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If $f\in C^1[0,1]$, show that

$$\int_0^1 |f(x)|dx\le\max\left\{\int_0^1 |f'(x)|\,dx,\;\bigg|\int_0^1 f(x)\,dx\bigg|\right\}.$$

I have tried to make connection between $|f|$ and $|f'|$ by using


integrate the equation with $t$ on $[0,1]$, it gets

$$|f(1)|\leq \bigg|\int_0^1 f(t)\,dt\bigg|+\int_0^1 |f'(t)|\,dt,$$

but it is not the desired inequality.

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up vote 6 down vote accepted

If $f$ is non-negative or non-positive, then

$$\int_0^1 |f(x)| \; dx \;\; = \;\; \left|\int_0^1 f(x) \; dx\right| \;\; \leq \;\; \operatorname{max}\left(\int_0^1 |f'(x)| \; dx , \left|\int_0^1 f(x) \; dx\right|\right).$$ Otherwise, by the intermediate value theorem, let $x_0$ be a member of $[0,1]$ such that $f(x_0) = 0$ .
By the extreme value theorem, let $x_m$ be a member of $[0,1]$ such that for all members $x$ of $[0,1]$, $|f(x)| \leq |f(x_m)|$ .

$\begin{align} \int_0^1 |f(x)| dx &\leq \int_0^1 |f(x_m)| dx \\ &= |f(x_m)| \\ &= |f(x_m)-f(x_0)|\\ &= \left|\int_{x_0}^{x_m} f'(x) dx\right| \\ &= \int_{x_0}^{x_m} |f'(x)| dx \\ & \leq \int_0^1 |f'(x)| dx \\ &\leq \operatorname{max}\left(\int_0^1 |f'(x)| dx , \left|\int_0^1 f(x) dx\right|\right). \end{align}$


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If $f$ has constant sign, then $\int_0^1|f(x)|dx=|\int_0^1f(x)dx|$.

If $f$ does not have constant sign, then there are points in $[0,1]$ where $f$ takes its positive maximum value $M$ and its negative minimum value $m$. You can show that the integral of $|f|$ is less than the max of $M$ and $-m$, and by using the fundamental theorem of calculus, you can show that the integral of $|f'|$ is at least $M-m$.

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Let's see what the various integrals measure, starting with $\int_0^1 |f(x)| \, dx$. Imagine the function $f$ on the interval $[0,1]$, split into the parts where it's positive, and the parts where it's negative. The total (unsigned) area of all the parts is exactly $\int_0^1 |f(x)| \, dx$.

Next, $\int_0^1 |f'(x)| \, dx$. We know that $\int_0^1 f'(x) \, dx = f(1) - f(0)$, since $f'$ measures the amount by which $f$ is increasing. Similarly, $|f'|$ measures the amount by which $f$ is changing. Divide $f$ into parts where it's increasing and parts where it's decreasing, and reflect each decreasing part starting at $x$ along $y = f(x)$. Now $\int_0^1 |f'(x)| \, dx$ measure how far the modified function reaches. Suppose now that $f(0) = 0$. In that case, we have the easy bound $|f| \leq \int_0^1 |f'(x)| \, dx$ on the interval $[0,1]$. Since $[0,1]$ has unit length, $\int_0^1 |f(x)| \, dx \leq \max_{x \in [0,1]} |f(x)|$, so in case $f(0) = 0$ we're done.

If $f(0) \neq 0$ then the bound $\int_0^1 |f'(x)| \, dx$ can be terribly wrong. The extreme situation is when $f$ is constant, and then the latter integral is equal to zero. So we need to consider the final integral $\left| \int_0^1 f(x) \, dx \right|$. Recall we split $f$ into positive and negative parts. Add their signed areas and return the magnitude to get $\left| \int_0^1 f(x) \, dx \right|$. In general we may get large cancellation this way - indeed, a simple symmetric construction (like the sine function) has $\left| \int_0^1 f(x) \, dx \right| = 0$ while $\int_0^1 |f(x)| \, dx$ can be arbitrarily large. However, if $f$ never changes sign, then it's easy to see that the two integrals are equal.

So we divide into two cases. If $f$ never changes sign then $\int_0^1 |f(x)| \, dx = \left| \int_0^1 f(x) \, dx \right|$. Otherwise, assume without loss of generality that $f(0) > 0$. The problem with $\int_0^1 |f'(x)| \, dx$ was that in its estimation of $\max_{x\in [0,1]} |f(x)|$ it was missing the contribution of $f(0)$. However, since $f$ crosses zero, there is somewhere a contribution of exactly $f(0)$, which does not contribute to $\max_{x\in [0,1]} |f(x)|$ (indeed, $f$ is rather diminishing its magnitude). So in that case it is again true that $\max_{x\in [0,1]} |f(x)| \leq \int_0^1 |f'(x)| \, dx$, and we're done.

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In the second sentence, is "Image" a typo? – Ricky Demer Aug 26 '11 at 9:48
Yep, corrected. Thanks. – Yuval Filmus Aug 26 '11 at 12:20

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