7
$\begingroup$

I have some doubts concerning the continuity of a function in many variables. I will present a specific example, which I worked out (hopefully in a good way), and I will then ask you some questions because I'm really struggling with the "deep" theoretical sense of these procedures.

Say I have the function

$$f(x, y, z) = \begin{cases} (x + 2y + 3z)\ln(x^2 + 2y^2 + 3z^2) & (x, y, z) \neq (0, 0, 0) \\\\\\ C & (x, y, z) = (0, 0, 0)\end{cases}$$

It's asked if/when the function is continuous at the origin.

First way

I thought I could restrict to a particular path, like $x = y$ and $x = z$, reducing to a one dimensional problem:

$$f(x) = \begin{cases} 6x\ln(6x^2) & x \neq 0 \\\\ C & x = 0 \end{cases}$$

Here I can clearly study the limits when $x\to 0^+$ and $x\to 0^-$ and conclude that the function is continuous at the origin iff $C = 0$.

Alternatively, I can take a sequence $x_n = 1/n$ and operate that way.

*Question 1: how can I be sure about this? I mean, how can I be sure that choosing some other path I wouldn't obtain something strange? This is a more general question actually, meant in general when applying the method of restriction to some path.

Second way

Our professor taught us the "distance method", that is: a function $f(x)$ is continuous at $x_0$ iff there exist a function $h(d)$, a function of a distance $d$ (distance meant as in the Euclidean way, I guess) such that

$$|f(x) - f(x_0)| \leq h(d)$$

with $h(d) \to 0$ for $d \to 0$.

This opened the abysmal world of majorizations and estimations, and I actually really need clarification from someone who got this well, in order to not sink in this oblivion.

For example, using this, here is what I did. Please, read until the end (question included).

$$|f(x, y, z) - \underbrace{f(0, 0, 0)}_{\text{supposed zero}}| \leq |x + 2y + 3z||\ln(x^2 + 2y^2 + 3z^2)|$$

Now I did: $|x + 2y + 3z| \leq |3x + 3y + 3z|$ and the same in the logaritm

$$\leq |3x + 3y + 3z||\ln(3x^2 + 3y^2 + 3z^2)|$$

At this point I did: $|3x + 3y + 3z| = 3(|x+y+z|) \leq 3(|x| + |y| + |z|)$ and also after that: $3(|x| + |y| + |z|) \leq 3(x^2 + y^2 + z^2)$

This allowed me to call a distance function $h(d^2) = x^2 + y^2 + z^2$ where $d^2$ would be the Euclidean distance squared ($d^2 = x^2+y^2+z^2)$, so in the end:

$$\leq 3h(d)|\ln(3h(d))| \to 0 \qquad \text{as}\ d\to 0$$

So apparently I have managed to find such a function that, when shrunk, compress the function into it's value at the origin, which is zero. Hence again $C$ must be zero.

And here I am stuck:

  • How to really majorize? I mean things like $x < x^2$ are true when $|x| > 1$ where as the counterpart $x > x^2$ are true when $|x| < 1$. Since I'm analyzing limits to zero, should I beware of such conditions from the beginning, or can I just think about finding a big ball $B$, hence majorizing as if $(x, y, z) >> 0$ (large enought) and in the end shrinking letting $(x, y, z) \to (0, 0, 0)$?

  • Isn't it always possible to compute such a majorization, even if a function is not continuous? How can I be sure with this method?

Sorry for this poem, but THANK YOU SO MUCH to whoever will reply to this.

$\endgroup$

2 Answers 2

6
$\begingroup$

Really good conceptual questions, but unfortunately, there are minor mistakes with inequalities all over, which I'll point out along the way. The correct way of solving the problem is method 2; method 1 is only for getting motivation. You compute along different paths to get some inspiration, or if you're trying to prove the limit does not exist (if you can find two paths with different limits then the full limit does not exist). The second method of upper and lower bounds is the right way to go, and so much of analysis relies on it; it's a very important skill to develop. As you may have already realized, inequalities are much more common than equalities.

Your first bullet point is a very good and crucial observation, but you made the wrong conclusion. Indeed, some inequalities only hold in certain regimes. For this question specifically, you're interested in what happens as $\xi:=(x,y,z)\to 0$. So, you just fix a small punctured ball $B_{\delta}(0)\setminus \{0\}$, and show that for all $\xi$ in this punctured ball, you have an upper bound $|f(\xi)-f(\xi_0)|\leq \phi(\|\xi-\xi_0\|)$, for some function $\phi$, such that $\lim\limits_{r\to 0^+}\phi(r)=0$. By the way, you don't have to, i.e it is not a forced requirement that you must restrict yourself to a small ball; rather, it is a freedom that you have which you may or may not decide to invoke. If for example you were interested in the limit as $\xi\to\infty$, this means you focus on the complement of a large ball, $B_R(0)^c$.

