Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Please help me prove this.

Suppose $f$ is defined in$\left [ a,b \right ]$ ,$f\geq 0$, $f$ is integrable in$ [a,b]$ and $\displaystyle\int_{a}^{b}fdx=0$

prove: $\displaystyle\int_{a}^{b}f(x)^2dx=0$

Thanks a lot!

share|improve this question
thanks for the tip –  Lilly Feb 4 '12 at 6:54
To accept an answer, you can click your name at the top of the page. Then you can find all the questions you have asked. For each answer to the question you have asked, there is a tick. By clicking the tick, you can accept the answer. Note that you can only accept one answer to each of your question. So choose the best answer which is the most helpful to you. Also look at here: math.stackexchange.com/faq#howtoask –  Paul Feb 4 '12 at 7:23
@Paul Thank you very much! –  Lilly Feb 4 '12 at 7:37
+1 for accepting answers! –  Paul Feb 4 '12 at 8:26

3 Answers 3

up vote 3 down vote accepted

Let's assume we're working with the Lebesgue integral.

Define $$ f_k(x)=\left\{\begin{array}{}f(x)&\text{if }k\le f(x)<k+1\\0&\text{otherwise}\end{array}\right. $$ Since $f$ is measurable, so is each $f_k$.

Then since $f=\sum\limits_{k=0}^\infty f_k$ and each $f_k\ge0$, we have by dominated convergence $$ \begin{align} \sum_{k=0}^\infty\int_a^bf_k(x)\;\mathrm{d}x &=\int_a^bf(x)\;\mathrm{d}x\\ &=0\tag{1} \end{align} $$ Since $f_k(x)\ge0$, we have $\int_a^bf_k(x)\;\mathrm{d}x\ge0$. $(1)$ says that the sum of non-negative terms is $0$; therefore, each $\int_a^bf_k(x)\;\mathrm{d}x=0$. Furthermore, by dominated convergence, $$ \begin{align} \int_a^bf^2(x)\;\mathrm{d}x &=\sum_{k=0}^\infty\int_a^bf_k^2(x)\;\mathrm{d}x\\ &\le\sum_{k=0}^\infty(k+1)\int_a^bf_k(x)\;\mathrm{d}x\\ &\le\sum_{k=0}^\infty(k+1)\cdot0\\ &=0\tag{2} \end{align} $$

Let's assume we're working with the Riemann integral.

To be integrable, $f$ must be bounded (otherwise, for the partition interval on which $f$ is unbounded, the upper sum is infinite). Let $M=\sup\limits_{[a,b]}f$. Since $f\ge0$, $$ \begin{align} \int_a^bf^2(x)\;\mathrm{d}x &\le M\int_a^bf(x)\;\mathrm{d}x\\ &=M\cdot0\\ &=0\tag{3} \end{align} $$

share|improve this answer

My first inclination would be to say that since $f\geq0$ and $\int_a^bfdx=0$, $f(x)=0$ for $x$ such that $a\leq x\leq b$. Because of this, $\int_b^af(x)^2dx=0$. You will probably need to fill in a fair amount of reasoning to make this hold up, though.

Also, suggestion - accept some answers to your questions. It makes people more likely to answer in the future.

share|improve this answer
Thanks a lot, how can i accept answers? –  Lilly Feb 4 '12 at 7:17

Intuitively $f$ must be zero almost everywhere (term from Lebesgue theory). Now, if $f$ is zero almost everywhere then so is $f^2$, and then $\int f^2=0$ too.

That this is really the case is in fact trivial because $$0=\int_a^b f(x) dx =\int_a^b |f(x)|dx=\|f\|_{L^1(a,b)}$$ in other words the $L^1$-norm of $f$ is zero and then (because it is a norm) $f$ must be the $L^1$-zero function, that is $f$ must be zero almost everywhere.

If we did not know that $f\mapsto\|f\|_{L^1(a,b)}$ is a norm, it is sufficient to prove that the set $A=\{x: \, f(x)>0\}$ has measure zero $\mu(A)=0$ (where $\mu$ is the Lebesgue measure). In order to do that we need some argument where we use $\int f = 0$ in simpler terms, one such method is to split the set into a countable number of disjoint sets that we can control (similar to robjohn's argument). Having said that, let us consider $$A= \bigcup_{n=1}^\infty A_n$$ where $A_n=\{x: 1/n\le f(x) < 1/(n-1)\}$ (where in the exceptional case $n=1$ we mean $A_1=\{x : \,f(x)\ge1\}$ ). Then since the sets are disjoint we have $$\mu(A)=\sum_{n=1}^\infty\mu(A_n)=\sum_{n=1}^\infty\int_{A_n}1dx$$ and since $f(x)\ge 1/n$ on $A_n$ we have $$\int_{A_n}1dx\le n\int_{A_n}f(x)dx\le n\int_a^bf(x)dx=n\cdot0=0$$ That is each $A_n$ has measure 0, and hence so has $A$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.