# Interchangeable limits

The following exercise 2.2.9 is borrowed from Terence Tao's Analysis II, page 33. While I understand the problem completely, I lack the technique to attack it. So I will really appreciate a small hint on how to start the proof.

Let $f : \mathbb{R}^2 \to \mathbb{R}$ be a function. Let $(x_0 , y_0 ) \in \mathbb{R}^2$ be a point. 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 $$\lim_{x\to x_0 } \liminf_{y \to y_0} f (x, y) = \lim_{y\to y_0 } \liminf_{x \to x_0} f (x, y) = f (x_0 , y_0 )$$ In particular, we have $$\lim_{x\to x_0 } \lim_{y \to y_0} f (x, y) = \lim_{y\to y_0 } \lim_{x \to x_0} f (x, y) = f (x_0 , y_0 )$$ whenever the limits on both sides exist.

We have defined $\limsup$ as $$\limsup_{x\to x_0} f(x,y) = \inf \left\{ \sup \{f(x,y): |x-x_0| <r \}: r>0\right\}$$

• What you done so far? Did you try to write down the $\epsilon - \delta$ definiton of continuity and take appropriate suprema etc? – Kavi Rama Murthy Aug 20 '18 at 8:29
• First, I wrote down $$\lim_{y \to y_0} \inf \left\{ \sup\{ f(x,y): |x-x_0|<r \} : r>0 \right\} = \inf \left\{ \sup\{ f(x,y_0): |x-x_0|<r \} : r>0 \right\}$$ Then I picked an arbitrary $\epsilon >0$. Using the $\epsilon-\delta$ assumption on continuity, I can find $\delta_\epsilon$ such that all $|f(x,y_0) - f(x_0,y_0) | < \epsilon$ whenever $|x-x_0| < \delta$ (using Euclidean metric on $\mathbb{R}^2$). But then the difference between infimum value and $f(x_0,y_0)$ is even less than the one we found above. I then recall that my choice of $\epsilon$ was arbitrary and squeeze the value of infimum – Cebiş Mellim Aug 20 '18 at 8:49
• You are fixing $y=y_0$ and using continuity in on evariable. Use continuity in two variables. Write $|f(x,y)-f(x_0,y_0)| <\epsilon$ as $f(x_0,y_0)-\epsilon< f(x,y)<f(x_0,y_0) +\epsilon$ and take supremum over $x,y$ such that $|x-y| <r$. You should now be able to complete the proof. – Kavi Rama Murthy Aug 20 '18 at 8:56
• Dear @KaviRamaMurthy, could you please review my proof below? I have rewritten it now. – Cebiş Mellim Aug 21 '18 at 8:42

Our goal is to show that $$\lim_{x\to x_0} \limsup_{y\to y_0} f(x,y) = f(x_0,y_0)$$ i.e., that $$\forall \epsilon>0 \quad \exists \delta_{\epsilon}>0 \quad \forall x \in \mathbb{R}: \quad |x-x_0|<\delta_{\epsilon} \implies |\limsup_{y\to y_0} f(x,y) - f(x_0,y_0)|<\epsilon$$
Pick an arbitrary $\epsilon>0$. Since $f(x,y)$ is continuous at $(x_0,y_0)$, we have $$\exists \gamma_{\epsilon}>0 \quad \forall (x,y) \in \mathbb{R}^2: \quad d((x,y),(x_0,y_0))<\gamma_{\epsilon} \implies d(f(x,y),f(x_0,y_0))<\epsilon$$ or, written alternatively, $$\exists \gamma_{\epsilon}>0 \quad \forall (x,y) \in \mathbb{R}^2: \quad \sqrt{(x-x_0)^2 + (y-y_0)^2}<\gamma_{\epsilon} \implies |f(x,y) - f(x_0,y_0)|<\epsilon$$ We set $\delta_{\epsilon} := \frac{\gamma_{\epsilon}}{2}$ and consider only $x,y \in \mathbb{R}$ such that $|x-x_0|<\delta_{\epsilon}$ and $|y-y_0|<\delta_{\epsilon}$. Due to the continuiuty assumption $$\forall (x,y) \in \mathbb{R}^2: \quad |x-x_0|<\delta_{\epsilon} \text{ and } |y-y_0|<\delta_{\epsilon}\implies f(x_0,y_0) - \epsilon < f(x,y) < f(x_0,y_0)+\epsilon$$ for all such $x,y$. Then $f(x_0,y_0)+\epsilon$ is an upper bound for $f(x,y)$ on this interval; hence, $$\forall (x,y) \in \mathbb{R}^2: \quad |x-x_0|<\delta_{\epsilon} \text{ and } |y-y_0|<\delta_{\epsilon}\implies f(x_0,y_0) - \epsilon < \sup\{f(x,y)\} \leq f(x_0,y_0)+\epsilon$$ or, written alternatively, $$\forall x \in \mathbb{R}: \quad |x-x_0|<\delta_{\epsilon}\implies f(x_0,y_0) - \epsilon < \sup\{f(x,y):|y-y_0|<\delta_{\epsilon}\} \leq f(x_0,y_0)+\epsilon$$ Taking the infimum with respect to $r>0$ yields $$\forall x \in \mathbb{R}: \quad |x-x_0|<\delta_{\epsilon}\implies f(x_0,y_0) - \epsilon \leq \inf\left\{ \sup\{f(x,y):|y-y_0|<\delta_{\epsilon}\}:r>0 \right\} \leq f(x_0,y_0)+\epsilon$$ Therefore, for our arbitrarily chosen $\epsilon>0$ we have found a $\delta_{\epsilon}>0$ such that whenever $|x-x_0|<\delta_{\epsilon}$ we have $f(x_0,y_0) - \epsilon \leq \limsup_{y\to y_0} f(x,y) \leq f(x_0,y_0)+\epsilon$, as desired.
• This is fine, but you should take $\delta_{\epsilon}=\frac {\gamma_{\epsilon}} {\sqrt 2}$ – Kavi Rama Murthy Aug 21 '18 at 8:48