For your example specifically, you say

Now I did: $|x + 2y + 3z| \leq |3x + 3y + 3z|$ and the same in the logaritm

$$|f(x,y,z)-f(0,0,0)|\leq |3x + 3y + 3z||\ln(3x^2 + 3y^2 + 3z^2)|$$

At this point I did: $|3x + 3y + 3z| = 3(|x+y+z|) \leq 3(|x| + |y| + |z|)$ and also after that: $3(|x| + |y| + |z|) \leq 3(x^2 + y^2 + z^2)$.

Unfortunately, this is full of mistakes. First of all, it is not true that $|x+2y+3z|\leq |3x+3y+3z|$ (think of $(x,y,z)=(-1,0,1)$, you'd be saying that $2\leq 0$). The correct way to argue is to directly apply the triangle inequality: \begin{align} |x+2y+3z|\leq |x|+2|y|+3|z|\leq 3(|x|+|y|+|z|). \end{align} Now at this stage, you CANNOT use the inequality $|x|\leq x^2$; this is only true for $|x|\geq 1$, whereas in our case we're interested in what happens as $\xi=(x,y,z)\to 0$. However, note that $|x|+|y|+|z|$ is known as the 'sum norm' or the $\ell^1$ norm; this can be bounded above by a multiple of the $\ell^2$ norm (the Euclidean norm) using the Cauchy-Schwarz inequality: \begin{align} \sum_{i=1}^n|\xi_i|=\sum_{i=1}^n|\xi_i|\cdot 1\leq \left(\sum_{i=1}^n|\xi_i|^2\right)^{1/2}\cdot \left(\sum_{i=1}^n1^2\right)^{1/2}=\sqrt{n}\cdot \|\xi\|, \end{align} where $\|\cdot\|$ denotes the Euclidean 2-norm. Note that the exact factor of $\sqrt{n}$ here is actually sharp, i.e the 'best possible', but for our purposes, it doesn't matter; all we need to know is there is some constant $C>0$ such that for all $\xi$, $\sum_{i=1}^n|\xi_i|\leq C\|\xi\|$. So, going back to the original problem, we now have \begin{align} |f(\xi)-f(0)|&=|x+2y+3z|\cdot |\ln(x^2+2y^2+3z^2)|\\ &\leq 3(|x|+|y|+|z|)\cdot |\ln(x^2+2y^2+3z^2)|\\ &\leq 3C\|\xi\|\cdot |\ln(x^2+2y^2+3z^2)|.\tag{$*$} \end{align} Next, we have another mistake of yours. You cannot replace the coefficients of the stuff inside $\ln$ with $3$. Why? Well, it is indeed true that \begin{align} x^2+2y^2+3z^2\leq 3(x^2+y^2+z^2)=3\|\xi\|^2, \end{align} and since $\ln$ is an increasing function, it is also true that \begin{align} \ln(x^2+2y^2+3z^2)\leq \ln(3\|\xi\|^2). \end{align} BUT, from here, it does not follow that \begin{align} |\ln(x^2+2y^2+3z^2)|\leq |\ln(3\|\xi\|^2)|. \end{align} This is because $\ln$ becomes negative on the interval $(0,1)$, so $\ln$ being increasing implies that $|\ln|$ is decreasing on $(0,1)$. Since we're interested in the limit $\xi\to 0$, we can restrict ourselves to a small punctured ball around the origin, $B_{\delta}(0)\setminus\{0\}$. So, we need a lower bound on the arguments: \begin{align} x^2+2y^2+3z^2\geq x^2+y^2+z^2=\|\xi\|^2, \end{align} and hence for small $\|\xi\|$, this implies \begin{align} |\ln (x^2+2y^2+3z^2)|\leq |\ln (\|\xi\|^2)|=2|\ln \|\xi\||.\tag{$**$} \end{align} So, combining $(*)$ with $(**)$, we see that \begin{align} |f(\xi)-f(0)|\leq 6C\|\xi\|\cdot |\ln \|\xi\||=\tilde{C}\cdot \left|\|\xi\|\log \|\xi\|\right|. \end{align} Again, I don't care about the exact constant in front. We now have our function, $h(r)=\tilde{C}|r\log r|$, which we know from elementary calculus approaches $0$ as $r\to 0^+$. Thus, we have proved that $f$ is continuous at the origin, if we define $f(0)=0$.

