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Suppose that $f$ is a real-valued function defined in an open set $E \subset \Bbb R^n$, and that the partial derivatives $D_1f, \ldots D_nf$ are bounded in $E$. Prove that $f$ is continuous in $E$.

So if $\textbf{x} = (x_1, x_2, x_3, ... , x_n)$ and $\textbf{y} = (y_1, y_2, y_3, ... , y_n)$, then we have to show that for all $\epsilon >0$, there exists a $\delta>0$ such that $d(\textbf{x}, \textbf{y}) < \delta \implies d(f(\textbf{x}), f(\textbf{y})) < \epsilon$.

I am reading a solution here and it says we can write

$f(x_1 + h_1, x_2 + h_2, x_3 + h_3, ... , x_n + h_n) - f(x_1, x_2, x_3, ... , x_n)$

as:

$f(x_1 + h_1, x_2 + h_2, x_3 + h_3, ... , x_n + h_n) - f(x_1, x_2 + h_2, x_3 + h_3, ... , x_n + h_n) + f(x_1, x_2 + h_2, x_3 + h_3, ... , x_n + h_n) - f(x_1, x_2, x_3 + h_3, ... , x_n + h_n) + \ldots + f(x_1, x_2, x_3, ... ,x_n + h_n) - f(x_1, x_2, x_3, ... , x_n)$

and then use the mean value theorem to get:

$D_1(x_1 + h_1t_1, x_2 + h_2, x_3 + h_3, ... , x_n + h_n)h_1$ + $D_2(x_1, x_2 + h_2t_2, x_3 + h_3, ... , x_n + h_n)h_2 + \ldots D_n(x_1, x_2, x_3, ... ,x_n + h_nt_n)h_n$

Since each $D_n$ is bounded, take the maximum of these bounds, call it $M$. Then we have that the expression directly above this sentence is $\leq M(|h_1| + |h_2| + \ldots + |h_n|)$, so:

$f(x_1 + h_1, x_2 + h_2, x_3 + h_3, ... , x_n + h_n) - f(x_1, x_2, x_3, ... , x_n) \leq M(|h_1| + |h_2| + \ldots + |h_n|)$

Then the proof just stops there and doesn't continue.

I don't understand what the $h_n$'s are supposed to represent. Are they real numbers? If so, why are we adding an arbitrary vector $\textbf{h}$ to $\textbf{x}$?

After we get $f(x_1 + h_1, x_2 + h_2, x_3 + h_3, ... , x_n + h_n) - f(x_1, x_2, x_3, ... , x_n) \leq M(|h_1| + |h_2| + \ldots + |h_n|)$

, from this how do we show that for all $\epsilon >0$, there exists a $\delta>0$ such that $d(\textbf{x}, \textbf{y}) < \delta \implies d(f(\textbf{x}), f(\textbf{y})) < \epsilon$

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mr eyeglasses
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    $\bf h = y-x$. The estimate you have is something resembling $d(f(x),f(y))\le M d(x,y)$... – Anthony Carapetis Nov 03 '15 at 02:35
  • @AnthonyCarapetis We aren't allowed to just assume the Euclidean metric, right? In my last inequality, I essentially have that $f(\textbf{y}) - f(\textbf{x}) \leq M(\textbf{y} - \textbf{x})$. So if we had the Euclidean metric, this would be true, but I'm not sure if we are allowed to assume it? – mr eyeglasses Nov 03 '15 at 03:24
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    This statement is not true for all metrics/topologies. The euclidean metric (or a topologically equivalent one) is almost certainly meant to be assumed. – Anthony Carapetis Nov 03 '15 at 03:27
  • @AnthonyCarapetis Thanks. There's one subtlety here that I am slightly confused about. In the inequality $$f(x_1 + h_1, x_2 + h_2, x_3 + h_3, ... , x_n + h_n) - f(x_1, x_2, x_3, ... , x_n) \leq M(|h_1| + |h_2| + \ldots + |h_n|)$$, this says that a vector is $\leq$ a scalar (since $M, h_1, \ldots h_n \in \Bbb R$), which doesn't make sense. What is the rigorous explanation of going from $M(|h_1| + |h_2| + \ldots + |h_n|) = M(|y_1 - x_1| + |y_2 - x_2| + \ldots + |y_n - x_n|)$ to $M(\textbf{y} - \textbf{x})$, since the former is a scalar and the latter is a vector? – mr eyeglasses Nov 03 '15 at 03:42
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    $f$ is real-valued, so both sides are scalar. You should only be writing inequalities between scalars. – Anthony Carapetis Nov 03 '15 at 03:43
  • @AnthonyCarapetis But $\textbf{y} - \textbf{x}$ is a vector, and so $M(\textbf{y} - \textbf{x})$ is a vector (with a scalar multiple $M$). Am I missing something? – mr eyeglasses Nov 03 '15 at 03:45
  • Why are you considering $M(\bf y - x)$ at all? The estimate is already in terms of distances (scalars), which is what you need for continuity. – Anthony Carapetis Nov 03 '15 at 03:49
  • @AnthonyCarapetis Oh, I was looking at the form of the inequality for Lipschitz continuity, and it was in the form $|f(x) - f(y)| \leq M|x - y|$. This is what we want, right? (Unless you are referring to a different type of continuity that I am missing) – mr eyeglasses Nov 03 '15 at 03:54
  • yes, Lipschitz continuity is what we get here. – Anthony Carapetis Nov 03 '15 at 03:56
  • @AnthonyCarapetis So for Lipschitz continuity, we need $|f(\textbf{y}) - f(\textbf{x})| \leq M|\textbf{y} - \textbf{x}|$. The inequality I got is: $$|f(\textbf{y}) - f(\textbf{x})| \leq M|(y_1 - x_1) + (y_2 - x_2) + \ldots + (y_n - x_n)| \leq M(|y_1 - x_1| + |y_2 - x_2| + \ldots + |y_n - x_n|)$$. However, $M|\textbf{y} - \textbf{x}| = M\sqrt{(y_1 - x_1)^2 + (y_2 - x_2)^2 + \ldots + (y_n - x_n)^2}$. – mr eyeglasses Nov 03 '15 at 04:07
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    Use $|y_1 - x_1| + \cdots + |y_n - x_n| \le \sqrt n |{\bf y - x}|$. – Anthony Carapetis Nov 03 '15 at 04:11
  • I think, you don't have the hypotheses to use the mean value theorem. I know that: Suppose $f$ maps a convex open set $E \subset \mathbb{R}^{n}$ into $\mathbb{R}^{m}$, $f$ differentiable in $E$, and there is a real number $M$ such that $$||f'(x)||\leq M$$ for every $x \in E$. Then $$||f(b)-f(a)||\leq M ||b-a||$$ for all $a,b \in E$. –  Aug 09 '20 at 11:01

1 Answers1

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One can also use $$f(\vec{x}+\vec{h}) - f(\vec{x}) \le M(|h_1|+|h_2| + \dots + |h_n|)\le n M\sqrt{\frac{ h_1^2+ \dots h_n^2}{n}} \\ = \sqrt{n} M\sqrt{h_1^2+ \dots h_n^2 } = \sqrt{n} M |\vec{h}|$$

By the QM-AM inequality. So using an equivalent definition of continuity, we have

$$|f(\vec{x}+\vec{h}) - f(\vec{x})| < \varepsilon $$ whenever $|\vec{h}|<\delta $ when we pick $\delta = \varepsilon / \sqrt n M$

meiji163
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