Prove $\sup_{0\le x\le 1}|f(x)|\le\int_0^1(|f(t)|+|f'(t)|)dt$ Let $f\in C^1([0,1])$. Prove the following:
$$\sup_{0\le x\le 1}|f(x)|\le\int_0^1(|f(t)|+|f'(t)|)dt$$
and
$$|f(1/2)|\le\int_0^1(|f(t)|+\frac12|f'(t)|)dt$$

Note that $e^{-x}(e^xf(x))'=f(x)+f'(x)$. So $\int_0^1(|f(t)|+|f'(t)|)dt\ge|\int_0^1(f(t)+f'(t))dt|\ge\int_0^1(e^tf(t))'dt=ef(1)-f(0)$
 A: The first inequality to be proved is $\sup_{0\leqslant x\leqslant 1}|f(x)|\leqslant\int_0^1(|f(x)|+|f'(x)|)\,\mathrm dx$, or
$$\sup_{0\leqslant x\leqslant 1}|f(x)|-\int_0^1|f(x)|\,\mathrm dx\leqslant\int_0^1|f'(x)|\,\mathrm dx.$$
This statement is essentially unchanged if $f$ is replaced by $-f$, and we do this just in the case when $\sup_{0\leqslant x\leqslant 1}|f(x)|= \sup_{0\leqslant x\leqslant 1}-f(x)$; otherwise, when
$$\sup_{0\leqslant x\leqslant 1}|f(x)|= \sup_{0\leqslant x\leqslant 1}f(x),$$ we keep the original $f$. Thus, either way, we may assume the above equality. Since $\int_0^1f(x)\,\mathrm dx\leqslant\int_0^1|f(x)|\,\mathrm dx$, it is therefore enough to show that
$$\sup_{0\leqslant x\leqslant 1}f(x)-\int_0^1f(x)\,\mathrm dx\leqslant\int_0^1|f'(x)|\,\mathrm dx.$$
Now observe that each side of this inequality is unchanged if the value of $f$ is shifted  by a constant, and so we reduce $f$ by its average value over the range of integration. Hence it is sufficient to prove that
$$\sup_{0\leqslant x\leqslant 1}f(x)\leqslant\int_0^1|f'(x)|\,\mathrm dx,$$ where $f$ satisfies the condition
$$\int_0^1f(x)\,\mathrm dx=0.$$
$\qquad$From this last condition, the function $f$, by its continuity, is either identically zero—a trivial case, which we will ignore—or takes both positive and negative values. Consequently there is a point $b$ in the range at which $f$ is positive and maximal:
$$\sup_{0\leqslant x\leqslant 1}f(x)=f(b).$$ Again by the continuity of $f$, there is an interval containing $b$ with an end point  $a$ such that $f(a)=0$ and $f(x)>0$ for either (A) $a<x<b$ or (B) $b<x<a$. In case A, because $\int_0^1f'(x)\,\mathrm dx\leqslant\int_0^1|f'(x)|\,\mathrm dx$, it follows that
$$\int_0^1|f'(x)|\,\mathrm dx\geqslant\int_a^b|f'(x)|\,\mathrm dx\geqslant\int_a^bf'(x)\,\mathrm dx=f(b)-f(a)=f(b),$$ which is the required result. In case B, the result follows similarly by interchanging $a$ and $b$ and replacing $f'$ by $-f'$.$$***$$
To show the second inequality, we proceed initially as in the first paragraph above, replacing $\sup_{0\leqslant x\leqslant 1}|f(x)|$ by $f(\frac12)$ and $|f'(x)|$ by $\frac12|f'(x)|$, and conclude that it is sufficient to prove the following inequality:
$$2f(\tfrac12)\leqslant\int_0^1|f'(x)|\,\mathrm dx,$$where (as before) $f$ satisfies the condition
$$\int_0^1f(x)\,\mathrm dx=0.$$ Now, arguing as above by continuity of $f$, there are $a$ and $b$, with $0\leqslant a<\frac12<b\leqslant1$, such that $f(a)=f(b)=0$. Then
$$\int_0^1|f'(x)|\,\mathrm dx\geqslant\int_a^{1/2}|f'(x)|\,\mathrm dx+\int_{1/2}^b|f'(x)|\,\mathrm dx$$
$$\geqslant\int_a^{1/2}f'(x)\,\mathrm dx-\int_{1/2}^bf'(x)\,\mathrm dx$$
$$=f(\tfrac12)-f(a)-f(b)+f(\tfrac12)$$
$$=2f(\tfrac12),$$ as required.
