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I am working on the following exercise, but I really don't know how to start this:

Let $X$ be a normed space, and let $W$ be a linear subspace. Fix a $f_W \in W'$. Prove that the following are equivalent:

a) $W$ is dense in $X$

b) there exists a unique extension of $f_W$ to a continuous linear functional on $X$.

I know that a linear functional $f_X$ is an extension of $f_W$ if $f_X(u)=f_W(u)$ for all $u \in W$.

And $W$ is dense if every point in $X$ is contained in $W$, or a limit point of $W$.

I am very thankful for help!

Community
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Yuhe
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  • Apparently your notation $W'$ means continuous (bounded) linear functionals on the subspace $W$. This is worth explicitly mentioning. – hardmath Nov 18 '16 at 13:02

2 Answers2

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Fuxuan's proof is fine. I will provide a simpler approach for $(a)\implies(b)$:

If there are two extentions $f_1, f_2\in X'$ of $f_W$, then $f_1-f_2\in X'$ and $(f_1-f_2)(W)=0$. Thus $W$ is in the null space of some linear functional, hence is contained in some closed proper subspace of $X$, and therefore cannot be dense.

Aweygan
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From (a) to (b): define $$f_X(u):=\lim_{u_n\rightarrow u} f_W(u_n)$$ for all $u\in X$, $\{u_n\}$ a sequence in $W$ that converges to $u$.

Conversely, assume $\bar{W}\neq X$. Define $Y=X/\bar{W}$. Then there is a nontrivial continuous linear functional $T$ on $Y$. Composite this map with the quotient map and we get a linear functional $T'$. Then $f_X+T'$ is another extension for $f_W$, a contradiction.

Fuhsuan Ho
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  • In fact, the question has been asked and answered in this post http://math.stackexchange.com/questions/332154/functional-analysis-continuous-extensions-of-linear-functionals?rq=1 – Fuhsuan Ho Nov 18 '16 at 12:36