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I am looking for a real valued continuous function $f$ on an interval $[-1,1]$ with the property that $\max_{[-1,1]} |f| = 1$ and $\left |\int_{-1}^{0} f - \int_{0}^{1} f \right|= 2$

Polynomials, trig functions, and exponential functions seem to all fail. I am out of ideas

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What makes you think that such a function exists? –  Calvin Lin Mar 16 '13 at 1:12
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3 Answers 3

Such a function $f$ cannot exist. Here is the reason.

Since $|f|\leq 1$, we have $|\int_{-1}^{0} f|\leq \int_{-1}^0|f|\leq 1$. From this we see that $\int_{-1}^{0} f=1$ when $f\equiv 1$ on $[-1,0]$, and $\int_{-1}^{0} f=-1$ when $f\equiv -1$ on $[-1,0]$. Here we use the fact that $f$ is continuous.

Similarly, we have $|\int_{0}^{1} f|\leq \int_{0}^1|f|\leq 1$. From this we see that $\int_{0}^{1} f=1$ when $f\equiv 1$ on $[0,1]$, and $\int_{0}^{1} f=-1$ when $f\equiv -1$ on $[0,1]$.

Therefore, if $|\int_{-1}^{0} f - \int_{0}^{1} f |= 2$, we must have:

Case (i). $\int_{-1}^{0} f=1\mbox{ and }\int_{0}^{1} f=-1$, or,

Case (ii). $\int_{-1}^{0} f=-1\mbox{ and }\int_{0}^{1} f=1$.

In Case (i), from the above reasoning, we have $f\equiv 1$ on $[-1,0]$ and $f\equiv -1$ on $[0,1]$, which is a contradiction. In case (ii), again from the above reasoning, we have $f\equiv -1$ on $[-1,0]$ and $f\equiv 1$ on $[0,1]$, which is again a contradiction.

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Actually why are we picking f = 1 and f = -1? Shouldn't f be arbitrary? –  sidht Mar 18 '13 at 18:10
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Hint:

$$ 2 = \left|\int_{-1}^0 f - \int_0^1 f \right| \leq \left| \int_{-1}^0 f \right| + \left| \int_0^1 f\right| \leq \int_{-1}^0 \left|f\right| + \int_0^1 \left|f\right| \leq 2 \max |f| = 2 $$

Technically not an answer.

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up vote 0 down vote accepted

Take $f = -1$ on $-1/n \leq t < 1/n$ and $nt$ on $-1/n \leq t \leq 1/n$ and $f = 1 $on $1/n < t \leq 1$

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