TITLE: Bounding the absolute value of a function with an integral
QUESTION [6 upvotes]: I am having trouble with the following problem in analysis: 
Suppose that $f, f^\prime \in C([0, 1])$. Prove that for all $x \in [0, 1]$
$$
|f(x)| \leq \int_0^1 (|f(t)| + |f^\prime (t)|) dt.
$$
Any pointers? I have tried writing this as a Riemann Sum (given arbitrary tagged partition) but am still not sure how to proceed.

REPLY [10 votes]: $\int_{0}^1|f(t)| dt$ is the average of $|f(t)|$ in the interval, and $\int_{0}^1|f'(t)|dt$ is the total variation of $f(t)$, if you think about this, it makes sense. By MVT, you have that $$\int_{0}^1|f(t)| dt=|f(a)|$$ for some $a\in[0,1]$. Let $x\in[0,1]$, WLOG say $x>a$, then \begin{align*}|f(x)|&\leq |f(a)|+|f(x)-f(a)|\\
&=\int_{0}^1|f(t)|dt+|\int_{a}^xf'(t)dt|\\
&\leq \int_{0}^1|f(t)|dt+\int_{a}^x|f'(t)|dt\\
&\leq\int_{0}^1|f(t)|dt+\int_{0}^1|f'(t)|dt  
\end{align*}