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 $X$.


1 Answer 1


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).

  • 1
    $\begingroup$ 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. $\endgroup$ Sep 16, 2022 at 13:18

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