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

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Here's my first challenge!

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

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

 

[I like Serena]

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! ;)

Spoiler

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.

 

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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

$\displaystyle \int_0^{1/2}f'(t)\, dt = 0 \neq 1 = f\left(\frac{1}{2}\right)$.

 

[Opalg]

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 \(\displaystyle \int_0^1f(t)\,dt\) corresponding to the dissection $\{0,\frac12,1\}$, so that \(\displaystyle \tfrac12\bigl(f(a) + f(b)\bigr) \leqslant \int_0^1f(t)\,dt.\)

Next, \(\displaystyle \int_{1/ 2}^bf'(t)\,dt - \int_a^{1/ 2}f'(t)\,dt = f(b) - f\bigl(\tfrac12\bigr) - f\bigl(\tfrac12\bigr) + f(a).\) 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]

 

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[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 \(\displaystyle \int_0^1f(t)\,dt\) corresponding to the dissection $\{0,\frac12,1\}$, so that \(\displaystyle \tfrac12\bigl(f(a) + f(b)\bigr) \leqslant \int_0^1f(t)\,dt.\)

Next, \(\displaystyle \int_{1/ 2}^bf'(t)\,dt - \int_a^{1/ 2}f'(t)\,dt = f(b) - f\bigl(\tfrac12\bigr) - f\bigl(\tfrac12\bigr) + f(a).\) 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!

 

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I will show two of my solutions, both leading to the same equation that implies the result.

Solution 1.

Spoiler

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.

Spoiler

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.