Is the Function \( f(x,y) \) Continuous on \( \mathbb{R}^2 \)?

[i_a_n]

Prove that
$f(x,y)=\left\{\begin{matrix}
e^{-1/|x-y|},x\neq y\\
0,x=y
\end{matrix}\right.$
is continuous on $\mathbb{R}^2$.

Can I conclude since $e^{-1/t}$ is continuous then for $x$,$y\in \mathbb{R}^2$, $x\neq y$,$e^{-1/|x-y|}$ is continuous on $\mathbb{R}^2$ since here $t=|x-y|$? And how to prove it when $x=y$, using the $\delta- \varepsilon$ way? Thank you a lot!

 

[tarantella]

Re: Prove a function is continuous

We have to show that the function is continuous at each point of $\Bbb R^2$. Your argument threats the case of points of the form $(x,y),x\neq y$. So now, fix $x_0$. We have to show that $\lim_{(x,y)\to (x_0,x_0)}=0$.

Hint: use $e^{-u}\leqslant \frac 1{1+u}$ for each non-negative real number $u$.

 

[Fernando Revilla]

Re: Prove a function is continuous

Hint: use $e^{-u}\leqslant \frac 1{1+u}$ for each real number $u$.

Surely is a typo, I suppose you meant for each real number $u$ close to $0$.

Another way: denoting $\Delta=\{(t,t):t\in\mathbb{R}\}$ (diagonal of $\mathbb{R}^2$), in a neighborhood $V$ of $(0,0)$ we have:

$\displaystyle\lim_{(x,y)\to (x_0,x_0)\;,\;(x,y)\in V-\Delta}f(x,y)=\displaystyle\lim_{(x,y)\to (x_0,x_0)\;,\;(x,y)\in V-\Delta}e^{-\dfrac{1}{|x-y|}}=e^{-\infty}=0$

$\displaystyle\lim_{(x,y)\to (x_0,x_0)\;,\;(x,y)\in V\cap\Delta}f(x,y)=\displaystyle\lim_{(x,y)\to (x_0,x_0)\;,\;(x,y)\in V\cap\Delta}0=0$

In a finite partition of $V$ all the limits coincide, so $\displaystyle\lim_{(x,y)\to (x_0,x_0)}f(x,y)=0$ .

 

[i_a_n]

The definition of continuity for a function is {lim x-> a} f(x) = f(a). For this function we only need to worry about when |x - y| -> 0, since the exponential function is continuous everywhere. So in other words we need to show that
{lim |x - y|-> 0} f(x,y) = f(x,x) = f(y,y)
Now f(x,x) = f(y,y) = 0 so we just need to show that
{lim |x - y| -> 0} f(x,y) = 0.
Since |x - y| > 0, -1/|x - y| -> infinity so f(x,y) -> 0 and we're done.Is a proof like this correct?

 

[blue_raver22]

Yes, you can conclude that $e^{-1/t}$ being continuous implies that $e^{-1/|x-y|}$ is continuous for $x,y \in \mathbb{R}^2$, $x \neq y$. This is because the function $e^{-1/|x-y|}$ is a composition of two continuous functions: the absolute value function $|x-y|$ and the exponential function $e^{-t}$.

To prove that $f(x,y)$ is continuous at points where $x=y$, we can use the $\delta-\varepsilon$ definition of continuity. Let $(x_0,y_0)$ be a point in $\mathbb{R}^2$ where $x_0=y_0$. We want to show that for any $\varepsilon > 0$, there exists a $\delta > 0$ such that for all points $(x,y)$ in the domain of $f$ satisfying $|x-x_0|<\delta$ and $|y-y_0|<\delta$, we have $|f(x,y)-f(x_0,y_0)|<\varepsilon$.

Since $x=y$ at $(x_0,y_0)$, we have $|x-y| = |x_0-y_0| = 0$. This means that $e^{-1/|x-y|} = e^{-1/0}$, which is undefined. However, we can see that as $|x-y|$ approaches 0, the function $e^{-1/|x-y|}$ approaches the limit $e^{-1/0} = e^{-\infty} = 0$. This can be verified by taking the limit as $|x-y|$ approaches 0:

$$\lim_{|x-y|\to 0} e^{-1/|x-y|} = e^{-1/\lim_{|x-y|\to 0}|x-y|} = e^{-1/0} = e^{-\infty} = 0$$

Therefore, for any $\varepsilon > 0$, we can choose $\delta = \frac{1}{-\ln(\varepsilon)}$ (note that this is a positive number since $\varepsilon < 1$) and we have:

$$|f(x,y)-f(x_0,y_0)| = |