11

Semi‐bounded quadratic functions over a sublevel setof another quadratic function
Huu‐Quang Nguyen1 and Reuy‐Lin Sheu2
1. Institute of Naturaı Science Education, Vinh University, Vinh, Nghe An, Vietnam
Department of Mathematics, National Cheng Kung University, Tainan, Taiwan

2. Department of Mathematics, National Cheng Kung University, Tainan, Taiwan

Abstract

Before attempting to solve an optimization problem, it is important to detect
whether the optimal value is finite.

This is called the boundedness problem.

Afunction f is called semi‐bounded if it is either bounded from below or boundedfrom above. In this article, we apply the separation peoperty developed by Nguyen

and Sheu [10] to study the semi‐boundedness of a quadratic fucntion f over thesublevel set \{x\in \mathbb{R}^{n}|g(x)\leq 0\} of another quadratic function g.

Key Words: S ‐lemma; S ‐lemma with equality; Separation property.

1Introduction
Given a pair of quadratic functions (f, g) in n real variables x=(x_{1}, x_{2}, \ldots , x_{n})^{T}with f(x)=x^{T}Ax+2a^{T}x+c and g(x)=x^{T}Bx+2b^{T}x+d , the theorem of S ‐lemma

with equality states the necessary and sufficient conditions under which the following two

statements are equivalent [i.e. (E_{1})\sim(E_{2}) ]:2(E)(\forall x\in \mathbb{R}^{n})g(x)=0\Rightarrow f(x)\geq 0.

(E)(\exists\lambda\in \mathbb{R})(\forall x\in \mathbb{R}^{n})f(x)+\lambda g(x) \geq 0.

The equivalence by (E_{1})\sim(E_{2}) means that both statements are true or false syn‐chronously.

Since (E_{2}) trivially implies (E_{1}) , it is clear that (E_{1})\sim(E_{2}) if and only

if (E_{1}) can impıy (E_{2}) . Geometrically, (E_{1})\Leftrightarrow\{x|g(x)=0\}\subseteq\{x|f(x)\geq 0\} . See
Figure 1 for an example. The question is, does there exist \lambda such that the non‐zero linear

Figure 1: Example for (E_{1}) : \{x|g(x)=0\}\subseteq\{x|f(x)\geq 0\}
combination f+\lambda g is non‐negative on the entire \mathbb{R}^{n} ?
The answer can be positive. Let f(x, y)=-x^{2}+y^{2}, g(x, y)=-x^{2} . Then,g(x, y)=0\Rightarrow x=0\Rightarrow f(0, y)=y^{2}\geq 0,
so (E_{1}) holds.
0,

On the other hand, choose\lambda=-1 .

Thenf(x, y)+\lambda g(x, y)=y^{2}\geq\forall(x, y)\in \mathbb{R}^{2} , so (E_{2}) holds as well.
The answer can be negative. Let f(x_{1}, x_{2})=-x_{1}^{2}+4x_{2}^{2}+2, g(x_{1}, x_{2})=x_{1}-x_{2} . Then,(E1) holds due tog(x_{1}, x_{2})= x ı— x 2 =0\Rightarrow f(x_{1}, x_{2})=3x_{2}^{2}+2\geq 0.

But for any \lambda\in \mathbb{R} , we consider (f+\lambda g) restricted on the x ‐axis:(f+\lambda g)(x_{1},0)=-x_{{\imath}}^{2}+2+\lambda x_{1}<0 ,asx_{1}arrow\pm\infty.

3Then, (E_{2}) fails for this pair of quadratics (f, g) and thus (E_{1})\not\simeq(E_{2}) .

