An inequality of $\int_0^1 |f(x)|dx$ 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.
 A: 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.
A: 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 $M-m$.
A: 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.
A: 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:
$$\int_0^1 |f'(x)|dx=\sum_{i \in I}|f(d_{i+1})-f(d_i)| \geq |f(d_q)-f(d_j)| \gt |f(d_q)|= $$
$$= f(x_0)= \max_{x \in [0,1]}{|f(x)|} \geq \int_{0}^{1} | f(x)|dx $$
$\textbf{Q.E.D.}$
