# Does point-wise weak convergence give weak convergence in $L^2(I;X)$?

Let $X$ be a separable reflexive Banach space, $F$ be a locally Lipschitz nonlinear operator on $X$ that is weakly continuous on $X$, and $u_n$ are $u_n$ weakly converges to $u$ on $L^2(0,T;X)$. Now, can we say $$F(u_n)\to F(u) \quad \text{weakly in} \quad L^2(0,T;X)$$

Note that we know
$$F(u_n(t))\to F(u(t)) \quad \text{weakly in }X\text{ for every }t\in[0,T]$$

The above problem was ill-defined, as the sequence $F(u_n)$ may not belong to $L^2(0,T;X)$ (see this example). Consider the same problem with added uniformly boundness, and continuity assumption:

Let $X$ be a separable reflexive Banach space, $F$ be a locally Lipschitz nonlinear operator on $X$ that is weakly continuous on $X$, and $u_n$ are uniformly bounded, and continuous functions over $[0,T]$ such that $u_n$ weakly converges to $u$ on $L^2(0,T;X)$. Now, can we say $$F(u_n)\to F(u) \quad \text{weakly in} \quad L^2(0,T;X)$$

Note that we know
$$F(u_n(t))\to F(u(t)) \quad \text{weakly in }X\text{ for every }t\in[0,T]$$

• Without further assumptions pointwise convergence des not imply $L^2$-convergence (even for $X=\mathbb R$ and the identity $u$). Aug 26, 2018 at 21:15
• This is more detailed. The weak convergence of $u_n$ has been assumed. Aug 27, 2018 at 1:02
• You are asking about $F(u_n)$ as members of $L^2(0,T;X)$, which I take to mean the operator from $L^2(0,T;X)$ which is pointwise $F$ (that is, $F(u)$ is the function which, evaluated at $t$, gives $F(u(t))$. Given the conditions on $F$ (locally Lipschitz and weakly continuous) it is not even clear to me that the pointwise $F$ operator is well-defined as an operator from $L^2(0,T;X)$ into itself. If $X=\mathbb{R}$, $T=1$, $F(x)=x^3$, then the pointwise $F$ operator should be $F(u)=u^3$. But this doesn't map $L_2$ into $L_2$, since $u=t^{-1/3}\in L_2$ but $u^3=t^{-1}\notin L_2$.
– user114263
Aug 27, 2018 at 11:32
• You are right. I did not mention that $u$ are continuous function on $[0,T]$. With this extra assumption, such cases won't happen. Aug 27, 2018 at 15:10
• I still do not get it, how do you know that the weak limit of $F(u_n(t))$ is $F(u(t))$ in $X$ without knowing that the weak limit of $F(u_n)$ in $L^2(0,T;X)$ is $F(u)$? IF this is correct, then the rest should just be dominated convergence at this stage, if I see it correctly. Aug 29, 2018 at 7:51

The answer is no. We will use $X=\mathbb{R}$. Along the lines of my comment that the pointwise $F$ function need not even map $L_2$ into $L_2$, we can find an example of a sequence $(u_n)_{n=1}^\infty$ in $L_2$ of continuous functions which converge in norm to $0$ in $L_2$ but such that the functions $F(u_n)$ have norms going to infinity. They will be peaky tent functions whose peaks are not too tall, but such that the sequence $F(u_n)$ has peaks which are too tall. We will let $F(t)=t^2$ and let $T=2$.

Fix $1<p<2$ and let $I_n=(0, 2n^{-p})$. We next define continuous $g_n$. On the left half of $I_n$, define $g_n$ to be linear, increasing from a value of $0$ at $0$ to a value of $2n$ at the midpoint of $I_n$. On the right half of $I_n$, let $g_n$ be the linear function which is $2n$ at the midpoint of $I_n$ and decreases to $0$ at $2n^{-p}$. Let $g_n$ be zero off of $I_n$. Let $u_n=\sqrt{g_n}$. Then $u_n$ is continuous. Furthermore, since (the non-zero part of) $g_n$ is a triangle with height $2n$ and base $2n^{-p}$, $$\|u_n\|_{L_2}= \sqrt{\|g_n\|_{L_1}} = \sqrt{2nn^{-p}}.$$ Since $1<p$, $\lim_n \|u_n\|_{L_2}=0$, so $(u_n)_{n=1}^\infty$ converges to zero not only weakly, but in norm, in $L_2$.

