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The problem is: Let a function $f$ be continuously differentiable on [0,1]. Prove that $$ \int\limits_0^1|f(x)|\,dx\le \max\left\{ \int\limits_0^1 |f'(x)| \, dx, \left| \int\limits_0^1 f(x) \, dx\right|\right\} $$ For $f$ nonpositive or nonnegative this is obvious. Intuitively I know, that it is true in the remaining case, where the first integral of the righthand side comes into play, which is a sum of $f$ changes between local maxima and minima. Please, help to prove this inequality strictly.

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  • $\begingroup$ Do you have your inequality turned around? Because $$\int_0^1|f(x)|dx \ge \left|\int_0^1f(x)dx\right|$$ is one of those inequalities that everyone should know. $\endgroup$ – Paul Sinclair Dec 5 '17 at 0:41
  • $\begingroup$ This is not my inequality, this is a problem from a book. I know the inequality your typed, therefore I was extremely accurate typing the problem, making sure the inequality looks exactly the same as in the book. $\endgroup$ – Andrei Petrov Dec 5 '17 at 1:32
  • $\begingroup$ I took a function $f(x)=x-0.5$ as an example, the inequality is true for it. $\endgroup$ – Andrei Petrov Dec 5 '17 at 2:02
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This is a very intriguing problem, when one sees that, as Paul Sinclair pointed out, $$\int_0^1|f(x)|dx \geq \left|\int_0^1f(x)dx\right|$$ This at first quite confuses us, but this insight is in fact integral to the solution. This tells us that the equality case here is important, for otherwise, having this in the maximum function would be meaningless. So, let us examine the equality case. What is important to see here is that iff $$\int_0^1|f(x)|dx = \left|\int_0^1f(x)dx\right|$$ Then for $x \in [0,1]$, we know that the sign of $x$ is constant. Now that we know this, we know that the inequality does indeed hold as long as the sign of $x$ is constant. Next we examine the other case.

If the sign of $x$ is not constant, we can make the following argument. Let, $a$ be the maximum of f over the range $[0,1]$ and let $-b$ be the minimum of f over the same range. We know, because there is a sign change, that $b$ and $a$ are positive values. Next, we know that from the fundamental theorem of calculus, $$\int_0^1 |f'(x)|dx \geq \left|\int_{f^{-1}(-b)}^{f^{-1}(a)} f'(x)dx \right| \geq f(f^{-1}(a)) - f(f^{-1}(-b)) = a + b$$ And finally, we also know that from the inequality $-b < f(x) < a$, we have $$\int_0^1 |f(x)dx| < \max(a,b) < a + b \leq\int_0^1 |f'(x)|dx$$ In this way, the inequality holds for both the case where the sign is constant as well as the case where it isn't, and thus we are done.

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  • $\begingroup$ I don't understand, how the inequality $\int\limits_0^1|f'(x)|dx \geqslant a+b$ follows from the fundamental theorem of calculus. $\endgroup$ – Andrei Petrov Dec 6 '17 at 6:21
  • $\begingroup$ I have added a little to explain this. $\endgroup$ – Isaac Browne Dec 6 '17 at 14:22
  • $\begingroup$ Ach, what an unexpected place to use the inequality for absolute values! This was the main hint, now I understand everything. Thank you! $\endgroup$ – Andrei Petrov Dec 6 '17 at 17:01

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