Proving continuity in multivariable calculus using $\epsilon - \delta$ definition

The textbook im using only has stated the definition of limit with examples explained intuitively and without full rigour, so im pretty lost on multivariable limit/continuity problems. I have the following problem at hand and I have to prove that its continuous using $$\epsilon - \delta$$ definition: given $$a_1, \ldots, a_n \in \mathbb{R^n}$$ and a map $$g: \mathbb{R^n} \rightarrow \mathbb{R}$$ defined as $$g(a_1, \ldots, a_n) = a_1x_1 + a_2x_2 + \cdots + a_nx_n$$ Now the right side of the equation looks like a polynomial and I know that they are always continuous everywhere. But I cannot use any other theorems to prove it since it wouldnt be $$\epsilon - \delta$$. Some theorem I couldve used to make it easier is by taking coordinate function wise limit, but that does not help with the goal.

Essentially I have to show that :

$$\forall \epsilon > 0, \exists \delta > 0$$ such that $$x \in \mathbb{R^n}$$ and $$0 < |x - a| \implies |g(a_1, \ldots, a_n) - (a_1x_1 + a_2x_2 + \cdots + a_nx_n)| < \epsilon$$

Would appreciate any insight or general approach to solving multivariable limit/continuity problems

3 Answers

I am assuming that the variable is a vector $$x=(x_1,x_2...x_n)$$ and you have to show continuity near $$x=x_0=(x_{01},x_{02}....x_{0n})$$. To show that we need to show that we can make $$|g(x)-g(x_0)|$$ as small as we want by making $$|x-x_0|$$ as small as we need to. Thus given $$\epsilon$$ we want to make the following to be small than $$\epsilon$$: $$|g(x)-g(x_0)|=|\sum_{i=1}^n a_ixi-\sum_{i=1}^n a_ix_{0i}|=\sum_{i=1}^na_i(x_i-x_{i0}|\le \sum_{i=1}^n|a_i||x_i-x_{i0}|$$ To achieve it we choose $$\delta\lt\frac{\epsilon}{n \max{|a_i|}}$$ and we get: $$|g(x)-g(x_0)|\lt\sum_{i=1}^n|a_i|\frac{\epsilon}{n \max{|a_i|}}=\frac{\epsilon}{n}\sum_{i=1}^n\frac{|a_i|}{ \max{|a_i|}}\le\epsilon$$

We want to show that $$g$$ is continuous everywhere, so pick some arbitrary point $$a = (a_1, ..., a_n)$$. Now consider some arbitrary $$\varepsilon > 0$$ and our goal becomes showing that there exists some $$\delta > 0$$ such that for all $$a' = (a'_1, ..., a'_n)$$ with $$\|a - a'\| < \delta$$ we have $$|g(a) - g(a')| < \varepsilon$$.

Expanding this, what we want is that $$|g(a) - g(a')| = |a_1 x_1 + ... + a_n x_n - (a'_1 x_1 + ... + a'_n x_n)| = |(a_1 - a'_1)x_1 + ... + (a_n - a'_n) x_n| < \varepsilon.$$ Define $$X = |x_1| + ... + |x_n|$$ and let $$\delta = \varepsilon / X$$. We know that the euclidean norm is bounded by the sum of absolute differences, i.e. if we have some $$a'$$ such that $$\|g(a) - g(a')\| < \delta$$ then we can conclude $$\|g(a) - g(a')\| \leq |a_1 - a'_1| + ... + |a_n - a'_n|$$, so we get that $$|a_i - a'_i| < \delta$$ for all $$i \in [1, n]$$. Using this and the triangle inequality, we find \begin{align*} |g(a) - g(a')| &= |(a_1 - a'_1)x_1 + ... + (a_n - a'_n) x_n| \\ &\leq |(a_1 - a'_1)| \cdot |x_1| + ... + |(a_n - a'_n)| \cdot |x_n| \\ &< \delta |x_1| + ... + \delta |x_n| \\ &= \delta \cdot (|x_1| + ... + |x_n|) \\ &= \delta \cdot X \\ &= \varepsilon \end{align*} and thus it follows that $$g$$ is continuous at $$a$$. As $$a$$ was arbitrary, $$g$$ is continuous everywhere.

