# Question about proof of Fundamental Theorem of Duality

I have a question about the proof of the following result (from Aliprantis and Border). Recall that $$ker f = \{x \in X: f(x)=0\}$$.

Theorem. Let $$f_1,...,f_n$$ be linear functionals on a vector space $$X$$. Then $$f$$ lies in the span of $$f_1,..., f_n$$ (that is, $$f = \sum_{i=1}^n \lambda_if_i$$ for some scalars $$\lambda_1,...,\lambda_n$$) if and only if $$\bigcap_{i=1}^n ker f_i \subset ker f$$.

Proof. If $$f = \sum^n_{i=1} \lambda_i f_i$$, then clearly $$\bigcap^n_{i=1} ker f_i \subset ker f$$. For the converse, assume that$$\bigcap^n_{i=1}ker f_i \subset ker f$$. Then $$T:X\to \mathbb R^n$$ via $$T(x)= (f_1(x),...,f_n(x))$$ is a linear operator. Since$$\bigcap^n_{i=1}ker f_i \subset ker f$$, if $$(f_1(x),..., f_n(x)) = (f_1(y),..., f_n(y))$$, then $$f_i(x − y) = 0$$ for each $$i$$ and so $$f(x) = f(y)$$. Thus the linear functional $$\phi:T(X) \to \mathbb R$$ defined by $$\phi(f_1(x),...,f_n(x) = f(x)$$ is well defined. Now note that $$\phi$$ extends to all of $$\mathbb R^n$$, so there exist scalars $$\lambda_1,...,\lambda_n$$ such that $$\phi(\alpha_1,...,\alpha_n) = \sum^n_{i=1} λ_i\alpha_i$$. Thus $$f(x) = \sum^n_{i=1} \lambda_i f_i(x)$$ for each $$x \in X$$, as desired.

I'm having trouble seeing why $$\phi$$ can be extended to all of $$\mathbb R^n$$. The way the proof is written, it seems like this step should be obvious, so I guess I'm missing something easy. Any hints would be appreciated.

• Do you want to use something like Hahn-Banach theorem? Commented Dec 14, 2019 at 19:28
• This here might be helpful: math.stackexchange.com/questions/3263817/… Commented Dec 15, 2019 at 20:06

In finite dimension, every subspace has a complement. Every linear map defined on a subspace can be extended to the whole space by defining the extended map as $$0$$ on such a complement subspace.

• What do you mean by "complement"?
• If $U$ is a subspace of $E$, I mean a subspace $V$ such that $U \oplus V = E$. Commented Dec 14, 2019 at 18:59
• The existence of a complement does not need the vector space to be finite-dimensional. If $U$ is a subspace of $E$, it has a basis $B$. This can be extended to a basis $B'$ of $E$, and then the linear span of $B' \setminus B$ is a complement to $U$.
• @grndl: This is explained in the answer. To define a linear map it suffices to define it on a basis and then extend linearly. So just define the extension of $\phi$ by $\phi'(x) = \phi(x)$ for $x$ in $T(X)$, and $\phi'(x)$ can be whatever you want for basis elements $x$ outside of $T(X)$, for example zero.