Understanding a multivariable $\epsilon - \delta$ proof of the Sum Law for limits

Question

Theorem

Let $f,g: A \to \mathbb R^m$ be well defined on $A \subset \mathbb R^k$.

If $$f(x) \to b, g(x) \to c, \text{ for } x \to a, x \in A$ $$

then

$$f+g \to b +c$$

Proof

$\forall x \in A$ is

$$||(f+g)-(b+c)|| \le ||f-b||+||g-c||$$

[Q1: Where does this inequality come from? I thought it is the triangular inequality but I can not see how it is applied ]

For an arbitrary $\epsilon > 0$ we can choose $\delta_1 > 0$ such that

$$||f-b|| < \epsilon /2$$ for all $x$ with $|x-a|| < \delta_1$

The same is written for $g,b \text{ and } \delta_2$

[Q2: Where does the above come from? Where did $...< \epsilon / 2$ come from?]

If we set $\delta = \min\{\delta_1,\delta_2\}$ then

$$||(f+g)-(b+c)|| < \epsilon \text{ for all } x\in A \text{ with } ||x-a||<\delta$$

I have literally used a half day on understanding this. Please clarify this for me.

Answer 1

Note that $(f+g)-(b+c) = (f-b)+(g-c)$, so that's how the triangle inequality was applied.

The $\epsilon/2$ arrived so that you would have $$\|(f+g)-(b+c)\|\le \|f-b\| + \|g-c\|<\epsilon/2 + \epsilon/2 = \epsilon.$$ In order to guarantee this, we need to make $\|x-a\|<\delta_1$ for the first inequality and $\|x-a\|<\delta_2$ for the second; requiring that $\|x-a\|<\min(\delta_1,\delta_2)$ ensures that both conditions are met.

(P.S. You might find some of my YouTube lectures on multivariable calculus/analysis helpful. See the link in my profile.)

Answer 2

1. This comes from first rewriting $(f + g) - (b + c)$ as $(f - b) + (g - c)$ and applying the triangle inequality.

2. This is using the fact that $f(x) \to b$. This precisely means that for any value of $\epsilon' > 0$, we can pick such a $\delta$. This is just this fact applied to $\epsilon' = \tfrac 12 \epsilon$.

   The reason we pick $\tfrac 12 \epsilon$ is because we're "looking ahead" at what's going to happen when we substitute things back into our equality, bearing in mind our ultimate goal: we want to make $\lVert (f + g) - (b + c) \rVert < \epsilon$, and since we're adding up two things, let's just make both of those be less than $\tfrac 12 \epsilon$.

Answer 3

You want to prove that $f(x)+g(x)$ is near $b+c$ when $x$ is near $a$. This means that you have to look at the quantity $$(f(x)+g(x))-(b+c)=(f(x)-b)+(g(x)-c))$$ when $x$ is near $a$. In the first place the triangle inequality gives $$|(f(x)+g(x))-(b+c)|\leq|f(x)-b|+|g(x)-c|\ .\tag{1}$$ Since we want that the LHS is $<\epsilon$ we have to make sure that the RHS is $<\epsilon$ when $x$ is near $a$. When both summands on the RHS are $<{\epsilon\over2}$ this is guaranteed. Now $\lim_{x\to a}f(x)=b$ implies that there is a $\delta_1>0$ such that $0<|x-a|<\delta_1$ guarantees $|f(x)-b|<{\epsilon\over2}$, and similarly, there is a $\delta_2>0$ such that $0<|x-a|<\delta_2$ guarantees that $|g(x)-c|<{\epsilon\over2}$.

Since we want both estimates $<{\epsilon\over2}$ to be fulfilled simultaneously we impose on $x$ the stronger condition $$0<|x-a|<\delta:=\min\{\delta_1,\delta_2\}>0\ .$$ When $x$ is so near to $a$ then the required error bound $$|(f(x)+g(x))-(b+c)|<\epsilon$$ is guaranteed.

Answer 4

I'll go ahead and reproduce the proof with arguments to answer your questions.

Assuming your assumptions, we get that for $x \in A$ $$ ||(f+g)(x)-(b+c)||=||f(x)+g(x)-(b+c)||=||f(x)-b+g(x)-c)|| $$ $$ \leq||f(x)-b||+||g(x)-c)|| \; \:\; \:\; \:\; \: (*) $$ This should clear up your first question. Now since $f$ and $g$ has a limit for $x\to a$ it means per definition that given some $\varepsilon >0$ we can find $\delta_f>0$, s.t. $\forall x \in B_{\delta_f}(a)$ it holds that $$ ||f(x)-b||<\varepsilon/2.\; \:\; \:\; \:\; \: (1) $$ By the same argument there exists a $\delta_g>0$, s.t. $\forall x \in B_{\delta_g}(a)$ it holds that $$ ||g(x)-b||<\varepsilon/2. \; \:\; \:\; \:\; \: (2) $$ The reason why we want the values to be less than $\varepsilon /2$ is due to the evaluation in equation $(*)$. Observe that if we choose $\delta = \text{min}\{\delta_f,\delta_g\}$ both equation $(1)$ and $(2)$ two hold for $x \in B_\delta(a)$. Thus we see from $(*)$ that $$ ||(f+g)(x)-(b+c)||\leq||f(x)-b||+||g(x)-c)||< \varepsilon/2 + \varepsilon/2 = \varepsilon. $$ Which by definition means that $(f+g)(x) \to (b+c)$ for $x \to a$.