Lower bound of an integral How might I find a constant $k>0$ s.t. $\int\limits_0^1|f(x)|dx\geq k\max\{|f(x)|:x\in[0,1]\}$ for all continuous $f$ defined on $[0,1]$? Thank you.
 A: For any positive integer $n$, let $f_n(x)=x^n$.  On $[0,1]$, the maximum of $f_n(x)$ is $1$, and $\int_0^1 f_n(x)\,dx=\frac{1}{n+1}$.  
Thus any $k$ with the desired property must be $\le \frac{1}{n+1}$ for all $n$, so cannot be positive. 
A: Posting my comment as an answer. It gives a different example that is useful to keep in mind as well. 
Fix an arbitrary $\epsilon > 0$ and imagine a "triangular spike" function with height of the triangle being $1$ and base $2\epsilon$. Precisely, I mean the function
$$
f(x) = 
\begin{cases}
\frac{x}{\epsilon}, &0\leq x \leq \epsilon,
\\
2 - \frac{x}{\epsilon}, &\epsilon \leq x \leq 2\epsilon,
\\
0, &2\epsilon \leq x \leq 1.
\end{cases}
$$
(Exercise: Verify that $f$ is continuous.) For this function, $\int_{0}^1 |f(x)| dx$ is just the area of the triangle, i.e., $\frac12 \cdot 1 \cdot 2\epsilon = \epsilon$. On the other hand, $\| f \|_{\infty} := \sup \{ |f(x)| \mid 0 \leq x \leq 1 \}$ is $1$. Hence, the quantity
$$
\frac{\int_{0}^1 |f(x)| dx}{\| f \|_{\infty}} 
$$ can be arbitrarily close to $0$, and so there does not exist any $k > 0$ satisfying your requirements. 