Finally, I'm not really sure what your confusion is with the last bullet point. We can always find a majorization, yes. For example, \begin{align} h(r):=\sup_{|\xi-\xi_0|=r}|f(\xi)-f(\xi_0)| \end{align} is a function which by construction satisfies $|f(\xi)-f(\xi_0)|\leq h(\|\xi-\xi_0\|)$. However, if the function $f$ is not continuous at $\xi_0$, then the limit of $h$ will not be $0$ (we have $f$ is continuous at $\xi_0$ if and only if $\lim\limits_{r\to 0^+}h(r)=0$).


Now, I just want to emphasize that you have to be very careful with the direction your inequalities point. Be careful with absolute values, and increasingness/decreasingness of the functions.

Also, I should stress that overall constants are unimportant here, since they do not affect whether or not a limit is $0$.

$\endgroup$
3
  • $\begingroup$ First of all, a big thank you for the work you have done here. I really appreciate, and I am carefully reading this. I'm not a student of mathematics but you made it really clear, also when talking about $\ell$ space, which we never faced but I understood (reading on Wikpedia too). At the moment I have just a clarification to ask: when you said "you don't have to [...] it is a freedom that you have which you may or may not decide to invoke", does this mean I could also take a big ball, and eventually shrink it, or that there are other ways to majorize? Sorry for all these questions! $\endgroup$
    – Heidegger
    Commented Mar 2, 2023 at 19:24
  • 1
    $\begingroup$ @Numb3rs look through the derivation again. The only place I used that $\xi$ lived in a small ball was when using the inequalities for $|\ln|$. Otherwise, I placed no restrictions on how big $\xi$ is. So, I'm just saying that since my end goal is to prove a statement about limits as $\xi\to 0$, and since my inequality for $|\ln|$ only holds for small $\|\xi\|$, there is no harm in me making the assumption later on that $\xi$ lives inside a small ball. $\endgroup$
    – peek-a-boo
    Commented Mar 2, 2023 at 20:08
  • $\begingroup$ Limpid and clear! $\endgroup$
    – Heidegger
    Commented Mar 2, 2023 at 22:10
2
$\begingroup$

In your work the inequality $$3(|x| + |y| + |z|) \leq 3(x^2 + y^2 + z^2)$$ does not hold near $(0,0,0)$ (take for example $x=y=z=1/2$), but you can replace it with
$$|x| + |y| + |z|= (1,1,1)\cdot (|x|, |y|,|z|)\leq \sqrt{3}\sqrt{x^2 + y^2 + z^2}$$ which follows by Cauchy-Schwarz inequality. Also $|\ln(x^2 + 2y^2 + 3z^2)|\leq |\ln(3x^2 + 3y^2 + 3z^2)|$ is not correct because $t\to |\ln(t)|$ is decreasing in $(0,1]$.

Actually the estimate of $\ln$ is not necessary. Indeed, again by Cauchy-Schwarz inequality, $$|x + 2y + 3z|=|(1,\sqrt{2},\sqrt{3})\cdot(x,\sqrt{2}y,\sqrt{3}z)|\leq \sqrt{1+2+3}\sqrt{x^2+2y^2+3z^2},$$ which implies $$|(x + 2y + 3z)\ln(x^2 + 2y^2 + 3z^2)|\leq \sqrt{6}\sqrt{h(x,y,z)}|\ln(h(x,y,z))|$$ where $h(x,y,z):=x^2+2y^2+3z^2$.

It follows that the limit of $f$ is zero as $(x,y,z)\to (0,0,0)$, because we have that $h(x,y,z)\to 0$ and, from calculus, $\lim_{t\to 0^+}\sqrt{t}\ln(t)=0$.

$\endgroup$
3
  • $\begingroup$ ignore my previous comment; I misread your definition of $h$ (though the fact that $h(x,y,z)\leq 3(x^2+y^2+z^2)$ is a red-herring since $\ln$ becomes negative near the origin, so we shouldn't use this inequality inside $\ln$). Now, there's only a small typo, which is you forgot the absolute values around $\ln$. $\endgroup$
    – peek-a-boo
    Commented Mar 2, 2023 at 19:05
  • $\begingroup$ @peek-a-boo Thanks for pointing out the typo! BTW the inequality $h(x,y,z)\leq 3(x^2+y^2+z^2)$ is used just to show that $h\to 0$. $\endgroup$
    – Robert Z
    Commented Mar 2, 2023 at 19:27
  • $\begingroup$ oh right, II thought $h\to 0$ was 'obvious' and jumped ahead a few steps with the $\ln$. $\endgroup$
    – peek-a-boo
    Commented Mar 2, 2023 at 20:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .