MHB Integral Inequality: Prove $\left|f\left(\frac{1}{2}\right)\right|$ Bound

AI Thread Summary
The discussion centers on proving the inequality $\left|f\left(\frac{1}{2}\right)\right| \le \int_0^1 |f(t)|\, dt + \frac{1}{2}\int_0^1 |f'(t)|\, dt$ for a continuously differentiable function $f$ defined on the interval [0,1]. Participants analyze various approaches, including considering the minimum values of $f$ on subintervals and using Riemann sums to establish bounds. A key point made is that replacing $f$ with $-f$ allows for the assumption that $f\left(\frac{1}{2}\right) \geq 0$. The discussion highlights the mathematical rigor involved in deriving the inequality through integration and properties of differentiable functions. The thread concludes with acknowledgment of effective solutions presented by participants.
Euge
Gold Member
MHB
POTW Director
Messages
2,072
Reaction score
245
Here's my first challenge!

Let $f : [0,1] \to \Bbb R$ be continuously differentiable. Show that

$\displaystyle \left|f\left(\frac{1}{2}\right)\right| \le \int_0^1 |f(t)|\, dt + \frac{1}{2}\int_0^1 |f'(t)|\, dt$.
 
Mathematics news on Phys.org
Euge said:
Here's my first challenge!

Let $f : [0,1] \to \Bbb R$ be continuously differentiable. Show that

$\displaystyle \left|f\left(\frac{1}{2}\right)\right| \le \int_0^1 |f(t)|\, dt + \frac{1}{2}\int_0^1 |f'(t)|\, dt$.

Nice challenge! ;)

We have:
$$f(x)=\int_0^x f'(t)\,dt \le \int_0^x |f'(t)|\,dt$$
If $f(x)<0$, we have:
$$-f(x)=\int_0^x -f'(t)\,dt \le \int_0^x |f'(t)|\,dt$$
Therefore:
$$|f(x)| \le \int_0^x |f'(t)|\,dt$$
It follows that:
$$\left|f\left(\frac 12\right)\right| \le \int_0^{1/2} |f'(t)|\,dt$$
By symmetry (or substitution of $u=1-t$):
$$\left|f\left(\frac 12\right)\right| \le \int_{1/2}^1 |f'(t)|\,dt$$
Adding them up and dividing by 2 tells us that:
$$\left|f\left(\frac 12\right)\right| \le \frac 12\int_0^1 |f'(t)|\,dt$$
And since $\int_0^1 |f(t)|\,dt \ge 0$, the requested result follows.
 
I'm glad you liked the challenge, ILS! You have a very good attempt, but you started out with an error in the first line. If you take $f(x) = 1$, then

$\displaystyle \int_0^{1/2}f'(t)\, dt = 0 \neq 1 = f\left(\frac{1}{2}\right)$.
 
Euge said:
Here's my first challenge!

Let $f : [0,1] \to \Bbb R$ be continuously differentiable. Show that

$\displaystyle \left|f\left(\frac{1}{2}\right)\right| \le \int_0^1 |f(t)|\, dt + \frac{1}{2}\int_0^1 |f'(t)|\, dt$.
[sp]Replacing $f$ by $-f$ if necessary, we may assume that $f\bigl(\frac12\bigr) \geqslant0.$

Suppost that the minimum value of $f$ in the interval $\bigl[0,\frac12\bigr]$ occurs at $a$, and the minimum value of $f$ in the interval $\bigl[\frac12,1\bigr]$ occurs at $b$. Then $\frac12\bigl(f(a) + f(b)\bigr)$ is the lower Riemann sum for $$\int_0^1f(t)\,dt$$ corresponding to the dissection $\{0,\frac12,1\}$, so that $$\tfrac12\bigl(f(a) + f(b)\bigr) \leqslant \int_0^1f(t)\,dt.$$

Next, $$\int_{1/ 2}^bf'(t)\,dt - \int_a^{1/ 2}f'(t)\,dt = f(b) - f\bigl(\tfrac12\bigr) - f\bigl(\tfrac12\bigr) + f(a).$$ It follows that $$\begin{aligned} f\bigl(\tfrac12\bigr) &= \tfrac12\!\bigl(f(a) + f(b)\bigr) + \tfrac12\!\!\int_a^{1/ 2}f'(t)\,dt - \tfrac12\!\!\int_{1/ 2}^bf'(t)\,dt \\ &\leqslant \int_0^1f(t)\,dt + \tfrac12\!\!\int_a^b|f'(t)|\,dt \\ &\leqslant \int_0^1|f(t)|\,dt + \tfrac12\!\!\int_0^1|f'(t)|\,dt. \end{aligned}$$[/sp]
 
