# Takahashi minimization theorem for lower pseudo-continuous functions on complete metric spaces

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 $$X$$.

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).

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).

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 $$f(x_{\gamma})\le f(x_\beta)-d(x_{\gamma},x_\beta). Then for any $$\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)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 $$n, the sequence $$(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 $$f(x_\gamma)\le f(x_{\alpha_n}) by the inductive condition $$(*_{\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).

• That's awesome. Nevertheless, I think that the standard proof (without transfinite induction goes through): One assumes that there is no minimizer and constructs recursively a sequence $x_n$ with $f(x_{n+1})\le (f(x_n)+c_n)/2$ where $c_n=\inf f(B_n)$ and $B_n=\{y\in X: f(y)\le f(x_n)-d(x_n,y)\}$. This sequence is Cauchy and lower semicontinuity is only used to show that the limit $x_*$ satisfies $f(x_*)\le \liminf\limits_{n\to\infty} f(x_n)$. But the sequence $f(x_n)$ is strictly decreasing so that the limit exists and the inequality follows from lower pseudocontinuity. Commented Sep 16, 2022 at 13:18