# Strictly positive extensions of linear functionals

There is an interesting exercise in Steven Roman’s book Advanced Linear Algebra:

(Page 226, Chapter 9) 25. Let $$f:S\rightarrow\mathbb{R}$$ be a strictly positive linear functional on a subspace $$S$$ of $$\mathbb{R}^n$$. Prove that $$f$$ has a strictly positive extension to $$\mathbb{R}^n$$. Use the fact that if $$U\cap\mathbb{R}_+^n=\{0\}$$, where

$$\mathbb{R}_+^n = \{(a_1,\dots,a_n) \mid a_1,\dots,a_n \ge 0\}$$

and $$U$$ is a subspace of $$\mathbb{R}^n$$, then $$U^\perp$$ contains a strongly positive vector.

Firstly, some notation and terminology. Write $$A \odot B$$ for the direct sum $$A \oplus B$$ where $$\langle a,b \rangle = 0$$ for all $$a \in A,b \in B$$. If $$f \in V^*$$ is a linear functional, then the Riesz vector $$R_f$$ is the unique vector for which $$f v = \langle v,R_f \rangle$$ for all $$v \in V$$. A vector $$v=(v_1,\dots,v_n)$$ is strictly positive, written $$v>0$$, if $$v_i \ge 0$$ for all $$i$$ but $$v \ne 0$$. The vector $$v$$ is strongly positive, written $$v \gg 0$$, if $$v_i>0$$ for all $$i$$. Let $$f:S \rightarrow \mathbb{R}$$ be a linear functional on a subspace $$S$$ of $$\mathbb{R}^n$$. If $$v>0 \Rightarrow f(v)>0$$ for all $$v \in S$$ then $$f$$ is strictly positive.

Notice that if $$S$$ does not contain any strictly positive vectors, then the fact that $$f$$ is strictly positive is useless. We therefore split the proof into two cases. Below, we write $$V$$ for $$\mathbb{R}^n$$.

# The proof for case 1

Suppose that $$S$$ contains a strictly positive vector $$v_1$$ and let $$K=\ker(f)$$. We have the decompositions $$S = \langle R_f \rangle \odot K$$ and

$$V = S^\perp \odot S = S^\perp \odot \langle R_f \rangle \odot K,$$

where $$R_f \in K^\perp$$ is the Riesz vector for $$f$$ (see Lemma C below).

Now if $$v \in \ker(f)$$ then $$f(v)=0$$, so $$K$$ does not contain any strictly positive vectors (everything there is sent to 0!). Applying the hint given by Roman, $$K^\perp = S^\perp \odot \langle R_f \rangle$$ contains some strongly positive vector $$z = s+a R_f$$ where $$s \in S^\perp, a \in \mathbb{R}$$. Since $$a \langle R_f, v_1 \rangle = \langle z, v_1 \rangle > 0$$, we must have $$a \ne 0$$. So $$z \in K^\perp \setminus S^\perp$$, and by Lemma A (below) there exists a $$\lambda$$ such that the linear functional $$g(x)=\langle x, \lambda z \rangle$$ extends $$f$$. But $$v_1 > 0$$ and $$z \gg 0$$, so

$$\lambda \langle v_1,z \rangle = g(v_1) = f(v_1) > 0$$

implies that $$\lambda > 0$$. Finally, applying Lemma B (below) shows that $$g$$ is strictly positive.

# The proof for case 2

Now suppose that $$S$$ does not contain any strictly positive vectors. I was unable to conjure up a purely algebraic proof for this case, but fortunately the analysis here is not too ugly. The basic idea is that the behavior of $$f$$ is irrelevant since $$S$$ contains no strictly positive vectors, so we try define an extension that “overrides” $$f$$ on strictly positive vectors. Of course, this requires us to bound the behavior of $$f$$ in some way; before we can proceed, we need a lemma.

# The lemma

Let $$p,s \in \mathbb{R}^n$$ where $$p$$ is strongly positive. Then there exists a $$C>0$$ such that

$$\displaystyle \left\vert\frac{\langle v,s \rangle}{\langle v,p \rangle}\right\vert < C$$

for all positive vectors $$v \in \mathbb{R}^n$$.

