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Let $\mathscr{L}(E,F)$ denote the space of all linear functionals from $E \to F$. Let $\mathscr{C}(E,F)$ denote the space of continuous linear functionals from $E \to F$. My question:

  • How to prove that $\mathscr{C}(E,F)$ is a subspace of $\mathscr{L}(E,F)$?

  • For $T \in \mathscr{C}(E,F)$ define $$||T||=\inf \bigl\{ k\in\mathbb{R} \ :\ \forall x \in E, \ ||T(x)||_{F} \leq k||x||_{E}\bigr\}$$ Then how to show that above defines a norm on $\mathscr{C}(E,F)$?

Proving subspace is easy, I think since, if $f,g \in \mathscr{C}(E,F)$ then $\alpha\cdot f + \beta \cdot g \in \mathscr{C}(E,F)$ this carries on from the definition of continuity itself. But what about proving norm.

Detailed explanations will help. Thanks.

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Note that the assumption of continuity implies that $\lVert\cdot\rVert$ is well-defined. We need to check the axioms.

Homogeneity ($\lVert aT\rVert=|a|\lVert T\rVert$) is a consequence of $\inf\{\lambda S\}=\lambda\inf S$ for $\lambda\geqslant 0$.

Let $T\in\mathcal C(E,F)$ be such that $\lVert T\rVert=0$. Then for all $k>0$ and all $x\in E$, $\lVert Tx\rVert_F\leqslant k\lVert x\rVert_E$, hence $T\equiv 0$.

For triangular inequality, let $S,T\in\mathcal C(E,F)$. For an integer $n$, consider $a_n,b_n\in\Bbb R$ such that $$\forall x\in E, \quad\lVert Sx\rVert\leqslant a_n\lVert x\rVert,\lVert Tx\rVert\leqslant b_n\lVert x\rVert,$$ and $a_n\leqslant \lVert S\rVert+1/n$, $b_n\leqslant \lVert T\rVert+1/n$. What about $\lVert (S+T)x\rVert$?

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with attention to $||T||=\inf \bigl\{ k\in\mathbb{R} \ :\ \forall x \in E, \ ||T(x)||_{F} \leq k||x||_{E}\bigr\} $ k>0 so$0\le ||T||$ it will be remain to show $$||T_1+T_2||\le ||T_1||+||T_2||$$ $||(T_1+T_2)(x)||_F\le ||T_1(x)||_F+||T_2(x)||_F\le( k_1+k_2)||x||_E$ ($k_1 $ and $ k_2$ are coresspond to inf of defintion of $T_1$and $T_2$) so k correspond to $T_1+T_2$ will be $k\le k_1+k_2$ so

||T_1+T_2||\le ||T_1||+||T_2||

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