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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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    $\begingroup$ $\bf h = y-x$. The estimate you have is something resembling $d(f(x),f(y))\le M d(x,y)$... $\endgroup$ – Anthony Carapetis Nov 3 '15 at 2:35
  • $\begingroup$ @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? $\endgroup$ – mr eyeglasses Nov 3 '15 at 3:24
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    $\begingroup$ This statement is not true for all metrics/topologies. The euclidean metric (or a topologically equivalent one) is almost certainly meant to be assumed. $\endgroup$ – Anthony Carapetis Nov 3 '15 at 3:27
  • $\begingroup$ @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? $\endgroup$ – mr eyeglasses Nov 3 '15 at 3:42
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    $\begingroup$ Use $|y_1 - x_1| + \cdots + |y_n - x_n| \le \sqrt n |{\bf y - x}|$. $\endgroup$ – Anthony Carapetis Nov 3 '15 at 4:11
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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 $\delta = \varepsilon / \sqrt n M$

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