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Question:

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

(tf(t))'=f(t)+tf'(t),

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.

4 Answers 4

6

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}

QED.

2

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 $M-m$.

2

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.

  • 0
    Yep, corrected. Thanks.2011-08-26
0

If $f$ doesn't change sign, the thesis follows, so we can assume that $f$ change sign.

$\int_{0}^{1} | f(x)|dx \leq \max_{x \in [0,1]}{|f(x)|}$

$f \in \textbf{$C^1$}$ so let $x_0$ be the point in $[0,1]$ such that the above maximum is reached, then let's consider the following set: $D=\left\{ x \in [0,1] \text{ such that $f'(x)=0$} \right\} \cup \left\{ 0,1 \right\}$ .

$\textbf{W.L.O.G. we can assume $f(x_0) \gt 0$}$

Let's order the elements of $D$ in increasing order, such that we have $d_{i} \lt d_{i+1} $ and let $I$ be the set of indexes.

Using the fundamental theorem of calculus and the definition of absolute value and studying a bit how $f'$ change sign we can say that:

$\int_0^1 |f'(x)|dx=\sum_{i \in I}|f(d_{i+1})-f(d_i)|$

$f \in \textbf{$C^1$}$ so $x_0$ is actually one of the $d_i$ ,let's say $x_0=d_q$, we have also assumed that $f$ change sign, so one of the $d_i$ is such that $f(d_i) \lt 0$,let's say $d_j$ is such that $f(d_j) \lt 0$.

So we can use the triangle inequality to say:

$\int_0^1 |f'(x)|dx=\sum_{i \in I}|f(d_{i+1})-f(d_i)| \geq |f(d_q)-f(d_j)| \gt |f(d_q)|= $

$= f(x_0)= \max_{x \in [0,1]}{|f(x)|} \geq \int_{0}^{1} | f(x)|dx $

$\textbf{Q.E.D.}$