Tag: Continuity

Find continuous functions satisfying given conditions

by RoRi

December 17, 2015

Find continuous functions which satisfy the given conditions for all $x \in \mathbb{R}$ .

$\displaystyle{ \int_0^x f(t) \, dt = e^x}$ .
$\displaystyle{ \int_0^{x^2} f(t) \, dt = 1 - 2^{x^2}}$ .
$\displaystyle{ \int_0^x f(t) \, dt = f^2(x) - 1}$ .

No such function can exist since for $x = 0$ we have
$\int_0^x f(t) \, dt = \int_0^0 f(t) \, dt = 0 \neq e^0 = 1.$
Taking derivatives of both sides of the given equation we have
$\begin{align*} &&\frac{d}{dx} \left(\int_0^{x^2} f(t) \, dt\right) &= \frac{d}{dx} \left( 1 - 2^{(x^2)} \right) \\[9pt] \implies && (2x)(f(x^2)) &= \frac{d}{dx} \left(1 - e^{x^2 \log 2} \right) \\[9pt] \implies && (2x)(f(x^2)) &= -(2x \log 2) e^{x^2 \log 2} \\[9pt] \implies && f(x^2) &= -(\log 2)(2^{(x^2)}) \\[9pt] \implies && f(x) &= - 2^x \log 2. \end{align*}$
Again, taking derivatives of both sides we have
$\begin{align*} &&\frac{d}{dx} \left( \int_0^x f(t) \, dt \right) &= \frac{d}{dx} \left( f^2(x) - 1 \right) \\[9pt] \impies && f(x) &= 2f(x)f'(x) \\[9pt] \implies && 1 &= 2f'(x) \end{align*}$

at all points $x$ such that $f(x) \neq 0$ . (Since $\int_0^x f(t) \, dt = f^2(x) - 1$ is not satisfied by the zero function $f(x) = 0$ , we know there are real $x$ such that $f(x) \neq 0$ .) Then, integrating

$f'(x) = \frac{1}{2} \quad \implies \quad f(x) = \frac{1}{2} x + C.$

Now, we can solve for $C$ by evaluating the given identity at $x = 0$ ,

$\begin{align*} && \int_0^x f(t) \, dt &= f^2 (x) - 1 \\[9pt] \implies && \int_0^0 f(t) \, dt &= f^2 (0) - 1 \\[9pt] \implies && 0 &= \left( \frac{0}{2} + C\right)^2 - 1\\[9pt] \implies && 0 &= C^2 - 1 \\[9pt] \implies && C &= \pm 1. \end{align*}$

Therefore, we have

$f(x) = \frac{1}{2} x \pm 1.$

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

by RoRi

October 16, 2015

Let $f$ be a function such that the integral

$A(x) = \int_a^x f(t) \, dt$

exists for all $x$ in an interval $[a,b]$ . Consider the following statements:

a. $f$ is continuous at $c$ .
b. $f$ is discontinuous at $c$ .
c. $f$ is increasing on $(a,b)$ .
d. $f'(c)$ exists.
e. $f'$ is continuous at $c$ .

$\alpha$ . $A$ is continuous at $c$ .
$\beta$ . $A$ is discontinuous at $c$ .
$\gamma$ . $A$ is convex on $(a,b)$ .
$\delta$ . $A'(c)$ exists.
$\varepsilon$ . $A'$ is continuous at $c$ .

Make a table to indicate by T whether the statement in row $(a)-(e)$ implies the statement in column $(\alpha) - (\varepsilon)$ . Leave the cell in the table blank if there is no implication.

The requested table is as follows:

$\begin{tabular}{c | c c c c c} & $\alpha$ & $\beta$ & $\gamma$ & $\delta$ & $\varepsilon$ \\ \hline a & T& & & T& \\ b & T& & & & \\ c & T& & T& & \\ d & T& & & T& \\ e & T& & & T&T \end{tabular}$

We know the first column has all T’s since the function $A(x) = \int_a^x f(t) \, dt$ is continuous for any function $f$ by the first fundamental theorem we know the derivative $A'(x)$ exists at every point in the open interval $(a,b)$ . Since differentiability implies continuity (by example 7 on page 163 of Apostol), we then know that $A(x)$ is continuous at any point $c \in (a,b)$ .

We know the second column can have no T’s anywhere since, by the same argument as above, the function $A(x)$ is continuous at every point $c \in (a,b)$ . Thus, it cannot be discontinuous at any point $c \in (a,b)$ .

The statement a,b,d, and e cannot imply that $A(x)$ is convex on $(a,b)$ 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 $(\gamma)$ since we know that a function is convex if its derivative is increasing. Since $f$ is the derivative of $A$ and it is increasing, we have that $A$ is convex.

