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I recently came across this problem, namely we are given a continuous function $f:\mathbb R\to\mathbb R$ such that

$\int_0^1f(u(x))\mathrm dx=0,\;\forall u\in C^0([0,1]):\int_0^1u(x)\mathrm d x=0.$

I am asked to prove that the same property holds for any $u\in L^\infty([0,1])$ such that $\int_0^1 u(x)\mathrm dx=0.$

My attempt is to exploit Lusin Theorem because my first thought is that an essentially bounded measurable function is nearly a continuous one and then use the property given for continuous function. However I' still having problems in figuring out the reasoning. Is my path correct or not? and how should I approach the problem? Thank you.

Edit This is a further thought that came to my mind. Is such an $f$ necessarily a linear map? Because of course linear maps do the jobs, but what about the converse? The thought came by noticing that if $u$ is a continuous function satisfying the hypothesis given, then so does $\lambda u$, and if $v$ is another such function, then $u+v$ is fine as well.

I didn't want to ask another question because it descends from the original problem i proposed. Hope it is ok to ask here. Bye.

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    It strike me that all you need is dominated convergence. Produce a sequence of continuous functions converging in $\mathbb L^1$ satisfying the condtions, which is easy, then a subsequence converges a.e., then f ( the subsequence) converges a.e. and boundedly.2012-05-03

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Indeed, any such function $f$ must be linear. Here's a sketch.

Fix some $a \ge 0$ and consider the function $v(x) = \begin{cases} a, & 0 \le x \le \frac{1}{1+a} \\ -1, &\frac{1}{1+a} < x \le 1.\end{cases}$ Then we have $\int_0^1 f(v(x))\,dx = \frac{1}{1+a} f(a) + \frac{a}{1+a} f(-1).$

Fix $\epsilon > 0$, and modify $v$ on some small neighborbood of $\frac{1}{1+a}$ to obtain a function $u$ which is continuous, satisfies $\int_0^1 u(x)\,dx = 0$, satisfies $-1 \le u \le a$, and is equal to $v$ except on a set of measure at most $\epsilon$. (An appropriate piecewise linear function would work.) Let $M = \sup_{[-1, a]} |f|$; then $f(v(x))$ and $f(u(x))$ are two functions bounded by $M$ which differ on a set of measure at most $\epsilon$, so $\left| \int_0^1 f(v(x))\,dx - \int_0^1 f(u(x))\,dx\right| \le \int_0^1 |f(u(x)) - f(v(x))|\,dx \le 2 M \epsilon.$ But $\int_0^1 f(u(x))\,dx = 0$ by assumption, and so we have $\left| \int_0^1 f(v(x))\,dx\right| \le 2 M \epsilon.$ $\epsilon$ was arbitrary, so in fact $\int_0^1 f(v(x))\,dx = 0$.

Combining this with our previous computation shows $f(a) = -a f(-1)$; in particular, with $a=1$ we have $f(1) = -f(-1)$. Repeating the argument using $-v$ instead of $v$ shows that $f(-a) = -a f(1) = a f(-1)$. So we have that $f(x) = (-f(-1))x$ for all $x \in \mathbb{R}$, i.e. $f$ is linear.

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    I just noticed that this argument doesn't really use the continuity of $f$. It would suffice to assume that $f$ is Borel measurable and locally bounded.2012-05-03