Let $f:\mathbb{R}^2\to \mathbb{R}$ be a function. Let $(x_0,y_0)\in \mathbb{R}^2$. If $f$ is continuous at $(x_0,y_0)$ show that
$\lim_{x\to x_0}\limsup_{y\to y_0}f(x,y)=\lim_{y\to y_0}\limsup_{x\to x_0}f(x,y)=f(x_0,y_0)$
and $\lim_{x\to x_0}\liminf_{y\to y_0}f(x,y)=\lim_{y\to y_0}\liminf_{x\to x_0}f(x,y)=f(x_0,y_0)$
I want to just focus on this first part $\lim_{x\to x_0}\limsup_{y\to y_0}f(x,y)=\lim_{y\to y_0}\limsup_{x\to x_0}f(x,y)=f(x_0,y_0)$
My main issue is that I don't know what $\limsup$ means in the context of a multivariable function.
My assumption is that $\limsup_{y\to y_0}f(x',y)=\inf_{r>0}(\sup\{f(x',y):y\in B_r(y_0)\}$ where $x'\in X$ is just held constant?
Since for any sequence $(x',y_n)->(x',y_0)$ then $f(x',y_n)\to f(x',y_0)$ by sequential continuity of $f$. So then $\limsup_{y\to y_0} f(x',y)=\limsup_{n\to \infty} (f(x',y_n))=\liminf_{n\to \infty} (f(x',y_n))=f(x',y_0)$.
I believe this makes sense because in single variable I have shown such things as if $a_n\to a_0$ as $n\to \infty$ then $\lim_{a\to a_0} f(a)=\lim_{n\to\infty} f(a_n)$ and also that a sequence $(a_n)$ converges to $a$ iff $\limsup a_n=\liminf a_n=a$ And since $f$ is real valued I think I can justify that these hold.
From here I believe I can say $\limsup_{y\to y_0} f(x,y)=f(x,y_0$ since $x'\in X$ is chosen arbitrarily?
If so then $\lim_{x\to x_0}f(x,y_0)=f(x_0,y_0)$ by continuity again.
My process would then be apply the same logic to show $\lim_{y\to y_0}\limsup_{x\to x_0}f(x,y)=f(x_0,y_0)$. The cases of $\liminf$ would be similar.
Does this work? The major potential problem I see is this part:
From here I believe I can say $\limsup_{y\to y_0} f(x,y)=f(x,y_0$ since $x'\in X$ is chosen arbitrarily?
Which I'm not sure actually makes sense.
Best Answer
$f$ is assumed to be continuous only at $(x_{0}, y_{0})$, not at $(x, y_0)$ for an arbitrarily chosen $x$.
Recall that $\limsup_{x \to x_{0}}f(x) := \inf_{r > 0}\sup_{|x - x_{0}| < r}f(x)$ and $\liminf_{x \to x_{0}}f(x) := \sup_{r > 0}\inf_{|x - x_{0}| < r}f(x)$.
Note that the exact choice of metric is irrelevant here since talking about continuity in $\mathbb{R}^{n}$, it suffices to know that the distance is determined by the quantity of the form $|x_{i} - y_{i}|$. We give $\mathbb{R}^{2}$ the Euclidean metric, say.
By condition, $\forall \varepsilon > 0$, $\exists \delta > 0$, such that when $\sqrt{(x - x_{0})^{2} - (y - y_{0})^{2}} < \delta$, $|f(x, y) - f(x_{0}, y_{0})| < \varepsilon$.
Fix some $x$ with $|x - x_{0}| < \frac{\delta}{2}$. $\forall |y - y_{0}| < \frac{\delta}{2}$, $\sqrt{(x - x_{0})^{2} - (y - y_{0})^{2}} < \delta$. Hence by definition, $f(x_{0}, y_{0}) - \varepsilon \leq \sup_{|y - y_{0}|< r}f(x, y) \leq f(x_{0}, y_{0}) + \varepsilon$, when $r < \frac{\delta}{2}$. Taking the infimum over $r > 0$, we then get: $f(x_{0}, y_{0}) - \varepsilon \leq \inf_{r > 0}\sup_{|y - y_{0}| < r}f(x, y) \leq f(x_{0}, y_{0}) + \varepsilon$. In particular, we have:
$\forall |x - x_{0}| < \frac{\delta}{2}$, $|\limsup_{y \to y_{0}}f(x, y) - f(x_{0}, y_{0})| < \varepsilon$. ie, $\lim_{x\to x_0}\limsup_{y\to y_0}f(x,y) = f(x_{0}, y_{0})$.
A similar argument shows that $\lim_{y\to y_0}\limsup_{x\to x_0}f(x,y)=f(x_0,y_0)$.