Prove the intermediate value theorem for derivatives

by RoRi

September 28, 2015

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

Assume $f$ is differentiable on an open interval $I$ . Let $a < b$ be two points in $I$ . Then, the derivative $f'$ takes every value between $f'(a)$ and $f'(b)$ somewhere in $(a,b)$ .

Define a function
$g(x) = \frac{f(x) - f(a)}{x-a} \qquad \text{if } x \neq a, \qquad g(a) = f'(a).$

Prove that $g$ takes every value between $f'(a)$ and $f'(b)$ in the interval $(a,b)$ . Then, use the mean-value theorem for derivatives to show $f'$ takes all values between $f'(a)$ and $g(b)$ somewhere in the interval $(a,b)$ .
Define a function
$h(x) = \frac{f(x) - f(b)}{x-b}, \qquad \text{if } x \neq b, \qquad h(b) = f'(b).$

Show that the derivative $f'$ takes on all values between $f'(b)$ and $h(a)$ in the interval $(a,b)$ . Conclude that the statement of the intermediate-value theorem is true.

Proof. First, since $f$ is differentiable everywhere on the interval $I$ , we know $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ . Thus, if $x \in [a,b]$ and $x \neq a$ then $g$ is continuous at $x$ since it is the quotient of continuous functions and the denominator is nonzero. If $x = a$ then
$\lim_{x \to a} g(x) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a} = f'(a) = g(a);$

hence, $g$ is continuous at $x = a$ as well. Therefore, $g$ is continuous on the closed interval $[a,b]$ . So, by the intermediate value theorem for continuous functions we know $g$ takes on every value between $g(a)$ and $g(b)$ somewhere on the interval $(a,b)$ . Since $g(a) = f'(a)$ this means $g$ takes on every value between $f'(a)$ and $g(b)$ somewhere on the interval $(a,b)$ .
By the mean-value theorem for derivatives, we then know there exists some $c \in (a,b)$ such that

$\begin{align*} &&f(x) - f(a) &= f'(c)(x-a) \\ \implies && f'(c) &= \frac{f(x) - f(a)}{x-a} \\ \implies && f'(c) &= g(x) \end{align*}$

for some $c \in (a,x)$ . Since $x < b$ , we then conclude there is some $c \in (a,b)$ such that $f'(c) = g(x)$ for any $x$ . Since $g$ takes on every value between $f'(a)$ and $g(b)$ , so does $f'$ .
Proof. This is very similar to part (a). By the same argument we have the function $h$ is continuous on $[a,b]$ ; thus, $h$ takes on every value between $h(a)$ and $f'(b)$ by the intermediate value theorem for continuous functions. Then, by the mean value theorem, we know there exists a $c \in (x, b)$ such that
$f'(c) = \frac{f(x) - f(b)}{x-b} = h(x) \qquad \text{for some } c \in (x,b).$

Thus, $f'$ takes on every value between $f'(b)$ and $h(a)$ . Since $h(a) = g(b)$ ; $f'$ takes on every value between $f'(a)$ and $f'(b). \qquad \blacksquare$

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

by RoRi

September 11, 2015

Let $f$ be a continuous function on the interval $[a,b]$ and assume

$\int_a^b f(x) g(x) \, dx = 0$

for every function $g$ which is continuous on the interval $[a,b]$ . Prove that $f(x) = 0$ for all $x \in [a,b]$ .

Proof. Since $\int_a^b f(x) g(x) \, dx = 0$ must hold for every function $g$ that is continuous on $[a,b]$ , it must hold for $f$ itself (since $f$ is continuous on $[a,b]$ by hypothesis). Therefore we must have,

$\int_a^b f(x) f(x) \, dx = \int_a^b (f(x))^2 \, dx = 0$

However, $(f(x))^2 \geq 0$ for all $x \in [a,b]$ . 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 $f$ is continuous at every point of $[a,b]$ ; hence, $f^2$ is also continuous at every point of $[a,b]$ (since the product of continuous functions is continuous). Therefore,

$(f(x))^2 = 0 \qquad \text{for all } x \in [a,b].$

Which implies,

$f(x) = 0 \qquad \text{for all } x \in [a,b]. \qquad \blacksquare$

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

by RoRi

September 11, 2015

Let $f$ be an integrable, non-negative function on the interval $[a,b]$ . Prove that if

$\int_a^b f(x) \, dx = 0$

then $f(x) = 0$ at each point $x \in [a,b]$ at which $f$ is continuous.

