# MATH 409 Lecture 10

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## Continuity

Given set , a function , and a point , the function is **continuous** at if for any there exists such that and imply

We say that the function is **continuous on a set** if is continuous at every point . The function is **continuous** if it is continuous on the entire domain .

Note:

- In the case , is continuous at if and only if

### Properties

**Theorem.** A function is continuous at if and only if for any sequence of elements of , as implies as .

**Theorem.** Suppose that functions are both continuous at . Then , and are also continuous at . If, additionally, , then is continuous at as well.

### Bounded Functions

A function is **bounded on a subset** if there exists such that for all . In the case , we say that is **bounded**.

The function is bounded above on if there exists such that for all .

It is bounded below on if there exists such that for all .

Equivalently, is bounded on if the image is a bounded subset of . Likewise, the function is bounded above on if the image is bounded above. It is bounded below on if is bounded below.

#### Example

, for

The function is unbounded. At the same time, it is bounded on and on .

**Theorem.** If is a closed, bounded interval of the real line, then any continuous function is bounded.

*Proof.* Assume that is unbounded. Then for every there exists a point such that . We obtain a sequence of elements of such that the sequence diverges to infinity.

Since the interval is bounded, the sequence has a convergent subsequence (due to the Bolano-Weierstrass Theorem). Let . Then (due to comparison theorem). Since the sequence is a subsequence of , it diverges to infinity. In particular, it does not converge to . It follows that is discontinuous at .

Thus any continuous function on has to be bounded.

## Discontinuities

A function is **discontinuous** at a point if it is not continuous at . There are various kinds of discontinuities including the following ones.

- has a
**jump discontinuity**at if both one-sided limits at exist, but they are not equal: . - has a
**removable discontinuity**at a point if the limit at exists, but it is different from the value at : . - If is continuous at , then it is locally bounded at , which means that is bounded on the set provided is small enough. Hence any function
**not locally bounded**at is discontinuous at .

## Examples

**Constant function.** for all and some .

Since for all , is continuous.

** Identity function.** for all .

Since for all , the function is continuous.

**Step function.**

Since and , the function has a jump discontinuity at . It is continuous on .

**Harmonic function.** and for

Since for all , the function is continuous on . It is discontinuous at because it is not locally bounded at 0.

**Harmonic sine.** and for .

Since does not exist, the function is discontinuous at 0. Notice it is neither jump nor removable discontinuous, and the function is bounded. The reason for discontinuity is oscillation.

and for .

Since , the function is continuous at 0.

**Dirichlet function.** .

Since never exists, the function has no points of continuity.

**Riemann function.**

Since for all , the function is continuous at irrational points and discontinuous at rational points. Moreover, all discontinuities are removable.

## Extreme Value Theorem

**Theorem. [Extreme Value Theorem].** if is a closed, bounded interval of the real line, then any continuous function attains its extreme values (maximum and minimum) on . To be precise, there exist points and in such that

*Proof.* Since is continuous, it is bounded on . Hence and are well-defined numbers. In different notation, and . Clearly, for all .

For any , the number is not an upper bound of th eset , while is not a lower bound of . Hence we can find points such that and . By construction, and as . The Bolzano-Weierstrass Theorem implies that the sequences and have convergent subsequences (or, in other words, they have limit points). Let be a limit point of and be a limit point of . Notice that . The continuity of implies is a limit point of and is a limit point of . We conclude that and .

We used the fact that the sequence is bounded when we applied the Bolzano-Weierstrass theorem.

We used the fact that the sequence is closed when we stated that the continuity of implies is a limit point of and is a limit point of .

Remarks:

- The theorem may not hold if is not closed. Counterexample: for . Neither maximum nor minimum is attained.
- The theorem may not hold if is not bounded. Counterexample: for . The maximal value is attained at , but the minimal value 0 is never attained.

## Intermediate Value Theorem

**Theorem. [Intermediate Value Theorem].** If a function is continuous, then any number that lies between and is a value of , that is, for some .

*Proof.* In the case , the theorem is trivial.

In the case , we notice that the function is continuous on , , and lies between and . Hence we acn assume without loss of generality that .

Further, if lies between and , then 0 lies between and . Moreover, the function is continuous on and if and only if . Hence it is no loss to assume that .

Now the theorem is reduced to a simpler case:

**Theorem.** If a function is continuous and , then for some .

Let . The set is nonempty (as ) and bounded (as . Therefore, exists. Observe that ( as ; as is a lower bound of ).

To complete the proof, we need the following lemma:

**Lemma.** If a function is continuous at a point and , then maintains its sign in a sufficiently small neighborhood of .

*Proof.* Since is continuous at and , there exists such that whenever . The inequality implies has the same sign as .

The lemma implies that . Indeed, if , then for some , the function maintains its sign in the interval .

- In the case , we obtain that and is not a lower bound of .
- In the case , we obtain that and is not the greatest lower bound of

Either way we arrive at a contradiction.

**Corollary.** If a real-valued function is continuous on a closed bounded interval , then the image is also a closed bounded interval.

*Proof.* By the Extreme Value Theorem, there exist points such that for all . Let denote the closed interval with endpoints and . Let denote the closed interval with endpoints and . We have that . The Intermediate Value Theorem implies that . Since , we obtain that .