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 $d(\textbf{x}, \textbf{y}) < \delta \implies d(f(\textbf{x}), f(\textbf{y})) < \epsilon$
Best Answer
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 $\delta = \varepsilon / \sqrt n M$