1
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ What you done so far? Did you try to write down the $\epsilon - \delta$ definiton of continuity and take appropriate suprema etc? $\endgroup$ – Kavi Rama Murthy Aug 20 '18 at 8:29
  • $\begingroup$ 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 $\endgroup$ – Cebiş Mellim Aug 20 '18 at 8:49
  • 1
    $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy Aug 20 '18 at 8:56
  • $\begingroup$ Dear @KaviRamaMurthy, could you please review my proof below? I have rewritten it now. $\endgroup$ – Cebiş Mellim Aug 21 '18 at 8:42
0
$\begingroup$

We only prove the first claim, others follow in a similar manner.

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.

$\endgroup$
  • 1
    $\begingroup$ This is fine, but you should take $\delta_{\epsilon}=\frac {\gamma_{\epsilon}} {\sqrt 2}$ $\endgroup$ – Kavi Rama Murthy Aug 21 '18 at 8:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.