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

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

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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

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 $g(y_0)\leq h(x_0,y_0) \leq f(x_0)$.

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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 $$g(y)\le \sup_{y\in Y}g(y)\le \inf f(x)$$

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@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\}$

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$f(r) = \infty + 0; f = (r \infty)/0 = \textrm{complex} \infty$

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Please edit this answer using MathJax (see the help center). As it is, it is very low quality and not clear what you have written. – Johanna Jan 29 '15 at 3:55
had to add (complex inf) – mark mcgrath Jan 29 '15 at 3:55