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 \(R_f \in \langle w \rangle\).