• Why are we allowed to assume g'(a)? usually in single variable calc, we defined a variable to be a limit and then went on with the work. But here we use g'(a) which is what is troublesome for me. Why are we allowed to do that and what makes it a legal move? – UnKnoWnZ Oct 22 '20 at 3:59
• I am not sure what you mean by "assuming $g(a')$", can you elaborate a bit more on that? – Watercrystal Oct 22 '20 at 4:03
• yes. I meant that since we arent given g'(a) explicitly in the question, how or why are we allowed to use g'(a)? (perhaps im missing/forgotten on an important concept/definition that allows us to use that) – UnKnoWnZ Oct 22 '20 at 4:04
• Well, in the $\varepsilon$-$\delta$ definition of continuity (at some point $a$) you want to show that for every point which is close to $a$ (in the sense that the euclidean distance from said point to $a$ is $< \delta$) we also have that the corresponding function value of that point is close to that of $a$. Hence what we do is make it concrete: Define $\delta$ and then take an arbitrary point $a'$ satisfying $\|a - a'\| < \delta$ and show that $g(a)$ is close to $g(a')$ as well (i.e. $|g(a) - g(a')| < \varepsilon$). Also note that there is a (huge) difference between $g(a)$ and $g'(a)$. – Watercrystal Oct 22 '20 at 4:09

Often for multivariable problems like this (needing to show continuity of $$f$$ at a point $$(b_1,b_2,\dots,b_n)$$), given an $$\epsilon$$, one can first find a $$\delta_i$$ for each coordinate, $$i$$, such that whenever $$|x_i-b_i|<\delta_i$$, then $$|f(b_1,b_2,\dots, x_i,b_{i+1},\dots b_n)-f(b_1,b_2,\dots, b_i,b_{i+1},\dots b_n)|<\textrm{some expression involving}\; \epsilon$$. Then, it often turns out that any $$\delta$$ small enough compared to all the $$\delta_i$$s (e.g. $$\leq \textrm{min}(\delta_1,\delta_2,\dots,\delta_n)$$) does the job for the entire function, i.e. for that suitably small $$\delta$$ we in fact have that $$|(x_1,x_2,\dots,x_n)-(b_1,b_2,\dots,b_n)|<\delta$$ implies $$|f(x_1,x_2,\dots,x_n)-f(b_1,b_2,\dots,b_n)|<\epsilon$$.

Regarding the question at hand - suppose you are given some $$\epsilon>0$$ and some point $$(b_1,b_2,\dots,b_n)$$. The way you can apply the idea in the above paragraph here, is to first, using the fact that each single variable function $$g(b_1,b_2,...,x_i,b_{i+1},\dots,b_n)=a_1b_1+a_2b_1+\dots+a_ix_i+a_{i+1}b_{i+1}+\dots+a_nb_n$$ is continuous, we can pick a $$\delta_i$$ such that $$|g(b_1,b_2,...,x_i,b_{i+1},\dots,b_n)-g(b_1,b_2,...,b_i,b_{i+1},\dots,b_n)|<\frac{\epsilon}{n}$$, i.e. $$|x_i-b_i|<\delta_i$$ implies that $$|a_ix_i-a_ib_i|<\frac{\epsilon}{n}$$. It is now very easy to show via the triangle inequality that, setting $$\delta=\textrm{min}(\delta_1,\delta_2,\dots,\delta_n)$$, whenever $$|(x_1,x_2,\dots,x_n)-(b_1,b_2,\dots,b_n)|<\delta$$, then $$|g(x_1,x_2,\dots,x_n)-g(b_1,b_2,\dots,b_n)|<\epsilon$$, completing the epsilon-delta proof of continuity.