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Consider following problem from Apostol Chap. 4: Apostol Chap 4. Continuity of max function

I did the following proof however later discovered the standard proof involves an auxiliary function, along the lines of the solution outlined here. I was wondering if I made any mistake. Here is the proof:

Without loss of generality let us assume $f_{k}(a)=\max(f_{1}(a),...,f_{m}(a))$

$\therefore f(a)=f_{k}(a)$

Each $f_{i}(.)$ is continuous at $a.$

Take any $\epsilon \gt 0. \therefore \exists \delta_{i} \gt 0 $ such that $|f_{i}(x)-f_{i}(a)|\lt\epsilon \forall x$ such that $|x-a|\lt \delta$. Let $\delta=min(\delta_{1},...,\delta_{m})$

$\therefore \forall x$ such that $|x-a|\lt \delta,$ $|f_{i}(x)-f_{i}(a)|\lt \epsilon, \forall i$.

i.e. $(f_{i}(a)-\epsilon)\lt f_{i}(x)\lt (f_{i}(a)+\epsilon), \forall i$

$f_{k}(a)=\max(f_{1}(a),...,f_{m}(a))$ $\therefore (f_{k}(a)+\epsilon) \ge (f_{i}(a)+\epsilon), \forall i$

$\forall x$ $ s.t. |x-a| \lt \delta, f_{k}(x)\lt(f_{k}(a)+\epsilon)$

$\therefore \forall x$ $ s.t. |x-a| \lt \delta, \max(f_{1}(x),...,f_{m}(x))\lt(f_{k}(a)+\epsilon)$ ... (A)

Also, $ \forall x$ $ s.t. |x-a| \lt \delta, f_{k}(x)\gt (f_{k}(a)-\epsilon)$

$\therefore \max(f_{1}(x),...,f_{m}(x))\gt(f_{k}(a)-\epsilon)$... (B)

Combining (A) and (B), $ \forall x$ $ s.t. |x-a|\lt \delta, (f(a)-\epsilon)\lt f(x)\lt (f(a)+\epsilon)$

$\therefore f(x)$ is continuous at $a$.

Is this correct? Thanks a lot in advance for your help.

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Canine360
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1 Answers1

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You haven't given a correct justification for equation (A). You don't know that $f_k(x)=\max(f_1(x),\dots,f_m(x))$, only that $f_k(a)=\max(f_1(a),\dots,f_m(a))$.

What you can say is that for all $x$ such that $|x-a| <\delta$, and for all $1\leq i\leq m$, $f_i(x)<f_i(a)+\epsilon\leq f_k(a)+\epsilon$. This allows you to conclude (A): for all $x$ such that $|x-a|<\delta$, $\max(f_1(x),\dots,f_m(x)) \leq f_k(a)+\epsilon$.

kccu
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  • I meant, since $f_{k}(x)$ itself is always $\gt$ $f_{k}(a)-\epsilon$, max of all $f_{i}(x)$ cannot be $\le$ $f_{k}(a)-\epsilon$ anywhere in the $\delta$-neighborhood of $a$, right? Of course at individual points of that neighbourhood there can be other $f_{i}(x)$ that's $\gt f_{k}(x)$. I'm afraid I'm still not clear why this argument is not true. Thanks again. – Canine360 Oct 09 '16 at 13:35
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    Ah I see your point... Your justification for equation (A) was still not correct, but your justification for equation (B) is. I will edit my answer. – kccu Oct 10 '16 at 22:13