The two examples indicate that to tell whether (Eı) \sim(E_{2}) is not a simple question.
It indeed was a long story. The first variant of the S ‐lemma with equality was proposed

by Finsler [5] in 1937 where (f, g) is a pair of homogeneous quadratic forms. The second
one was proposed by Fradkov and Jakubovich [6] in 1979 where they claimed that B\neq 0
implies that (E_{1})\sim(E_{2}) . The result was unfortunately wrong with the following counterexample f(x, y)=-x^{2}+y^{2}, g(x, y)=(x-y)^{2} where we notice that g does not satisfy
the two‐side Slater condition. The third variant was proposed by Luo, Sturm, and Zhang[8] in 2004 where g is assumed to be strictly concave (or strictly convex) satisfying thetwo‐side Slater condition. The fourth variant was proposed by Hoang Tuy and Hoang

Duong Tuan [ı5] in 2013 for a quadratic function f and a quadratic form g . The fifth
variant was a complete necessary and sufficient condition for (E_{1})7^{6}(E_{2}) by Xia, Wangand Sheu [14] in 2016. There are other different special cases of the S ‐lemma with equality

such as those in [3, 11] and in an early survey paper by Polik and Terlaky [12] in 2007.
Recently, Nguyen and Sheu [10] defined a notation of “separation property” andproved that it is indeed equivalent to the S ‐lemma with equality. Suppose the sublevel
set \{x|f(x)<0\} consists of two connected components, denoted by L_{-} and L_{+} . We also
define various level sets of f by L_{\alpha}^{*}(f)=\{x|f(x)*\alpha\}, where *\in\{\leq, <, =, >, \geq\}.

The hypersurface \{x\in \mathbb{R}^{n}|g(x)=0\} is said to separate \{x\in \mathbb{R}^{n}|f(x)<0\} if thetwo connected components L_{-} and L_{+} of \{x\in \mathbb{R}^{n}|f(x)<0\} lie in the opposite sides of\{x\in \mathbb{R}^{n}|g(x)=0\} such that L_{-}\cup L_{+}=\{x\in \mathbb{R}^{n}|f(x)<0\} andg(L_{-})g(L_{+})=g(a^{-})g(a^{+})<0, \forall a^{-}\in L_{-};\forall a^{+}\in L_ {+}.
By proper coordinate change, a non‐constance quadratic function f(x) must adopt one

4of the following five canonical forms:-x_{1}^{2}-\cdots-x_{k}^{2}+\delta(x_{k+1}^{2}+\cdots+x_{m}^{2})+\theta ;(1)-x_{1}^{2}-\cdots-x_{k}^{2}+\delta(x_{k+{\imath}}^{2}+\cdots+x_{m}^{2})-1 ;(2)-x_{1}^{2}-\cdots-x_{k}^{2}+\delta (x_{k+1}^{2}+ +x_{m}^{2})+x_{m+1} ;(3)x_{{\imath}}^{2}+ +x_{m}^{2}+\delta x_{m+1}+c' .

(4)\delta x_{1}+c' .

(5)where k\geq 1 is the number of negative eigenvalues of f(x) and \delta, \theta\in\{0,1\} . The following

theorems proved in [10] will be used for studying the semi‐boundedness property of thearticle.

Theorem 1 [10] If f(x) is a non‐constance quadratic function then L_{0}^{<}(f)=\{x\in\mathbb{R}^{n}|f(x)<0\} contains exactly two connected components if and only if f has form
(1) with k=1 . Furthermore, the two connected components are

L_{0-}^{<}(f)=\{x\in L_{0}^{<}(f)|x_{1}<0\} and L_{0^{+}}^{<}(f)=\{x\in L_{0}^{<}(f)|x_{1}>0\} .