Proof. The proof is by contradiction. Let $p \in [a,b]$ be a point at which $f$ is continuous, and suppose $f(p) \neq 0$ . This implies $f(p) > 0$ since $f$ is non-negative by hypothesis. Since $f$ is continuous at $p$ , we know by the sign-preserving property of continuous functions (Theorem 3.7 in Apostol), that there is some neighborhood about $p$ , say $(p - \delta, p + \delta)$ such that $f(x)$ has the same sign as $f(p)$ for all $x \in (p - \delta, p + \delta)$ , i.e., such that $f(x) > 0$ for all $x \in (p - \delta, p + \delta)$ . But then by the monotone property of the integral we know,

$\int_{p - \delta}^{p + \delta} f(x) \, dx > \int_{p - \delta}^{p + \delta} 0 \,dx = 0.$

But then,

$\int_a^b f(x) \,dx = \int_a^{p - \delta} f(x)\, dx + \int_{p - \delta}^{p + \delta} f(x) \, dx + \int_{p + \delta}^b f(x) \, dx = 0.$

This is a contradiction since $f$ is nonnegative we know

$\int_a^{p - \delta} f(x) \, dx \geq 0 \qquad \text{and} \qquad \int_{p + \delta}^b f(x) \, dx \geq 0.$

Since $\int_{p - \delta}^{p + \delta} f(x) \, dx$ is strictly positive, the sum of the three pieces of the integral $\int_a^b f(x) \, dx$ cannot equal $0. \qquad \blacksquare$

Find points at which a given composite function is continuous

by RoRi

September 4, 2015

Define functions $f$ and $g$ as follows:

$f(x) = \frac{x+|x|}{2}, \text{ for all } x, \qquad g(x) = \begin{cases} x & \text{for } x < 0,\\ x^2 &\text{for } x \geq 0. \end{cases}$

Find a formula for $h(x) = g(f(x))$ . Find the values at which $h(x)$ is continuous.

If $x < 0$ , we have $|x| = -x$ and we compute the composition,

$h(x) = g(f(x)) = g \left( \frac{x - x}{2} \right) = g(0) = 0.$

If $x \geq 0$ , then $|x| = x$ and we compute the composition,

$h(x) = g(f(x)) = g \left( \frac{x+x}{2} \right) = g(x) = x^2.$

Hence,

$h(x) = \begin{cases} 0 & \text{if } x < 0, \\ x^2 & \text{if } x \geq 0. \end{cases}$

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

by RoRi

September 3, 2015

Define functions $f$ and $g$ as follows:

$f(x) = \begin{cases} 1 & \text{if } |x| \leq 1 \\ 0 & \text{if } |x| > 1. \end{cases}, \qquad g(x) = \begin{cases} 2-x^2 & \text{if } |x| \leq 2 \\ 2 & \text{if } |x| > 2. \end{cases}$

Find a formula for $h(x) = f(g(x))$ and find the values at which $h(x)$ is continuous.

If $|x| > 2$ then $g(x) = 2$ and so $|g(x)| > 1$ . Therefore, by the definition of $f$ ,

$h(x) = f(g(x)) = 0.$

If $\sqrt{3} < x \leq 2$ , then $|g(x)| = |2 - x^2| > 1$ and so again

$h(x) = f(g(x)) = 0.$

Next, if $1 \leq x \leq \sqrt{3}$ then $|g(x)| = | 2 - x^2| \leq 1$ and so

$h(x) = f(g(x)) = 1.$

Finally, if $|x| < 1$ , then $|g(x)| = |2 - x^2| > 1$ and we have

$h(x) = f(g(x)) = 0.$

Putting this all together we have

$h(x) = \begin{cases} 1 & \text{if } 1 \leq |x| \leq \sqrt{3} \\ 0 & \text{otherwise}. \end{cases}$

From this expression we have $h(x)$ is continuous everywhere except at $|x| = 1$ and $|x| = \sqrt{3}$ .

Find points at which a given composite function is continuous

by RoRi

September 3, 2015

Define functions $f$ and $g$ as follows:

$f(x) = \frac{x+|x|}{2}, \text{ for all } x, \qquad g(x) = \begin{cases} x & \text{for } x < 0,\\ x^2 &\text{for } x \geq 0. \end{cases}$

Find a formula for $h(x) = f(g(x))$ . Find the values at which $h(x)$ is continuous.

If $x < 0$ , we have $|x| = -x$ and we compute the composition,

$h(x) = f(g(x)) = f(x) = \frac{x-x}{2} = 0.$

If $x \geq 0$ , then $|x| = x$ and we compute the composition,

$h(x) = f(g(x)) = f(x^2) = \frac{2x^2}{2} = x^2.$

Putting these together,

$h(x) = \begin{cases} 0 & \text{if } x < 0, \\ x^2 & \text{if } x \geq 0. \end{cases}$

Certainly $h$ is continuous for all $x \neq 0$ since the constant function $h(x) =0$ and the polynomial $h(x) = x^2$ are continuous. Further, $h$ is continuous at $x = 0$ since

$\lim_{x \to 0^-} h(x) = \lim_{x \to 0^+} h(x) = \lim_{x \to 0} h(x) = 0 = h(0).$

Prove some continuity properties of a function

by RoRi

August 31, 2015

Given a function $f$ and a closed interval $[a,b]$ such that

$|f(u) - f(v)| \leq |u-v|$

for all $u,v \in [a,b]$ .

