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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.

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No such constant $k > 0$ exists. Imagine a "triangular spike" of height $1$ and base $2\epsilon$; The integral of this function is $\epsilon$. Since $\epsilon > 0$ can be picked arbitrarily small, the left hand side can be arbitrarily small compared to the right. – Srivatsan Nov 9 '11 at 11:37
I don't think that's possible. ''smooth out'' the functions $f_n(x)=\cases{n, &0<x<1/n\cr 0&x\ge 1/n}$ – David Mitra Nov 9 '11 at 11:37
Thank you, guys! – max Nov 9 '11 at 13:38
The maximum is a local property, while the area under the curve is a global property. As indicated in responses and comments, those two characterisations of a continuous integrable function are unrelated. – Xi'an Nov 9 '11 at 15:43
up vote 1 down vote accepted

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.

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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.

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