The Takahashi Theorem holds also for lower pseudo-continuous functions.

To derive a contradiction, assume that $f:X\to [0,+\infty]$ a lower pseudo-continuous function such that for any point $x\in X$ 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 $\;f(x_\beta)<f(x_\alpha)$.

Proof of Claim. We start an indutive constuction choosing any point $x_0\in X$. 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).

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).

The Takahashi Theorem holds also for lower pseudo-continuous functions.

To derive a contradiction, assume that $f:X\to [0,+\infty]$ a lower pseudo-continuous function such that for any point $x\in X$ 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 $\;f(x_\beta)<f(x_\alpha)$.

Proof of Claim. We start an indutive constuction choosing any point $x_0\in X$. 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 $\alpha\le \beta$ 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).

The Takahashi Theorem holds also for lower pseudo-continuous functions.

To derive a contradiction, assume that $f:X\to [0,+\infty]$ a lower pseudo-continuous function such that for any point $x\in X$ 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 $\;f(x_\beta)<f(x_\alpha)$.

Proof of Claim. We start an indutive constuction choosing any point $x_0\in X$. 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).