# Epsilon-delta definition in proving the continuity of $\frac{1}{2}x^2$

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.

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 $$|x-a|<\delta\implies\left|\frac{x^2-a^2}2\right|<\varepsilon.$$

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.