Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ and $Y$ be two nonempty sets and let $h:X\times Y\rightarrow \mathbb{R}$ have a bounded range in $\mathbb{R}$.Let $f:X\rightarrow \mathbb{R}$ and $g:Y\rightarrow \mathbb{R}$ defined by $$f(x)=\sup\{h(x,y):y\in Y\}$$ and $$g(y)=\inf\{h(x,y):x\in X\}$$Then can we prove that $$\sup\{g(y):y\in Y\} \leq \inf\{f(x):x\in X\}?$$

share|cite|improve this question
This is related to the weak duality inequality in convex optimization, with $ h $ corresponding to the Lagrangian and $ f $ and $ g $ corresponding to the primal and dual objective functions. – littleO Nov 7 '15 at 22:07
up vote 4 down vote accepted

If $A$ and $B$ are bounded subsets of $\mathbb R$, then $\sup A\leq \inf B$ is equivalent to the statement that for all $a\in A$ and $b\in B$, $a\leq b$. Thus, it suffices to show that for each $x\in X$ and $y\in Y$, $g(y)\leq f(x)$.

Let $x_0\in X$ and $y_0\in Y$ be fixed but arbitrary. Then $g(y_0)\leq h(x_0,y_0) \leq f(x_0)$.

share|cite|improve this answer

Clearly $$g(y)=\inf_{x\in X} h(x,y)\le h(x,y)\le \sup_{y\in Y}h(x,y)\le f(x)\ \forall x\in X,y\in Y$$ Then $$g(y)\le \inf_{x\in X} f(x)\le f(x)\ \forall y\in Y$$ and so $$g(y)\le \sup_{y\in Y}g(y)\le \inf f(x)$$

share|cite|improve this answer
@Marvis How about now? – Nameless Dec 27 '12 at 9:24
Yes. +1 now. ${}$ – user17762 Dec 27 '12 at 9:27

$g(y)\leq f(x)$ $\forall$ $x\in X,y\in Y\implies g(y)\leq \inf\{f(x):x\in X\}$ $\forall$ $y\in Y\implies$$\sup\{g(y):y\in Y\} \leq \inf\{f(x):x\in X\}$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.