**Takahashi minimization theorem** says : Let $(X,d)$ is a complete metric space, $f:X\to \mathbb{R}\cup\{+\infty\}$ is a proper(not constantly +$\infty$) lower semi continuous function, which is bounded from below, $Z=\{x \in X: f(x)=\inf f\}$. Let for all $x \in X \setminus Z$, there exists $y\in X\setminus \{x\}$ such that $f(y)+d(x,y)\leq f(x).$ Then $Z\not= \emptyset.$


My question is that if we replace "lower pseudo-continuous" instead of "lower semi continuous" in the above theorem, whether or not the result hold?


$f$ is said to be lower pseudo-continuous on $X$, if for all $y \in X$, the set $\{x \in X : f(x) \leq f(y)\}$ would be a closed subset of $X$. 



Takahashi minimization theorem for lower pseudo-continuous functions on complete metric spaces
The Takahashi Theorem holds also for lower pseudo-continuous functions.


To derive a contradiction, assume that $f:X\to [0,+\infty]$ a proper lower pseudo-continuous function such that for any point $x\in X$ with $f(x)<+\infty$ there exists a point $y\in X\setminus\{x\}$ such that $f(y)\le f(x)-d(x,y)<f(x)$.


**Claim.** There exists a transfinite sequence of points $(x\_\alpha)\_{\alpha\in\omega\_1}$ of the complete metric space $(X,d)$ such that for any countable ordinals $\alpha<\beta$ the following condition holds:


$(\*\_{\alpha,\beta})$ $\;\;d(x\_\beta,x\_\alpha)\le f(x\_{\alpha})-f(x\_\beta)\;$ and $\;f(x\_\beta)<f(x\_\alpha)<+\infty$.


*Proof of Claim.* We start an inductive constuction choosing any point $x\_0\in X$ with $f(x\_0)<+\infty$. Assume that for some countable ordinal $\gamma$ we have constructed points $x\_\alpha$, $\alpha<\gamma$, satisfying the conditions $(\*\_{\alpha,\beta})$ for all $\alpha<\beta<\gamma$.


If $\gamma=\beta+1$ for some ordinal $\beta$, then by the property of $f$, there exists a point $x\_\gamma\in X\setminus\{x\_\beta\}$ such that $f(x\_{\gamma})\le f(x\_\beta)-d(x\_{\gamma},x\_\beta)<f(x\_\beta)$.
Then for any $\alpha<\gamma$ we get 
$$d(x\_\alpha,x\_\gamma)\le d(x\_\alpha,x\_\beta)+d(x\_\beta,x\_\gamma)\le f(x\_\alpha)-f(x\_\beta)+f(x\_\beta)-f(x\_\gamma)=f(x\_\alpha)-f(x\_\gamma)$$  

and $$f(x\_\gamma)<f(x\_\beta)\le f(x\_\alpha),$$which means that the condition $(\*\_{\alpha,\gamma})$ holds for any $\alpha<\gamma$.


Next, assume that the ordinal $\gamma$ is limit. Choose any strictly increasing sequence of ordinals $(\alpha\_n)\_{n\in\omega}$ with $\sup\_{n\in\omega}\alpha\_n=\gamma$. By the inductive conditions $(\*\_{\alpha\_n,\alpha\_m})$ for $n<m$, the sequence $(f(x\_{\alpha\_n}))\_{n\in\omega}$ is decreasing and being lower bounded, is Cauchy. Then the sequnece $(x\_{\alpha\_n})\_{n\in\omega}$ also is Cauchy (by the properties $(\*\_{\alpha\_n,\alpha\_m})$). Since the metric space $(X,d)$ is complete, the sequence $(x\_{\alpha\_n})\_{n\in\omega}$ has a limit point $x\_\gamma\in X$. 


The lower pseudo-continuity of $f$ guarantees that for every $n\in\mathbb N$ the set $F\_n=\{x\in X:f(x)\le f(x\_{\alpha\_n})\}$ is closed in $X$. The conditions $(\*\_{\alpha\_n,\alpha\_m})$ for $m\ge n$ ensure that $\{x\_m\}\_{m\ge n}\subset F\_n$ and hence $x\_\gamma\in\bigcap\_{n\in\mathbb N}F\_n$, which means that $f(x\_\gamma)\le f(x\_{\alpha\_n})$ for all $n\in\mathbb N$. 


It remains to check that the point $x\_\gamma$ satisfies the condition $(\*\_{\alpha,\gamma})$ for every $\alpha<\gamma$. Since $\alpha<\gamma=\sup\_{n\in\mathbb N}\alpha\_n$, we can choose $n\in\mathbb N$ such that $\alpha\_n>\alpha$. Then for any $m\ge n$ the condition $(\*\_{\alpha,\alpha\_m})$ yields
$$d(x\_{\alpha\_m},x\_{\alpha})\le f(x\_{\alpha})-f(x\_{\alpha\_m})\le f(x\_{\alpha})-f(x\_\gamma).$$
Passing to the limit at $m\to\infty$ we get the inequality $$d(x\_\gamma,x\_{\alpha})\le f(x\_\alpha)-f(x\_\gamma)$$which coincides with the first part of $(\*\_{\alpha,\gamma})$.


To see the second part, observe that $f(x\_\gamma)\le f(x\_{\alpha\_n})<f(x\_\alpha)$ by the inductive condition $(\*\_{\alpha,\alpha\_n})$.
This completes the proof of Claim.


Now the contradiction follows from the fact that $(f(x\_\alpha))\_{\alpha\in\omega\_1}$ is a strictly decreasing transfinite sequence of real numbers. But the real line does not contain so long strictly decreasing sequences (by the first countability).


