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Here's the framework that I am situated in and I'm having some concerns. Let $X$ be a compact metric space and $f:X\times X\times X\to \mathbb{R}$ a continuous function that can be composed as a sum of continuous functions $f_{1},f_{2}:X\times X\to \mathbb{R}$ so that $f(x,y,z)=f_{1}(x,y)+f_{2}(y,z)$.

Is it then true that: \begin{align*} \inf_{x\in X}f_{1}(x,y)+\inf_{z\in X}f_{2}(y,z)=\inf_{x,z\in X}(f_{1}(x,y)+f_{2}(y,z))=\inf_{x,z\in X}f(x,y,z)\,\, ? \end{align*} And are the functions $\inf_{x\in X}f_{1}(x,\cdot):X\to \mathbb{R}$ continuous?

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It seems to me that $y$ plays no role at all, except for the last sentence (where $f$ plays no role at all). –  André Caldas Apr 18 '12 at 11:47
    
You're right that $y$ does not play any role in the first question: my question is exactly whether the equality is true for all $y\in X$. And the second question concerns only $f_{1}$ and similarly $f_{2}$, not the function $f$. –  Thomas E. Apr 18 '12 at 11:53
    
Fix $y$. Call $g_1(x) = f_1(x,y)$ and do the same for $g_2$. Then, unless I am missing some thing, you are simply asking if $\inf (g_1(x) + g_2(z)) = \inf g_1(x) + \inf g_2(z)$. You can probably easily find a counter-example for this equality. I guess the $y$ was getting in your way... :-) –  André Caldas Apr 18 '12 at 13:56
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for each $x,y,z \in X$ we have \[ \inf_x f_1(x,y) + \inf_z f_2(y,z) \le f_1(x,y) + f_2(y,z) \] so the left hand side is a lower bound for $\{f(x,y,z)\mid x,z \in X\}$. As the infimum is the greatest lower bound, we obtain \[ \inf_x f_1(x,y) + \inf_z f_2(y,z) \le \inf_{x,z} \bigl(f_1(x,y) + f_2(y,z) \bigr). \] On the other hand, we have for each $x,y,z \in X$ \[ \inf_{x,z} \bigl(f_1(x,y) + f_2(y,z) \bigr) \le f_1(x,y) + f_2(y,z) \] As this holds (fixing $z$ for the moment) for each $x$ we get \[ \inf_{x,z} \bigl(f_1(x,y) + f_2(y,z) \bigr) \le\inf_x\bigl( f_1(x,y) + f_2(y,z) \bigr) = \inf_x f_1(x,y) + f_2(y,z) \] This holds for each $z$ and therefore \[ \inf_{x,z} \bigl(f_1(x,y) + f_2(y,z) \bigr) \le\inf_z\bigl( \inf_x f_1(x,y) + f_2(y,z) \bigr) = \inf_x f_1(x,y) + \inf_z f_2(y,z). \] So the inequality in question holds always, we don't need coninuity or compactness.

Regarding your second question: Since $X^2$ is compact, $f_1$ is uniformly continuous. Now let $\epsilon > 0$. By uniform continuity there is $\delta > 0$ s. t. $|f_1(x',y') - f_1(x,y)| < \epsilon$ for $d(x',x), d(y',y) < \delta$. Now let $y,y' \in X$ with $d(y',y) < \delta$. Since $X$ is compact, infima of continuous functions are attained and therefore there are $x, x' \in X$ with $g(y) = f_1(x,y)$ and $g(y') = f_1(x',y')$. Assume wlog $g(y) < g(y')$, we have \begin{align*} g(y') &= f_1(x',y')\\\ &= \inf_a f_1(a,y')\\\ &\le f_1(x,y')\\\ &\le f_1(x,y) + \epsilon \quad\text{as $d(y,y') < \delta$}\\\ &= g(y) + \epsilon\\\ \iff g(y') - g(y) &\le \epsilon \end{align*} So $g$ is (uniformly) continuous.

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thanks a lot martini. –  Thomas E. Apr 18 '12 at 13:06
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