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I am looking at the exercise: Let $f:[0,1] \to \mathbb{R}$ integrable such that $|f(x)| \leq \int_0^x |f(t)|dt, \forall x \in [0,1]$.

Show that $f=0$.

At the solution,it is taken that $f$ is bounded. Why??Because of the fact that $f$ is integrable?

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The integrability of $f$ implies that $$\int_0^x \lvert f(t)\rvert\,dt \leqslant \int_0^1 \lvert f(t)\rvert\,dt < \infty,$$ and from that the boundedness of $f$ follows by the given inequality. – Daniel Fischer Apr 21 '14 at 13:47
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Generally,if a function is Riemann integrable, then it's bounded.

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But only if it is Riemann integrable in a bounded and closed interval,right? – evinda Apr 21 '14 at 15:09
From the start, Riemann integral is defined on bounded function in closed, finite interval. – Hanna Khalil Apr 21 '14 at 16:50
I understand..Thank you!!! $\checkmark$ – evinda Apr 21 '14 at 17:48

The function $F(x)=\int_0^x |f(t)|dt$ is continuous and is defined on the compact set $[0,1]$; hence, $F$ is bounded. Since $|f(x)|\leq F(x)$ for each $x\in[0,1]$, $f$ is also bounded.

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$f$ is bounded because $f$ is integrable and satisfies your inequality. Notice that the integral is monotonically increasing in $x$. In other words:


For some finite $C$.

As a hint for the exercise, try integrating both sides of the inequality from $0$ to $x$ and notice that if $f$ is bounded, then $\int_0^y\int_0^xf(t)dtdx\leq yC$. On the other hand, $|f(y)|\leq\int_0^y|f(x)|dx$. What happens if you do three integrals...four integrals...

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@evinda: the first assertion is correct. The conclusion would not necessarily be true for $x\in[0,\infty)$. – Alex R. Apr 21 '14 at 13:58
@Alex, you are not describing a function on $[0,1]$, it is more properly only a function on $(0,1]$. It is a well-known theorem that a function (using the rigorous definition of the word) $f\colon[a,b]\to\mathbb{R}$ is Riemann integrable if and only if it is bounded and continuous almost everywhere (in the Lebesgue sense). – Dan Rust Apr 21 '14 at 15:35
@AlexR. A function that is Riemann integrable over an interval is most certainly bounded. $1/\sqrt x$ is not Riemann-integrable near $0$. – Mario Carneiro Apr 21 '14 at 15:35
Classically Riemann integrable functions are bounded on closed intervals. Improper Riemann integrals allow for unbounded functions. Pardon the confusion. In the original question, it says $f$ is integrable, which in the Lebesgue sense would allow for $1/\sqrt{x}$. – Alex R. Apr 21 '14 at 15:37
@DanielRust Even $1/\sqrt x$ on $(0,1]$ is not Riemann-integrable because every upper sum is infinite. – Mario Carneiro Apr 21 '14 at 15:37

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