This question comes from Linear and Nonlinear Functional Analysis with Applications (Philippe G. Ciarlet), Chapter 5, Problem 5.9-2.
5.9-2 Let $X$ be a normed vector space, let $Y$ be a strict subspace of $X$, and let $l: Y→ \mathbb{K}$ be a continuous linear functional. Show that there exist continuous linear functionals $\hat{l}:X→ \mathbb{K}$ that satisfy $\hat{l}(y)=l(y)$ for all $y\in Y$ and $\|\hat{l}\|_{X'} > \|l\|_{Y'}$.
I try to construct a continuous linear functional $l_1$ on subspace $Z=\{ax_0+y:a\in \mathbb{K},y\in Y\}(x_0\in X-Y)$ such that $\|{l}_1\|_{Z'} > \|l\|_{Y'}$ and use Hahn-Banach Theorem, but failed. Any help is appreciated!
Best Answer
Let $x_0\notin \bar{Y}$ and $\|x_0\|=1.$ Define $l_0$ on $Y\oplus \mathbb{K}x_0$ by $$l_0(y+ax_0)=l(y)+(1+\|l\|)ax_0$$ Then $l_0$ is a bounded linear functional and $$\|l_0\|\ge |l_0(x_0)|=1+\|l\|>\|l\|$$ Indeed, assume that $y_n+a_nx_0\underset{n\to \infty}{\longrightarrow}y+ax_0.$ Then $$0\underset{n\to \infty}{\longleftarrow} \|(y_n-y)+(a_n-a)x_0\|\ge |a_n-a|\operatorname{dist}(x_0,Y)$$ Hence $a_n\underset{n\to \infty}{\longrightarrow} a.$ Thus $y_n\underset{n\to \infty}{\longrightarrow} y.$ We obtain $$l_0(y_n+a_nx_0)=l(y_n)+(1+\|l\|)a_n\\ \underset{n\to \infty}{\longrightarrow} l(y)+(1+\|l\|)a=l_0(y+ax_0)$$ By Hahn-Banach theorem the functional $l_0$ can be extended to a bounded linear functional $\hat{l}$ on $X.$
The conclusion is not true if $\bar{Y}=X.$ In this case the functional $l$ has a unique continuous extension $\hat{l}$ from $Y$ to $X$ and the norm of $\hat{l}$ is equal to the norm of $l.$