Prove $f$ is continuous at every point $x \in [a,b]$ .
If $f$ is integrable on $[a,b]$ prove
$\left| \int_a^b f(x) \, dx - (b-a)f(a) \right| \leq \frac{(b-a)^2}{2}.$
If $c \in [a,b]$ is any point prove
$\left| \int_a^b f(x) \, dx - (b-a) f(c) \right| \leq \frac{(b-a)^2}{2}.$

Proof. Let $p$ be any point in $[a,b]$ . Then, for any $\varepsilon > 0$ let $\delta = \varepsilon$ . Then
$\begin{align*} |x-p| < \delta &&\implies && |x-p| &< \varepsilon \\ && \implies && |f(x) - f(p)| &< \varepsilon. \end{align*}$

Thus, for any $\varepsilon > 0$ , we have $|f(x) - f(p)| < \varepsilon$ whenever $|x-p| < \delta$ . Hence,

$\lim_{x \to p} f(x) = f(p).$

Therefore, $f$ is continuous at every point $p \in [a,b]. \qquad \blacksquare$
Proof. Since $f(a)$ is a constant (the value of the function $f$ evaluated at the constant $a$ ), we have
$(b-a) f(a) = \int_a^b f(a) \, dx.$

Therefore, we can compute,

$\begin{align*} \left| \int_a^b f(x) \, dx - (b-a) f(a) \right| &= \left| \int_a^b f(x) \, dx - \int_a^b f(a) \, dx \right| \\ &= \left| \int_a^b (f(x) - f(a)) \, dx \right|\\ &\leq \int_a^b |f(x) - f(a)| \, dx. \end{align*}$

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

$\begin{align*} - |f(x)| \leq f(x) \leq |f(x)| && \implies && \int_a^b -|f(x)| \, dx \leq \int_a^b f(x) \, dx \leq \int_a^b |f(x)| \, dx \\ && \implies && - \int_a^b |f(x)| \, dx \leq \int_a^b f(x) \, dx \leq \int_a^b |f(x)| \, dx \\ && \implies && \left| \int_a^b f(x) \, dx \right| \leq \int_a^b |f(x)| \, dx. \end{align*}$

Then, since $|f(x) - f(a)| \leq |x-a|$ by our hypothesis on $f$ , we have

$\begin{align*} \int_a^b |f(x) - f(a)| \, dx &\leq \int_a^b |x-a| \, dx \\ &= \int_a^b (x-a) \, dx &(x-a\geq 0 \text{ since } x \in [a,b]) \\ &= \frac{b^2}{2} - \frac{a^2}{2} - ab + a^2 \\ &= \frac{b^2-2ab+a^2}{2} \\ & = \frac{(b-a)^2}{2}. \qquad \blacksquare \end{align*}$
Proof. The proof proceeds similarly to that of part (b),
$\begin{align*} \left| \int_a^b f(x) \, dx - (b-a) f(c) \right| &= \left| \int_a^b f(x) \, dx - \int_a^b f(c) \, dx \right|\\ &= \left| \int_a^b (f(x) - f(c)) \, dx \right| \\ &\leq \int_a^b | f(x) - f(c) | \, dx \\ &\leq \int_a^b |x-c| \, dx. \end{align*}$

Then, since $c \in [a,b]$ , we break the integral into two pieces (to deal with the fact that $|x-c| = -(x-c)$ for $x < c$ and $|x-c| = x-c$ for $x \geq c$ ):

$\begin{align*} \int_a^b |x-c| \, dx &= \int_a^c -(x-c) \, dx + \int_c^b (x-c) \, dx \\ &= \left. -\frac{x^2}{2} \right|_a^c + c \cdot x \biggr \rvert_a^c + \left. \frac{x^2}{2} \right|_c^b - c \cdot x \biggr \rvert_c^b \\ &= \frac{a^2-c^2}{2} + c(c-a) + \frac{b^2-c^2}{2} - c(b-c) \\ &= \frac{a^2 - 2c^2 + b^2}{2} + 2c^2-ac-bc \\ &= \frac{a^2+b^2}{2} + c^2 - c(b+a). \end{align*}$

But now, we know $c \in [a,b]$ implies $c \leq b$ , so we have the inequality,

$\begin{align*} \frac{a^2+b^2}{2} + c^2 - c(b+a) &= \frac{a^2+b^2}{2} + c(c-(b+a)) \\ &\leq \frac{b^2+a^2}{2} + b(b-b-a) \\ &= \frac{b^2+a^2}{2} - ab \\ &= \frac{b^2 - 2ab + a^2}{2} \\ &= \frac{(b-a)^2}{2}. \qquad \blacksquare \end{align*}$