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Let $F$ be a continuously differenable function over $\mathbb{R}$. Let $\Omega$ be a bounded subset of $\mathbb{R}^2$. Assume that for every $w\in L^2(\Omega)$ then $v(x)=F(w(x))$, $x\in \Omega$, is also in $L^2(\Omega)$. Can we say that $$\|F(w_1(\cdot)+\eta w_2(\cdot))\|_{L^2(\Omega)}$$ is continuous in $\eta$?


What if we relax the assumption to: for every $w\in H^1(\Omega)$ then $F(w)$ is in $L^2(\Omega)$? You can assume now $F$ is twice continuously differenable function over $\mathbb{R}$.

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Yes.

Let $\mu$ be the Lebesgue measure. We start with

Lemma: function $F$ admits an estimate $$|F(x)| \le C_1|x| + C_2$$ for some constants $C_1, C_2 > 0$.

Proof: Suppose not. Then there is a sequence of points $x_n \to \infty$ such that $$\frac{|F(x_n|)}{|x_n|} \to \infty.$$ Consider a function $u$ which takes value $x_1$ on the set of measure $m_1$, value $x_2$ on the set of measure $m_2$, etc. Then $$||u||^2_{L^2} = m_1x_1^2 + m_2x_2^2 + \ldots, \\ ||F(u)||^2_{L^2} = m_1F(x_1)^2 + m_2F(x_2)^2 + \ldots.$$ Now choose $m_i$ in such a way that only the first of these series is converging. Thus, $F(u) \not \in L^2$ which is a contradiction. End proof

Let us now prove that $||F(w_1 + \eta w_2)||_{L^2}$ is continuous at $\eta = 0$.

Take $n \in \mathbb{N}$ and let $A_n = \{x \in \Omega \mid |w_1(x)| \le n, |w_2(x)| \le n\}$. We have $\mu(\Omega \setminus A_n) \to 0$. On the set $A_n$ function $w_1 + \eta w_2$ is bounded by $2n$ for $\eta < 1$. Let $$M = \max\limits_{|y| \le 2n}|F^{\prime}(y)|.$$

We have, using Lemma, $$||F(w_1) - F(w_1 + \eta w_2)||^2_{L^2(\Omega)} = ||F(w_1) - F(w_1 + \eta w_2)||^2_{L^2(A_n)} + \\ ||F(w_1) - F(w_1 + \eta w_2)||^2_{L^2(\Omega\setminus A_n)} \le M^2\eta^2||w_2||^2_{L^2(A_n)} + \\ ||C_1(2|w_1| + \eta|w_2|) + 2C_2||^2_{L^2(\Omega\setminus A_n)}.$$ The second term tends to $0$ as $n \to \infty$ and the first term is continuous in $\eta$.

This shows that the function $\eta \mapsto F(w_1 + \eta w_2)$ is continuous as a function from $\mathbb{R}$ to $L^2(\Omega)$. In particular, $\eta \mapsto ||F(w_1 + \eta w_2)||_{L^2}$ is also continuous.

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  • $\begingroup$ Excellent. If we relax the assumption to: for every $w\in H^1(\Omega)$ then $F(w)\in L^2(\Omega)$; would $F(\cdot)$ still be linearly dominated? $\endgroup$
    – Saj_Eda
    Sep 25, 2018 at 23:04
  • $\begingroup$ @SaraWinslet, no. By the Sobolev Embedding, $H^1(\Omega) \hookrightarrow L^4(\Omega)$, for example. This means that for a function $w \in H^1$ function $F(w) = w^2$ lies in $L^2$. $\endgroup$ Sep 25, 2018 at 23:21
  • $\begingroup$ But it may still have that continuity property. Right? Do you know of any counter example? $\endgroup$
    – Saj_Eda
    Sep 25, 2018 at 23:30
  • $\begingroup$ @SaraWinslet, I don't know the answer to your second question, unfortunately. The closest thing I can think of is Orlicz spaces but that's still far off, probably. $\endgroup$ Sep 26, 2018 at 21:09
  • $\begingroup$ That's what I am thinking too. I believe in your answer, the second part still works if we had any other upper bound for $|F(x)|$ not just affine-linear, it only needs to have less growth than $e^{x^2}-1$ near infinity. $\endgroup$
    – Saj_Eda
    Sep 26, 2018 at 21:15

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