📂Linear Algebra

Necessary and Sufficient Condition for a Linear Functional to Be Continuous

Theorem¹

A linear functional $f$ is continuous. $\iff$ $\ker(f)$ is a closed set in $X$.

Here, $\mathcal{N} (f) = \ker (f) = \left\{ x \in X \ | \ f(x) = 0 \right\}$ is the kernel of the linear transformation $f$.

Proof

Strategy: $(\implies)$ We deduce it directly from the definition of the kernel. $(\impliedby)$ By the properties of linear operators, the necessary and sufficient condition for continuity is boundedness. Showing that $f$ is bounded is relatively easy.

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• $(\implies)$

  If $x \in \overline { \ker (f) }$, then there exists a sequence $\left\{ x_{n} \right\}_{n \in \mathbb{N}}$ in $\ker (f)$ such that $x_{n} \to x$ as $n \to \infty$. Since $f$ is continuous, $\displaystyle \lim_{n \to \infty} f(x_{n}) = f(x)$, and since $x_{n} \in \ker (f)$,

  $$ 0 = \lim_{n \to \infty} f(x_{n}) = f(X) $$

  In other words, $x \in \ker (f)$, so

  $$ \overline{ \ker (f) } \subset \ker (f) $$

  Of course, $\ker (f) \subset \overline { \ker (f) }$, so $\ker (f) = \overline { \ker (f) }$, and therefore $\ker (f) $ is a closed set in $X$.

• $(\impliedby)$

  If we assume $\| f \| = \infty$, then

  $$ \| y_{n} \| = 1 $$

  $$ \lim_{n \to \infty } |f(y_{n} ) | = \infty $$

  there exists a sequence $\left\{ y_{n} \right\}_{n \in \mathbb{N}}$ in $X$ satisfying the above.

  Meanwhile, since $f \ne 0$, there must exist some $x_{0} \in X$ such that $f( x_{0 } ) \ne 0$. Now, if we set $\displaystyle e: = {{x_{0}} \over { f ( x_{0} ) }}$, then

  $$ f(e) = 1 $$

  Defining a new sequence $\displaystyle z_{n} : = f(e) - {{ y_{n} } \over { f(y_{n}) }}$ gives

  $$ f(z_{n} ) = f(e) - {{ f ( y_{n} ) } \over { f(y_{n}) }} = 0 $$

  so $z_{n } \in \ker (f)$, and since $\displaystyle \| z_{n} - e \| = \left\| {{ y_{n} } \over { f(y_{n}) }} \right\| = \| {{ 1 } \over { f(y_{n}) }} \| \left\| y_{n} \right\| \to 0$ as $n \to \infty$,

  $$ z_{n} \to e $$

  Therefore $e \in \ker (f)$, and $1 = f(e) = 0$, which is a contradiction.

  Properties of Linear Operators

  $T$ is continuous $\iff$ $T$ is bounded

  Saying $\| f \| < \infty$ means precisely that $f$ is bounded, and therefore $f$ is continuous.

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