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Let $f: \mathbb{R}^n \rightarrow \mathbb{R}$.

Here is what I have done:

We want to see, $\lim_{\vec x \rightarrow x_0} f(x) =^? f(x_0)$. I started with the two variable case, $f: (x,y) \mapsto f(x,y)$.

We have $|f(x_0, y_0) - f(x, y)| \leq |f(x_0, y_0) - f(x, y_0)| + |f(x, y_0) - f(x, y)|$. Let $\epsilon >0$, since $f$ is continuous on each variable we can find $\delta_1(\epsilon, x_0, y_0)$ and $\delta_2(\epsilon, x_0, y_0)$ such that:

$|x_0 - x| < \delta_1(\epsilon, \hat y) \Rightarrow |f(x_0,\hat y) - f(x,\hat y)| < \epsilon/2$ $\forall \hat y $

$|y_0 - y| < \delta_2(\epsilon, \hat x) \Rightarrow |f(x_0,y_0) - f(\hat x,y)| < \epsilon/2$ $\forall \hat x$

Let $(x_n)_n \uparrow x$, then $|f(x_0,y_0) - f(x_n,y_0)| =_{(1)} f(x_0,y_0) - f(x_n,y_0) > f(x_0,y_0) - f(x_m,y_0)$ if $x_m > x_n$. Let $n_0$ be such that $|x_0 - x_{n_0}| < \delta_1$, so it is $f(x_0,y_0) - f(x_{n_0},y_0) < \epsilon/2$

Now let, $(y_n)_n \rightarrow y$ and $n \geq n_0$. Then, $|f(x_0, y_0) - f(x_n, y_n)| \leq |f(x_0,y_0) - f(x_{n}, y_0)| + |f(x_{n},y_0) - f(x_{n}, y_n)| \leq |f(x_0,y_0) - f(x_{n_0}, y_0)| + |f(x_{n},y_0) - f(x_{n}, y_n)| $

Now let $n_1(\delta_2)$ be such that $\forall m \geq n_1$, $|y_0-y_m| < \delta_2$, we conclude $|f(x_0,y_0) - f(x_{n_0}, y_0)| + |f(x_{n},y_0) - f(x_{n}, y_n)| < \epsilon/2 + \epsilon/2$.

We can apply the same reasoning for $(x_n)_n \downarrow x$ if we change the order of the terms on $(1)$

So we conclude that $lim_n f(x_n, y_n) = f(x_0, y_0)$.

I have not used once that $f$ is nondecreasing on variable $y$. Is this correct? A

I'll be the first to admit that my analysis skills are a bit rusty! So any input/ cleaner proof/explanation is wholeheartly welcome.

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Continuity in each variable is not sufficient.

Counter-example: let f(0,0) = 0, and elsewhere f(x,y) = $xy/(x^2+y^2)$ (plot on Wolfram Alpha), then this function is continuous in both x and y, but not continuous in the general sense. (There is a discontinuity along the x=y axis, or any other straight line than the coordinate axes for that matter.)

The problem in your proof is in the specification of $\delta$ when you apply the hypothesis of continuity in each variable. You have to be careful about which functions you are applying the hypothesis to.

$|f(x_0,y_0) - f(x,y)| \leq |f(x_0, y_0) - f(x,y_0)| + |f(x,y_0)-f(x,y)|$.

  1. Continuity of $(x \mapsto f(x,y_0))$ at $x_0$: there exists $\delta_1(\epsilon, x_0, y_0)$ such that $|x_0-x|\leq \delta_1 \implies |f(x_0,y_0)-f(x,y_0)| \leq \epsilon/2$.

  2. Continuity of $(y \mapsto f(x,y))$ at $y_0$ (note that the first coordinate is fixed at $x$ rather than $x_0$): there exists $\delta_2(\epsilon,x,y_0)$ such that $|y_0-y| \leq \delta_2(x) \implies |f(x,y_0)-f(x,y)|\leq \epsilon/2$.

It is important to note that in the second item, you are applying continuity to a function that depends on $x$. Thus for every $x$ there may be a different $\delta_2$.

We thus have the following (failed attempt):

$$|x_0-x|\leq\delta_1 \wedge |y_0-y|\leq\delta_2(x) \implies |f(x_0,y_0)-f(x,y)|\leq\epsilon$$

The problem is that $\delta_2$ depends on $x$, when it should not.

The monotonicity hypothesis allows you to fix that.


On a separate note, the epsilon-delta definition of continuity/limits will be more convenient here than the sequential characterization you are using.

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