(6)

Theorem 2 [10] Hypersurface L_{0}^{=}(g) separates L_{0}^{<}(f) if and only if(i) f(x) has the form-x_{1}^{2}+\delta(x_{2}^{2}+\cdots+x_{m}^{2})+\theta, \delta, \theta\in\{0,1\} ;(ii) With the same basis and \delta as in (i), g(x) has the form b_{1}x_{1}+\delta(b_{2}x_{2}+\cdots+b_{m}x_{m})+b_{0}, b_{1}\neq 0 ;(iii)\mathb {R}^{n-1}f|_{L_{0}^{=}(g)}(x)=-(\delta\frac{b_{2} {b_{1} x_{2}+\cdots+ \delta\frac{b_{m} {b_{1} x_{m}+\frac{b_{0} {b_{1} )^{2}+\delta(x_{2}^{2}+\cdots+ x_{m}^{2})+\theta\geq0, \foral (x_{2}, \cdots, x_{n})^{T}\in

Theorem 3 [1\theta] If g(x) takes both positive and negative values, then ((E_{1}) holds, but(E2) fails” ( S‐lemma with equality fails) if and only if L_{0}^{=}(g) separates L_{0}^{<}(f) .

52Main results
We first observe that f(x) is bounded from below on L_{0}^{\leq}(g) if, and only if \exists\alpha_{0}\in \mathbb{R}such that L_{\alpha_{0} ^{<}(f)\cap L_{0}^{\leq}(g)=\emptyset . If g(x) satisfies the Slater condition, then(\exists\alpha_{0}\in \mathbb{R})L_{\alpha 0}^{<}(f)\cap L_{0}^{\leq}(g)= \emptyset\Leftrightar ow (\exists\alpha_{0}\in \mathbb{R})L_{0}^{\leq}(g)\subseteq L_{\alpha_{0} ^{\geq}(f)\Leftrightarrow (\exists\alpha_{0}\in \mathbb{R})g(x)\leq 0\Rightarrow(f- \alpha_{0})(x)\geq 0\Leftrightar ow(\exists\alpha_{0}\in \mathbb{R})(\lambda\geq 0)(\forall x\in \mathbb{R}^{n}) f(x)+\lambda g(x)\geq\alpha_{0} , (By S ‐lemma)

where the last expression can be determined by solving an SDP program.

Otherwise,

when there is no Slater point so that g(x)\geq 0, \forall x\in \mathbb{R}^{n}, L_{0}^{\leq}(g)=L_{0}^{=}(g) is subspace of\mathbb{R}^{n} .

Whether f(x) is bounded from below on L_{0}^{\leq}(g) can be determined by restricting f(x)on the subspace L_{0}^{=}(g) to see whether the restriction is convex or not. If we want to test

whether\sup_{L_{0}^{\leq}(g)}f(x)<\infty , the same can be done by verifying whether \inf_{L_{0}^{\leq}(g)}-f(x)>-\infty.

Below we apply the separation theorem for the level set of quadratic functions (The‐orem 2) to combine the two tests (test for the lower bound and for the upper bound) of
a semi‐bounded function into a simple necessary and sufficient criterion. First, we havethe following property:
Lemma 1 Assume that g(x)\leq 0 satisfies the Slater condition. Then, f(x) is boundedfrom below on L_{0}^{<}(g) if and only if f(x) is bounded from below on L_{0}^{\leq}(g) .
Proof

Proof. Let \overline{x} be a boundary point of L_{0}^{<}(g) such that g(\overline{x})=0 . We claim that

there exists a sequence in L_{0}^{<}(g) converging to \overline{x} . By a change coordinate s=x-\overline{x},g(s)=S^{T}BS+2(\overline{x}^{T}B+2b^{T})s and

s=0 is a boundary point of

L_{0}^{<}(g) . Therefore, we

may simply assume that the boundary point \overline{x}=0 and that g(x)=x^{T}Bx+2b^{t}x . If

\nabla g(0)=2Bx+2b=0 , then b=0 and g(x)=x^{T}Bx is a quadratic form. Since g(x)\leq 0
satisfies the Slater condition,

B has a negative eigenvalue associated with an eigenvector

6w. Then, the sequence \{\frac{w}{n}\}_{n\in \mathb {N} converges to \overline{x}=0 and g( \frac{w}{n})<0 . On the other hand, let\nabla g(0)\neq 0 . By Gateaux differentiability of g at 0 along the direction -\nabla g(0) , we have\lim_{harrow 0+}\frac{g(0+h(-\nabla g(0) -g(0)}{h}=harrow 0^{+}1\dot{ \imath} m\frac{g(h(-\nabla g(0) }{h}=\langle\nabla g(0), -\nabla g(0)\rangle<0.
It shows that, with h small enough, g(-h\nabla g(0))<0 and \{-h\nabla g(0)\} converges to 0 as
harrow 0^{+} . Now suppose thatf(x) is bounded from below on L_{0}^{<}(g) such that g(x)<0\Rightarrow

f(x)\geq\alpha_{0} for some \alpha_{0}\in \mathbb{R} . Since every point \overline{x}\in L_{0}^{=}(g) has a convergent sequence in

L_{0}^{<}(g) that approaches to it, by the continuity of f, f(\overline{x})\geq\alpha_{0} . Therefore, f is boundedfrom below on L_{0}^{\leq}(g) . Conversely, if f(x) is bounded from below on L_{0}^{\leq}(g) and g(x) has
a Slater point, it trivially implies that f(x) is bounded from below on L_{0}^{<}(g) .

\squareWith Lemma 1, we can pose the semi‐boundedness theorem equivalently on L_{0}^{<}(g) .

Theorem 4 Function f is semi‐bounded on L_{0}^{<}(g) if and only if\{ begin{ar ay}{l (\exist \alpha_{0})g(x)<0\Rightar owf(x)\neq\alpha_{0}(equivalently, (\exist \alpha_{0})f(x)=\alpha_{0}\Rightar owg(x)\geq0); L_{\alpha0}^{=}(f)doesnotsepar teL_{0}^{<}(g). \end{ar ay}
Proof

Proof for necessity: If, on the contrary, \forall\alpha\in \mathbb{R} , there exists some \overline{x}\in \mathbb{R}^{n}such that g(\overline{x})<0 and f(\overline{x})=\alpha . Then, f(x) is unbounded from below and also fromabove on L_{0}^{<}(g) . It cannot be not semi‐bounded. Secondly, for \alpha_{0} such that g(x)<0\Rightarrow

f(x)\neq\alpha_{0} , if L_{\alpha_{0} ^{=}(f) separates L_{0}^{<}(g) , by theorem 2, there exists a basis of \mathbb{R}^{n} such thatg(x) is of form (1) that g(x)=-x_{1}^{2}+\delta(x_{2}^{2}+\cdots+x_{m}^{2})+\theta ; and f-\alpha_{0} is affine function
with form blxı +\delta (b2x2+\cdot\cdot\cdot +b_{rn}x_{m})+b_{0}, b_{1}\neq 0 . Then, it is easy to see that points(x_{1},0, \cdots , 0)\in L_{0}^{<}(g) for |x_{1}| sufficiently large, on which f(x) is unbound from below andalso form above.

Proof for Sufficiency: If there exists \alpha_{0} such that g(x)<0\Rightarrow f(x)\neq\alpha_{0} and L_{\alpha_{0} ^{=}(f)does not separate L_{0}^{<}(g) , we need to prove that f is semi‐bounded on L_{0}^{<}(g) . Indeed,if f(x) is semi‐bounded on \mathbb{R}^{n} , it is also semi‐bounded on L_{0}^{<}(g) . If f(x) is not semi‐bounded on \mathbb{R}^{n} , it implies that f(x)-\alpha_{0} takes both positive and negative values on\mathbb{R}^{n} .

Due to g(x)<0\Rightarrow f(x)\neq\alpha_{0} , we have f(x)-\alpha_{0}=0\Rightarrow g(x)\geq 0 and thus7(E1) holds for the pair (g, f-\alpha_{0}) . From the assumption that L_{\alpha_{0} ^{=}(f) does not separate

L_{0}^{<}(g) , the S ‐lemma with equality holds by theorem 3, It implies that \exists\lambda\in \mathbb{R} such thatg(x)+\lambda(f(x)-\alpha_{0})\geq 0,\forall x\in \mathbb{R}^{n}.
If \lambda\geq 0 , by (classical)

S ‐lemma,(\exists\alpha_{0})f(x)-\alpha_{0}\leq 0\Rightarrow g(x)\geq 0 . Therefore,

L_{\alpha 0}^{\leq}(f)\cap L_{0}^{<}(g)=\emptyset and f(x) is bounded from below by \alpha_{0} on L_{0}^{<}(g) . If \lambda<0, (\exists-\lambda\geq0)g(x)+(-\lambda)(-f(x)+\alpha_{0})\geq 0, \forall x\in \mathbb{R}^{n} . Again by S ‐lemma, -f(x)+\alpha_{0}\leq 0\Rightarrow g(x)\geq 0.
Then, g(x)<0\Rightarrow f(x)<\alpha_{0} and f(x) is bounded from above by \alpha_{0} on L_{0}^{<}(g) . It meansthat

f is semi‐bounded on L_{0}^{<}(g) .

\squareExample: Let g(x, y)=x^{2}+y^{2}+3xy+x+y+1, f(x, y)=x^{2}+y^{2}-2xy-x-y . Bya change of variables x=u+v, y=-u+v , we find that g(u, v)=-u^{2}+5v^{2}+2v+1 isof form (1) so L_{0}^{<}(g) consists of two connected components. Under thesystem,u-v

coordinate

f(u, v)=4u^{2}-2v . It is easy to see that f(u, v)=0\Leftrightarrow v=2u^{2} on whichg(u, 2u^{2})>0 so that f=0\Rightarrow g\geq 0 . On the other hand, since f(u, v)=4u^{2}-2v has
a non‐vanishing quadratic term 4u^{2} , by Theorem 2, L_{0}^{=}(f) cannot separate L_{0}^{<}(g) . By
Theorem 4, f is semi‐bounded on L_{0}^{<}(g) .

Now let g(x, y)=x^{2}+y^{2} + 3xy + x + y + ı, f(x, y)=x-y . With x=u+v, y=-u+v,

g(u, v)=-u^{2}+5v^{2}+2v+1, f(u, v)=2u . Then, f(u, v)=0\Leftrightarrow u=0 on whichg(0, v)=5v^{2}+2v+1>0 . By Theorem 2, L_{0}^{=}(f) separates L_{0}^{<}(g) , so f is not semi‐bounded on L_{0}^{<}(g) . We can check that, with X_{n}(-2n, n), Y_{n}(n, -2n) , we have g(X_{n})=g(Y_{n})=-n^{2}-n+1 , which indicates that X_{n}, Y_{n}\in L_{0}^{<}(g) for n large enough. However,f(X_{n})=-3n^{n}\vec{arrow}^{\infty}-\infty, f(Y_{n})=3n^{n}\vec{arrow}^{\infty}+\infty . The function f is not semi‐bounded on
L_{0}^{<}(g) . The example justifies Theorem 4.
Theorem 5 Assume degree of g is two then function f is semi‐boud’ed on L_{0}^{=}(g) if andonly if\{ begin{ar ay}{l (\exist \alpha_{0})g(x)=0\Rightar owf(x)\neq\alpha_{0}; Whenrank(B)\geq2thenL_{\alpha_{0}^{=}(f)doesnotsepar teL_{0}^{*}(g) wher *\in\{<,>\}. \end{ar ay}
Proof

Proof for necessity: If, on the contrary, \forall\alpha\in \mathbb{R} , there exists some \overline{x}\in \mathbb{R}^{n} such8
that g(\overline{x})=0 and f(\overline{x})=\alpha . Then, f(x) is unbounded from below and also from aboveon L_{0}^{=}(g) . It cannot be not semi‐bounded.

Secondly, with rank (B)\geq 2 , for a_{0} such that g(x)=0\Rightarrow f(x)\neq\alpha_{0} , and consider two
following cases:

Case 1:

L_{\alpha_{0} ^{=}(f) separates L_{0}^{<}(g) . Then by Theorem 2, there exists a basis of \mathbb{R}^{n} suchthat g(x) is of form (1) that g(x)=-x_{1}^{2}+\delta(x_{2}^{2}+ +x_{m}^{2})+\theta where \delta\in\{0,1\} ;and f-\alpha_{0} is affine function with form b_{1}x_{1}+\delta (b_{2}x_{2}+ +b_{m}x_{m})+b_{0}, b_{1}\neq 0 . By

rank (B)\geq 2, \delta must be 1 (since if \delta=0 then g(x)=-x_{1}^{2}+\theta, rank(B)=1 ). Ifb_{2}==b_{m}=0 then, it is easy to see that points(\pm\sqrt{x_{2}^{2}+\theta}, x_{2}, \cdots , 0)\in L_{0}^{=}(g) for|x_{2}| sufficient large, on which f(x) is unbound from below and also form above. If b_{i}\neq 0
for some i\in\{2, \cdot\cdot\cdot , m\} , without loss of generality, we assumei=2 .

We consider asubset of L_{0}^{=}(g) form \{(x_{1}, x_{2},0, \cdots , 0)|g(x_{1}, x_{2},0, \cdots , 0)=-x_{1}^{2}+x_{2} ^{2}+\theta=0\} , on whichf(x_{1}, x_{2},0, \cdots, 0)=b_{1}x_{1}+b_{2}x_{2}+b_{0}\{ begin{ar y}{l b_{1}x_{1}+b_{2}x_{2}+b_{0}=\alpha -(\frac{\alpha-b_{0}-b_{2}x_{2} b_{1})^{2}+x_{2}^ +\thea=0 \end{ar y}.

It is easy to see that\{begin{ar y}{l b{\imath}x_{1}+b_{2}x +b_{0}=\alph -x_{\imath}^{2+x_{2}^ +\thea=0 \end{ar y}has solution for all large enough |\alpha| . (Indeed, this system equations is equivalence to, and it has solution if and only if \triangle=b_{1}^{2}(\alpha-b_{1})^{2}+\theta b_{1}^{2}-\theta b_{2}^{2}\geq 0 , obviously, with |\alpha| is large enough, \triangle>0 ). It means f is unbounded frombelow and also from above on L_{0}^{=}(g) .

Case 2:

L_{\alpha_{0} ^{=}(f) separates L_{0}^{>}(g) . Note that L_{0}^{>}(g)=L_{0}^{<}(-g) which implies that L_{\alpha_{0} ^{=}(f)separates L_{0}^{<}(\overline{g}) where \overline{g}=-g . Apply the result in Case 1 for pair (\overline{g}, f) , we therefore
conclude that f is unbounded from below and also from above on

L_{0}^{=}(\overline{g})L_{0}^{=}(\overline{g})=L_{0}^{=}(-g)=.

Proof for the sufficiency: If there exists \alpha_{0} such that g(x)=0\Rightarrow f(x)\neq\alpha_{0} and L_{\alpha_{0} ^{=}(f)does not separate L_{0}^{<}(g) and L_{0}^{>}(g) when rank (B)\geq 2 , we need to prove that f is semi‐bounded on L_{0}^{=}(g) . Indeed, if f(x) is semi‐bounded on \mathbb{R}^{n} , it is also semi‐bounded on

9L_{0}^{<}(g) . If f(x) is not semi‐bounded on \mathbb{R}^{n} , it implies that\overline{f}=f(x)-a_{0} takes both positive and negative values on \mathbb{R}^{n}(7)

