# Representation of a linear functional in vector space

In the book Functional Analysis,Sobolev Spaces and Partial Differential Equations of Haim Brezis we have the following lemma:

Lemma. Let $X$ be a vector space and let $\varphi, \varphi_1, \varphi_2, \ldots, \varphi_k$ be $(k + 1)$ linear functionals on $X$ such that $$[\varphi_i(v) = 0 \quad \forall\; i = 1, 2, . . . , k] \Rightarrow [\varphi(v) = 0].$$

Then there exist constants $\lambda_1, \lambda_2, \ldots, \lambda_k\in\mathbb{R}$ such that $\varphi=\lambda_1\varphi_1+\lambda_2\varphi_2+\ldots+\lambda_k\varphi_k$.

In this book, the author used separation theorem to prove this lemma. I would like ask whether we can use only knowledge of linear algebra to prove this lemma.

Thank you for all helping.

## 3 Answers

Your assumption is that $$\ker{\varphi} \supseteq \bigcap_{i=1}^k \ker{\varphi_i}$$.

Consider the linear map $$\ell \colon X \to \mathbb{R}^k$$ given by $$\ell(x) = (\varphi_1(x),\dots,\varphi_k(x))$$ and let $$V = \operatorname{im}\ell = \{\ell(x):x \in X\} \subseteq \mathbb{R}^k$$ be the image. We have $$\ker{\ell} = \bigcap_{i=1}^k \ker{\varphi_{i}} \subseteq \ker\varphi$$. Therefore $$\varphi = \tilde{\varphi} \circ \ell$$ for some linear functional $$\tilde{\varphi}\colon V \to \mathbb{R}$$ [explicitly, $$\tilde{\varphi}(v) = \varphi(x)$$ where $$x$$ is such that $$\ell(x) = v$$. This is well-defined and linear.]

Every linear functional defined on a subspace $$V$$ of $$\mathbb{R}^k$$ can be extended to a linear functional on all of $$\mathbb{R}^k$$ (write $$\mathbb{R}^k = V \oplus V^{\bot}$$ and set the extension to be zero on $$V^{\bot}$$) and every linear functional on $$\mathbb{R}^k$$ is of the form $$\psi(y) = \sum_{i=1}^k a_i y_i$$. Thus, there are $$\lambda_1,\dots,\lambda_k \in \mathbb{R}$$ such that $$\tilde\varphi(v) = \sum_{i=1}^k \lambda_i v_i$$ for all $$v \in V$$. In other words, $$\varphi = \sum_{i=1}^k \lambda_i \varphi_i$$.

Look at Rudin's "Functional Analysis" Lemma 3.9. The only issue I see is that the proof requires the extension of a functional from a subspace of a finite dimensional space to the entire finite dimensional space, but this is purely algebraic as far as I can see.

Here is a repackaging of linearalgebraist's answer:

Let $$L: X\to\mathbb{R}^k\\ Lx:=(\phi_1(x),\dots, \phi_k(x))$$ Then $${L}^*: (\mathbb{R}^k)^* \to X^* \\ {L}^*f:=f\circ {L}= f_1\phi_1+\dots+f_k\phi_k ,$$ so the conclusion you seek is equivalent to the general algebraic fact that $$\text{im} L^* = (\ker L)^{\bot}:=\{\phi \in X^* : \phi (x) = 0\; \forall x \in \ker L\}$$

More intuitively,
$$L$$ is injective on $$X/\ker L$$, which means that knowing $$Lx$$ amounts to knowing $$x$$ up to an element of $$\ker L$$. Hence, knowing $$Lx$$ implies knowing $$\phi(x)$$, because the element of $$\ker L$$ doesn't affect that. Clearly, all of this "knowing things" is linear, so you can write $$\phi(x)$$ as a linear combination of $$\phi_1(x),\dots,\phi_k(x)$$.