Opalg said:
[sp]Replacing $f$ by $-f$ if necessary, we may assume that $f\bigl(\frac12\bigr) \geqslant0.$

Suppost that the minimum value of $f$ in the interval $\bigl[0,\frac12\bigr]$ occurs at $a$, and the minimum value of $f$ in the interval $\bigl[\frac12,1\bigr]$ occurs at $b$. Then $\frac12\bigl(f(a) + f(b)\bigr)$ is the lower Riemann sum for $$\int_0^1f(t)\,dt$$ corresponding to the dissection $\{0,\frac12,1\}$, so that $$\tfrac12\bigl(f(a) + f(b)\bigr) \leqslant \int_0^1f(t)\,dt.$$

Next, $$\int_{1/ 2}^bf'(t)\,dt - \int_a^{1/ 2}f'(t)\,dt = f(b) - f\bigl(\tfrac12\bigr) - f\bigl(\tfrac12\bigr) + f(a).$$ It follows that $$\begin{aligned} f\bigl(\tfrac12\bigr) &= \tfrac12\!\bigl(f(a) + f(b)\bigr) + \tfrac12\!\!\int_a^{1/ 2}f'(t)\,dt - \tfrac12\!\!\int_{1/ 2}^bf'(t)\,dt \\ &\leqslant \int_0^1f(t)\,dt + \tfrac12\!\!\int_a^b|f'(t)|\,dt \\ &\leqslant \int_0^1|f(t)|\,dt + \tfrac12\!\!\int_0^1|f'(t)|\,dt. \end{aligned}$$[/sp]

This is an excellent solution. Thanks Opalg for participating!
 
I will show two of my solutions, both leading to the same equation that implies the result.

Solution 1.
By integration by parts,

$\displaystyle \int_0^1 f(t)\, dt = \int_0^{1/2} f(t)\, dt + \int_{1/2}^1 f(t)\, dt = \left(\frac{f\bigl(\tfrac{1}{2}\bigr)}{2} - \int_0^{1/2} tf'(t)\, dt\right) + \left(\frac{f\bigl(\tfrac1{2}\bigr)}{2} + \int_{1/2}^1 (1 - t)f'(t)\, dt\right) = f\bigl(\tfrac1{2}\bigr) - \int_0^{1/2} tf'(t)\, dt + \int_{1/2}^1 (1 - t)f'(t)\, dt.$

So,

$\displaystyle f\bigl(\tfrac1{2}\bigr) = \int_0^1 f(t)\, dt + \int_0^{1/2} tf'(t)\, dt - \int_{1/2}^1 (1 - t)f'(t)\, dt$,

and thus

$\displaystyle |f\bigl(\tfrac1{2}\bigr)| \le \int_0^1 |f(t)|\, dt + \frac{1}{2}\int_0^{1/2} |f'(t)|\, dt + \frac1{2}\int_{1/2}^1 |f'(t)|\, dt = \int_0^1 |f'(t)|\, dt + \frac{1}{2}\int_0^1 |f'(t)|\, dt.$

Solution 2.
For all $x \ge \tfrac1{2}$,

$\displaystyle f(x) = f\bigl(\tfrac1{2}\bigr) + \int_{1/2}^x f'(t)\, dt$.

Therefore

$\displaystyle \int_{1/2}^1 f(t)\, dt = \frac{f\bigl(\tfrac1{2}\bigr)}{2} + \int_{1/2}^1 \int_{1/2}^x f'(t)\, dt\, dx = \frac{f\bigl(\tfrac1{2}\bigr)}{2} + \int_{1/2}^1 \int_t^1 f'(t)\, dx\, dt$
$\displaystyle = \frac{f\bigl(\tfrac1{2}\bigr)}{2} + \int_{1/2}^1 (1 - t)f'(t)\, dt.$

Now we can write

$\displaystyle \frac{f\bigl(\tfrac1{2}\bigr)}{2} = \int_{1/2}^1 f(t)\, dt - \int_{1/2}^1 (1 - t)f'(t)\, dt.$

By a symmetric argument,

$\displaystyle \frac{f\bigl(\tfrac1{2}\bigr)}{2} = \int_0^{1/2} f(t)\, dt + \int_0^1 tf'(t)\, dt$.

Adding the latter two equations,

$\displaystyle f\bigl(\tfrac1{2}\bigr) = \int_0^1 f(t)\, dt + \int_0^{1/2} tf'(t)\, dt - \int_{1/2}^1 (1 - t)f'(t)\, dt.$

This leads to the result in exactly the same way as Solution 1.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top