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# Find continuous functions satisfying given conditions

Find continuous functions which satisfy the given conditions for all .

1. .
2. .
3. .

1. No such function can exist since for we have

2. Taking derivatives of both sides of the given equation we have

3. Again, taking derivatives of both sides we have

at all points such that . (Since is not satisfied by the zero function , we know there are real such that .) Then, integrating

Now, we can solve for by evaluating the given identity at ,

Therefore, we have

# Fill in a table of implications for the integral of a function

Let be a function such that the integral

exists for all in an interval . Consider the following statements:

a. is continuous at .
b. is discontinuous at .
c. is increasing on .
d. exists.
e. is continuous at .

. is continuous at .
. is discontinuous at .
. is convex on .
. exists.
. is continuous at .

Make a table to indicate by T whether the statement in row implies the statement in column . Leave the cell in the table blank if there is no implication.

The requested table is as follows:

We know the first column has all T’s since the function is continuous for any function by the first fundamental theorem we know the derivative exists at every point in the open interval . Since differentiability implies continuity (by example 7 on page 163 of Apostol), we then know that is continuous at any point .

We know the second column can have no T’s anywhere since, by the same argument as above, the function is continuous at every point . Thus, it cannot be discontinuous at any point .

The statement a,b,d, and e cannot imply that is convex on since they are statements regarding continuity and existence of derivatives. None of these properties have anything to do with convexity. Then, statement (c) does imply statement since we know that a function is convex if its derivative is increasing. Since is the derivative of and it is increasing, we have that is convex.

# Prove the intermediate value theorem for derivatives

Consider the following statement of the intermediate value theorem for derivatives:

Assume is differentiable on an open interval . Let be two points in . Then, the derivative takes every value between and somewhere in .

1. Define a function

Prove that takes every value between and in the interval . Then, use the mean-value theorem for derivatives to show takes all values between and somewhere in the interval .

2. Define a function

Show that the derivative takes on all values between and in the interval . Conclude that the statement of the intermediate-value theorem is true.

1. Proof. First, since is differentiable everywhere on the interval , we know is continuous on and differentiable on . Thus, if and then is continuous at since it is the quotient of continuous functions and the denominator is nonzero. If then

hence, is continuous at as well. Therefore, is continuous on the closed interval . So, by the intermediate value theorem for continuous functions we know takes on every value between and somewhere on the interval . Since this means takes on every value between and somewhere on the interval .
By the mean-value theorem for derivatives, we then know there exists some such that

for some . Since , we then conclude there is some such that for any . Since takes on every value between and , so does .

2. Proof. This is very similar to part (a). By the same argument we have the function is continuous on ; thus, takes on every value between and by the intermediate value theorem for continuous functions. Then, by the mean value theorem, we know there exists a such that

Thus, takes on every value between and . Since ; takes on every value between and

# Prove a property of the integral of the product of continuous functions

Let be a continuous function on the interval and assume

for every function which is continuous on the interval . Prove that for all .

Proof. Since must hold for every function that is continuous on , it must hold for itself (since is continuous on by hypothesis). Therefore we must have,

However, for all . We know from the previous exercise (Section 3.20, #7) that a non-negative function whose integral is zero on an interval must be zero at every point at which it is continuous. By hypothesis is continuous at every point of ; hence, is also continuous at every point of (since the product of continuous functions is continuous). Therefore,

Which implies,

# Prove that if the integral of nonnegative function is zero then the function is zero

Let be an integrable, non-negative function on the interval . Prove that if

then at each point at which is continuous.

Proof. The proof is by contradiction. Let be a point at which is continuous, and suppose . This implies since is non-negative by hypothesis. Since is continuous at , we know by the sign-preserving property of continuous functions (Theorem 3.7 in Apostol), that there is some neighborhood about , say such that has the same sign as for all , i.e., such that for all . But then by the monotone property of the integral we know,

But then,

This is a contradiction since is nonnegative we know

Since is strictly positive, the sum of the three pieces of the integral cannot equal

# Find points at which a given composite function is continuous

Define functions and as follows:

Find a formula for . Find the values at which is continuous.

If , we have and we compute the composition,

If , then and we compute the composition,

Hence,

This is the same function as in this previous exercise, and it is continuous everywhere.

# Find points at which a given composite function is continuous

Define functions and as follows:

Find a formula for and find the values at which is continuous.

If then and so . Therefore, by the definition of ,

If , then and so again

Next, if then and so

Finally, if , then and we have

Putting this all together we have

From this expression we have is continuous everywhere except at and .

# Find points at which a given composite function is continuous

Define functions and as follows:

Find a formula for . Find the values at which is continuous.

If , we have and we compute the composition,

If , then and we compute the composition,

Putting these together,

Certainly is continuous for all since the constant function and the polynomial are continuous. Further, is continuous at since

# Prove some continuity properties of a function

Given a function and a closed interval such that

for all .

1. Prove is continuous at every point .
2. If is integrable on prove

3. If is any point prove

1. Proof. Let be any point in . Then, for any let . Then

Thus, for any , we have whenever . Hence,

Therefore, is continuous at every point

2. Proof. Since is a constant (the value of the function evaluated at the constant ), we have

Therefore, we can compute,

The final inequality follows since, using the monotone property of the integral,

Then, since by our hypothesis on , we have

3. Proof. The proof proceeds similarly to that of part (b),

Then, since , we break the integral into two pieces (to deal with the fact that for and for ):

But now, we know implies , so we have the inequality,

# Find a way to define f(0) so that the function x sin (1/x) is continuous at 0

For define

Give a value for to make continuous at .

We claim that if we define , then the function with this additional point defined is continuous at .

Proof. Since for all we know

Then since

we apply the squeeze theorem (Theorem 3.3 in Apostol) to conclude

Therefore, by defining , we have extended to a function continuous at