If $a_n \to a$ in $L^2$ and $F:\mathbb{R} \to \mathbb{R}$ is continuous, does $\int F(a_{n_j}) \to \int F(a)$?

Question

Suppose $a_n \to a$ in $L^2(\Omega)$. Let $F:\mathbb{R} \to \mathbb{R}$ be continuous with $F(0) = 0$. We have that $F(b) \in L^1(\Omega)$ if $b \in L^2(\Omega)$ and $|F'(x)| \leq C_1 + C_2|x|$.

I want to show$$\int_{\Omega}F(a_{n_j}) \to \int_{\Omega}F(a)$$ for a subsequence.

Is it possible?

Does the domain $\Omega$ have finite measure? If not, $C_1$ is a problem: $F(x)$ could be $x$ for small $x$, and then $\int F(a)$ is not controlled by $\int a^2$. — , Jun 10 '14 at 16:32
Usually one uses $F\circ b$ or $Fb$ (especially for linear operators) for the composition of 2 functions. $F(b)$ normally denotes the value of $F$ at $b$, but considering what $F$ and $b$ are in your case, that's probably not what you were trying to say. — cQQkie, Jun 10 '14 at 19:25

score 0 · Accepted Answer · edited Apr 13 '17 at 12:20

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Since my attempt to close the question as a duplicate failed, I'll post an answer: the statement follows from a general theorem on the continuity of Nemytskii operator, which is stated and proved here.

edited Apr 13 '17 at 12:20

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answered Jul 17 '14 at 19:02

If $a_n \to a$ in $L^2$ and $F:\mathbb{R} \to \mathbb{R}$ is continuous, does $\int F(a_{n_j}) \to \int F(a)$?

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