By L_{0}^{=}(\overline{f})=L_{\alpha 0}^{=}(f) has no any common point with L_{0}^{=}(g) (due to assumption g(x)=0\Rightarrow f(x)\neq\alpha_{0}) , which implies that \overline{f}(x)=f(x)-\alpha_{0}=0 then g(x)>0 or -g(x)>0.
Therefore

The condition (E_{1}) holds for pair (g, \overline{f}) or pair (-g, \overline{f})(8)When rank (B)\geq 2 . Combine assumption that L_{\alpha_{0} ^{=}(f) does not separate L_{0}^{*}(g) where *\in\{<, >\} with (7), (8) and by theorem 3, S ‐lemma with equality holds for pair (g,\overline{f}) or
pair(-g,\overline{f}) .

It implies that \exists\lambda\in \mathbb{R} such that g(x)+\lambda(f(x)-\alpha_{0})\geq 0\forall x\in \mathbb{R}^{n} or-g(x)+\lambda(f(x)-\alpha_{0})\geq 0\forall x\in \mathbb{R}^{n}.
If \lambda\geq 0 , By (classical) S ‐lemma, (\exists\alpha_{0})f(x)-\alpha_{0}\leq 0\Rightarrow g(x)\geq 0 or -g(x)\geq 0.
Therefore,

L_{\alpha_{0}-1}^{\leq}(f)\cap L_{0}^{\leq}(g)=\emptyset or L_{\alpha_{0}-1}^{\leq}(f)\cap L_{0}^{\leq}(-g)=\emptyset . It means f(x) is boundedfrom below on L_{0}^{\leq}(g) or L_{0}^{\leq}(-g) , it implies f is semi‐bounded on L_{0}^{=}(g) .
If \lambda<0 , it means that \exists-\lambda\geq 0 such that g(x)+(-\lambda)(-f(x)+\alpha_{0})\geq 0\forall x\in \mathbb{R}^{n} or-g(x)+(-\lambda)(-f(x)+\alpha_{0})\geq 0\forall x\in \mathbb{R}^{n} . Again by S ‐lemma, -f(x)+\alpha_{0}\leq 0\Rightarrow g(x)\geq 0
or -g(x)\geq 0 . Therefore, L_{\alpha 0+1}^{\geq}(f)\cap L_{0}^{\leq}(g)=\emptyset or L_{\alpha_{0}+1}^{\geq}(f)\cap L_{0}^{\leq}(-g)=\emptyset . It means f(x)is bounded from above on L_{0}^{\leq}(g) or L_{0}^{\leq}(-g) , it implies that f is semi‐bounded on L_{0}^{=}(g) .
\squareNote that with rank (B)=1 , if L_{\alpha 0}^{=}(f) does not separate L_{0}^{<}(g) and L_{0}^{>}(g) , by the same
argument above, f is semi‐bounded on L_{0}^{=}(g) . If L_{\alpha_{0} ^{=}(f) separates L_{0}^{<}(g) or L_{0}^{>}(g) , bytheorem 2, there exists a basis of \mathbb{R}^{n} such that g(x) is of form (1) that g(x)=-x_{1}^{2}+\theta or-g(x)=-x_{1}^{2}+\theta (since rank (B)=1 ); and f-\alpha_{0} is affine function with form b_{1}x_{1}+b_{0}, b_{1}\neq
0. It is easy to see that f(x) is bound on L_{0}^{=}(g) since on L_{0}^{=}(g) , where x_{1} is constant.

103Conclusions
This is a short article which introduces the concept of semi‐boundedness of a quadratic

function f over the sublevel set of another quadratic function g .
property and the

S ‐lemma with equality,

By the separation

we have, in Theorem 4, given necessary andsufficient condition for the semi‐boundedness property. The image set of f over \{g\leq 0\}
is fundamentally important in mathematics. Being able to characterize the boundednessof the image set of f over \{g\leq 0\} is the first step to study the problem. We hopethat more results, such as the connectedness of the image set of f over \{g\leq 0\} , can beobtained following the separation property in the future.

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