Proof. Let $$E = \{ v \in \mathbb{R}^n \mid \Vert v \Vert = 1,v>0 \}$$, which is a compact set. It is well known that the maps $$v \mapsto \langle v,s \rangle$$ and $$v \mapsto \langle v,p \rangle$$ are continuous. Additionally, $$\langle v,p \rangle \ne 0$$ for any $$v \in E$$ since $$v>0$$ and $$p \gg 0$$. Therefore $$f(v)=\langle v,s \rangle / \langle v,p \rangle$$ is continuous on $$E$$, and since $$E$$ is compact, $$f(E)$$ is compact (and therefore bounded). Choose a $$C>0$$ such that $$|f(v)|<C$$ whenever $$v \in E$$. Then

$$\displaystyle |f(v)| = \left\vert\frac{\langle v/\Vert v \Vert,s \rangle}{\langle v/\Vert v \Vert,p \rangle}\right\vert < C$$

for all $$v>0$$, noting that $$v/\Vert v \Vert \in E$$.

# The proof for case 2 (continued)

Since $$S$$ contains no strictly positive vectors, we can apply Roman’s hint to conclude that $$S^\perp$$ contains a strongly positive vector $$v_1$$. Choose a subspace $$T$$ of $$S^\perp$$ such that $$S^\perp = \langle v_1 \rangle \odot T$$, and write

$$V = S^\perp \odot S = S^\perp \odot \langle w \rangle \odot K$$

where $$K = \ker(f)$$ and $$w \in K^\perp$$. By the lemma above, there exists a $$C>0$$ such that

$$\displaystyle \vert \langle v,w \rangle / \langle v,v_1 \rangle \vert < C$$

for all $$v>0$$. Let $$M=C|f(w)|$$ and define $$g:\mathbb{R}^n\rightarrow\mathbb{R}$$ by

$$a_1 v_1 + t + a w + k \mapsto a_1 M + a f(w)$$

where $$a_1,a \in \mathbb{R}$$, $$t \in T$$ and $$k \in K$$. (The sum on the left is just the decomposition of the vector according to the big direct sum $$V = \cdots$$ above.) If $$v = a_1 v_1 + t + a w + k$$ is a positive vector then $$a_1 = \langle v,v_1 \rangle > 0$$ and $$a = \langle v,w \rangle$$, so

$$\displaystyle \left\vert \frac{a}{a_1} f(w) \right\vert < C|f(w)| = M$$.

Therefore

$$\displaystyle g(v) = a_1 M + a f(w) = a_1 \left( M + \frac{a}{a_1} f(w) \right) > 0$$.

This shows that $$g$$ is a strictly positive extension of $$f$$.

# The lemmas

These lemmas are quite easy to prove.

### Lemma A

(Page 225, Chapter 9) 23. Let $$f$$ be a nonzero linear functional on a subspace $$S$$ of a finite-dimensional inner product space $$V$$ and let $$K = \ker(f)$$. Show that if $$g \in V^*$$ is an extension of $$f$$, then $$R_g \in K^\perp \setminus S^\perp$$. Moreover, for each vector $$u \in K^\perp \setminus S^\perp$$ there is exactly one scalar $$\lambda$$ for which the linear functional $$g(x) = \langle x, \lambda u \rangle$$ is an extension of $$f$$.

### Lemma B

A linear functional $$f$$ on $$\mathbb{R}^n$$ is strictly positive if and only if $$R_f \gg 0$$.

### Lemma C

Let $$V$$ be a finite-dimensional inner product space and let $$f:V\rightarrow F$$ be a nonzero linear functional. Write $$V = \langle w \rangle \odot \ker(f)$$ where $$w \in \ker(f)^\perp$$ is nonzero (note $$\dim(\ker(f)) = \dim(V) – 1$$). Then $$R_f \in \langle w \rangle$$.

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