for all $k\in \mathbb N$, i.e., $\bar x_n \to \bar a$ pointwise. Second, $\bar a, \bar x_n$ are non-negative. By Fatou lemma, we get $$ |\bar a|_{\ell_1} \le \liminf_n |\bar x_n|_{\ell_1} \quad \text{and thus} \quad f(\bar a) \le \liminf_n f(\bar x_n). $$

Notice that $f(\bar a) = f(a)$ and $f(\bar x_n) = f(x_n)$. The claim then follows.

  1. $f$ is not continuous.

We need to find a sequence $(x_n) \subset c_{00}$ and $a \in c_{00}$ such that $x_n \to a$ in $|\cdot|_\infty$ but $f(x_n) \not\to f(a)$. Define $x_{nk} = 1/n$ for all $k\le n$ and $0$ otherwise. Define $a_k = 0$ for all $k$. Clearly, they satisfy our requirement.

Akira
  • 18,439