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How can I prove that the function $$ f:\mathbb R\rightarrow \mathbb R$$ $$ x\mapsto\frac{1}{2}x^2$$ is continuous?

I currently have: $$ \epsilon > 0, \delta > 0$$ $$ |x-a| < \delta \rightarrow \left|\frac{1}{2}x^2-\frac{1}{2}a^2\right| < \epsilon$$ $$ \left|\frac{x^2}{2}-\frac{a^2}{2}\right|<\epsilon $$ $$ \left|\frac{x^2-a^2}{2}\right| < \epsilon $$

But at this point im not sure what to do.

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  • $\begingroup$ Please introduce properly your variables. $\endgroup$
    – Lelouch
    May 15 at 11:23

2 Answers 2

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Use the fact that $$ x^2-a^2=(x-a)(x+a)=(x-a)\bigl((x-a)+2a\bigr). $$ As a consequence, $$ \left|\frac{x^2-a^2}2\right|\le\frac12\bigl(|x-a|\bigl(|x-a|+2|a|\bigr)\bigr) $$ And if $|x-a|<1$, $|x-a|+2|a|<1+2|a|$. Therefore, $$ \left|\frac{x^2-a^2}2\right|\le|x-a|\frac{1+2|a|}2. $$ So, take $\delta=\min\left\{1,\frac{2\varepsilon}{1+2|a|}\right\}$, and then $$ |x-a|<\delta\implies\left|\frac{x^2-a^2}2\right|<\varepsilon. $$

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Show that $f(x) = ~\displaystyle \frac{1}{2}x^2$ is continuous at any $x = a ~: a \in \Bbb{R}$.

Different people use different approaches to constructing an $\epsilon, \delta$ demonstration.

My approach is to start with the assumption that $x$ is in a neighborhood around $a$, derive the necessary relationship between $\delta$ and $\epsilon$, and then (from scratch) verify that the relation accomplishes what I want it to accomplish.

I am going to assume that the value of $a$ for which continuity is being established is $> 0$. This assumption will simplify the proof. The demonstrations when $a = 0$, or $a < 0$ would be similar.


I will start with the idea that $0 < |x - a| < \delta. $

This implies that $-\delta < x - a < \delta \implies $

$$ a - \delta < x < a + \delta. \tag1 $$

My goal will be to show that for any $\epsilon > 0$, I can choose $\delta > 0$ so that:

$$ \left|\frac{1}{2} \left(x^2 - a^2\right)\right| < \epsilon $$

$$\iff ~ \frac{a^2}{2} - \epsilon < \frac{x^2}{2} < \frac{a^2}{2} + \epsilon. \tag2 $$

Since this answer is a derivation, you are seeing my step-by-step thinking.

Care is needed here. I know that if $r,s$ are any two positive values, then

$$r < s \iff r^2 < s^2. \tag3 $$

I know that $a$ is positive. I plan to impose some artificial constraints on $\delta.$ One of them will be that $\delta \leq (a/2)$. This will guarantee that if $x$ is in a neighborhood around $(a)$, of radius $\delta$, that this entire neighborhood will be positive. This will allow me to use the idea in (3) above.


Examining the RHS of (1) and (2),

I know that $\displaystyle ~\frac{x^2}{2} < \frac{a^2}{2} + a\delta + \frac{\delta^2}{2}.$

What I want is for $\displaystyle ~\frac{x^2}{2} < \frac{a^2}{2} + \epsilon.$

One way to accomplish this is to somehow have

$$ a\delta + \frac{\delta^2}{2} < \epsilon. \tag4 $$


Examining the LHS of (1) and (2),

I know that $\displaystyle ~\frac{a^2}{2} - a\delta + \frac{\delta^2}{2} < ~\frac{x^2}{2}.$

What I want is for $\displaystyle ~\frac{a^2}{2} - \epsilon < \frac{x^2}{2}.$

One way to accomplish this is to have

$$ - \epsilon < -a\delta + \frac{\delta^2}{2}. \tag5 $$


So, under the assumption that $a > 0$, and that $\delta < (a/2)$, I need to derive a relationsip between $\delta$ and $\epsilon$ so that the inequalities in (4) and (5) will be implied.

Then, with the derived relationship in place, I will (from scratch) verify that the relationship does what I want.


As an arbitrary personal choice, when I attack (4) above, I want to stay linear. So, I will impose the additional constraint on $\delta$ that $\delta \leq 1.$ This implies that $~\displaystyle \frac{\delta^2}{2} < \delta.$

This implies that the LHS in (4) above is now strictly less than $(a+1)\delta.$

So, tentatively, my constraints on $\delta$ will be

$$\delta \leq (a/2), ~~\delta \leq 1, ~~\delta \leq \frac{\epsilon}{a+2}.$$

This is accomplished by specifying that

$$\delta = \min\left(\frac{a}{2}, 1, \frac{\epsilon}{a+2}\right). \tag6 $$

Note that under the assumption that $a > 0,$ you must have that $(a+2) > 0.$

Also,

$~\displaystyle \delta \leq \frac{\epsilon}{a+2} \implies (a+2) \delta \leq \epsilon \implies $

$~\displaystyle -\epsilon \leq - (a+2)\delta < -a\delta.$

So, the specification in (6) above also works for the inequality in (5) above.


Opinions vary whether I can construe the analysis as done, at this point. In my opinion, I have to demonstrate that when $a > 0$, the combined premises in (1) and (6) above imply that the conclusion in (2) is reached.

I have shown that the assumption that $~\displaystyle \delta \leq \frac{a}{2}~$ implies that it is sufficient to show that the results in (4) and (5) are implied. These combined results will then imply that the result in (2) above is established.

$~\displaystyle \delta \leq 1 \implies \frac{\delta^2}{2} < \delta.$

Therefore,

$$a\delta + \frac{\delta^2}{2} < (a + 1)\delta < (a + 2)\delta \leq \epsilon. \tag7 $$

Thus, I have verified that the constraint in (6) implies the constraint in (4).

Further, the analysis in (7) implies that

$$a\delta < \epsilon \implies -\epsilon < -a\delta.$$

Therefore, the constraint in (5) has also been implied.

Therefore, the assumption that $a > 0$, coupled with the constraints in (1) and (6) imply that the constraint in (2) above is satisfied, as required.

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