let $L$ be a linear form, is $L$ continuous?

Question

Let $L$ be a linear form, goes from $E$ normed vectorial space, in $K=\mathbb R$ or $\mathbb C$.

Is it continuous?

My idea, is that $K$ is a banach space, so the space of linear forms is a banach space, so that affects the continuity of $L$?

Answer 1

Not necessarily. Let $E$ be the space of all sequences $(x_n)_{n\in\mathbb N}$ of complex numbers such that $x_n=0$ if $n$ is large enough and define $\bigl\|(x_n)_{n\in\mathbb N}\bigr\|=\max_{n\in\mathbb N}|x_n|$ and $L\bigl((x_n)_{n\in\mathbb N}\bigr)=\sum_{n=1}^\infty x_n$. Then $L$ is linear, but not continuous.

Answer 2

If $E$ has finite dimension, then yes, it is necessarily continuous.

If $E$ has infinite dimension, then there is always a linear functional that is not continuous. Pick a non-zero vector $x_0$. Then choose $x_1$ such that $\|x_0-x_1\|<1$ but $x_0,x_1$ are linearly independent.

When $x_0,x_1,...,x_n$ has been chosen, choose $x_{n+1}$ such that $\|x_{n+1}-x_0\|<1/n$ and $x_{n+1}$ is outside the linear span of $x_0,x_1,...,x_n$.

Complete $\{x_0,x_1,...\}$ to a basis $\{x_i\}_{i\in I}$ in any way you want, and define $f(x_0)=1$ and $f(x_i)=0$ for $i\neq 0$. Extend by linearity.

This functional is necessarily discontinuous, since $x_1,x_2,...$ tends to $x_0$, but $f(x_i)=0$ for $i\neq 0$ and $f(x_0)=1$.

Answer 3

It can be shown that a linear functional $f$ is continuous if and only if it is bounded, i.e. there exists $M > 0$ such that $$|f(x)| \le M\|x\|, \quad\forall x\in E$$

Any infinite-dimensional vector space $E$ admits a discontinuous linear functional. Namely, let $B$ be a Hamel basis for $E$ and let $(b_n)_{n=1}^\infty$ be a countable subset of $B$. We can assume that $\|b_n\| = 1$ for all $n \in \mathbb{N}$. Set $f(b_n) = n, \forall n \in \mathbb{N}$ and define $f(b)$ arbitrarily on the rest of $B$.

Extending by linearity gives a linear functional $f$ with $|f(b_n)| = n$ and $\|b_n\| = 1$ so $f$ cannot be bounded.