A semicontinuous diagonal function?

Question

Let $X$ be a compact Hausdorff space and let $(x_\alpha)_\alpha$ be a net in $X$ on the directed set $(A, >)$ that converges to $x$.

Let $f: X \times X \to [0, \infty]$. Suppose $f$ has the following properties:

(1) $f(x,x) \in \mathbb R$ for all $x \in X$.

(2) $f(y,y) \leq f(y',y)$ for all $y,y' \in X$.

(3) For all $\beta \in A$, the net $(f(x_\alpha,x_\beta))_\alpha$ converges to $f(x, x_\beta)$.

Does it follow that $\limsup_\alpha f(x_\alpha, x_\alpha) \leq f(x,x)$?

I've tried to start by using (2) to extract a subnet from $(f(x_\alpha, x_\alpha))_\alpha$ that converges to $f(x,x)$, but I've been unable to see that this can be done, and even if it can I'm not sure whether it will imply the result.

Answer 1

Here is a counterexample.

Let $X=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$ with the usual topology; $\left\langle\frac1n:n\in\Bbb Z^+\right\rangle$ is a net converging to $0$. For $n,m\in\Bbb Z^+$ let

$$f\left(\frac1n,\frac1m\right)=\begin{cases} 1,&\text{if }n=m\\ 2,&\text{otherwise.} \end{cases}$$

For $n\in\Bbb Z^+$ let $f\left(\frac1n,0\right)=f\left(0,\frac1n\right)=2$, and let $f(0,0)=0$. Then

1. $f$ takes non-negative real values on $X\times X$;
2. $f(x,x)\le f(y,x)$ for all $x,y\in X$; and
3. for each $m\in\Bbb Z^+$ the sequence $\left\langle f\left(\frac1n,\frac1m\right):n\in\Bbb Z^+\right\rangle$ converges to $2=f\left(0,\frac1m\right)$,

so the conditions are satisfied. But

$$\limsup\limits_{n\in\Bbb Z^+}f\left(\frac1n,\frac1n\right)=1>0=f(0,0)\,.$$