However, $F(u_n)=g_n$. Again thinking about $g_n$ as a triangle, we see that $g_n\geq n$ on a subset of $I_n$ which has measure $n^{-p}$ (half the measure of $I_n$). That is, on the left half of the triangle, since $g_n$ is linear there, half the time it is above half its maximum value, and the same on the right half. Therefore $g_n^2\geq n^2$ on a set of measure $n^{-p}$, and $$\|g_n\|_{L_2} \geq \sqrt{n^2 n^{-p}}.$$ Since $p<2$, this increases to infinity as $n$ does. Thus $(u_n)_{n=1}^\infty$ converges to $0$ (weakly and in norm) in $L_2$ while $(F(u_n))_{n=1}^\infty$ blows up as $n\to \infty$.

• Thanks. We can just add uniform continuity of $u_n(t)$ and $F(u_n)$ to get around such cases. I keep restating the question to make you not think about the case $F(u_n)$ is not in $L_2$. A "uniformly" has been added to the question statement. Aug 28, 2018 at 14:08
• It is good policy to add a note in the body of your question (instead of in the comments of the question) when you edit it, to avoid having those who gave a correct answer to the question as it was previously stated appearing not to have read the question thoroughly. It is also a good policy to include all hypotheses you want include in your question before people spend time answering them. I do not understand your most recent comment, as in my example, $F(u_n)$ IS in $L_2$ for each $n$.
– user114263
Aug 28, 2018 at 16:09
• Yes, although not uniformly bounded. Aug 28, 2018 at 16:29
• Nowhere in your question or your comments or revisions did you ask for anything to be uniformly bounded in the supremum norm.
– user114263
Aug 28, 2018 at 16:34

Since the question has now changed multiple times, I am editing in a new answer after some discussion in the comments.

Let $X$ be a separable and reflexive Banach space. Let $u_n$ be continuous functions $[0,T] \to X$ which are uniformly bounded, so there exists $C > 0$ such that $\|u_n(t)\|_X \leq C$ for all $t \in [0,T]$ and all $n$. Suppose that $(u_n)$ converges to some $u$ weakly in $L^2(0,T;X)$. If $F \colon X \to X$ is locally Lipschitz and weakly continuous on $X$, does $(F(u_n))$ converge weakly to $F(u)$ in $L^2(0,T;X)$?

This is wrong in general. Consider $L^2(0,2\pi)$, so $T = 2\pi$ and $X = \mathbb{R}$. Set $F(x) = x^2$ as a real function. It is locally Lipschitz continuous on $\mathbb{R}$, in particular weakly continuous.

The functions $$u_n(t) := \frac1{\sqrt{\pi}} \sin \left(\frac{nt}2\right)$$ are continuous and uniformly bounded. They form an orthonormal basis in $L^2(0,2\pi)$, thus they converge weakly to $u = 0$ there. But testing $F(u_n)$ against the constant $1$ function $\mathbf{1} \in L^2(0,2\pi)$ gives $$1 = \|u_n\|^2_{L^2(0,2\pi)} = \int_0^{2\pi} u_n(t)^2 \, dt = \int_0^{2\pi} F(u_n(t)) \, dt = \bigl(F(u_n),\mathbf{1}\bigr)_{L^2(0,2\pi)}$$ for all $n$, hence $F(u_n)$ does not converge weakly to $F(u) = 0$ in $L^2(0,2\pi)$.

Let $X$ be a separable and reflexive Banach space. Suppose that $u_n$ converges to some $u$ weakly in $L^2(0,T;X)$. If $F \colon X \to X$ is locally Lipschitz and weakly continuous on $X$, does $F(u_n)$ converge weakly to $F(u)$ in $L^2(0,T;X)$?

Weakly continuous Nemytskii operators on Lebesgue (or function) spaces are a rare breed and this is already the case for $X = \mathbb{C}$ or $\mathbb{R}$. More specifically, a weakly continuous Nemytskii operator mapping, say, $L^p(0,1)$ into itself is necessarily already affine linear. See this question and the answers there.

Old query by myself to a remark in the question:

As a sidenote, I do not quite understand how you get $F(u_n(t))$ that converges weakly to $F(u(t))$ in $X$ for (almost?) every $t \in (0,T)$ from your assumptions, maybe there is a misunderstanding here?

This claim is also false as shown in the comments to the OP, using the ONB in $L^2(0,2\pi)$ and $F(x) = x$.

• For every $t$, $u_n(t)$ is bounded sequence in $L^p(0,T;X)$. So, it converges weakly to some element say $u(t)$. Now, from weak continuity of $F$ on $X$, we can say $F(u_n(t))$ also weakly converges to an element in $X$. Later, we can show that it weakly converges to $F(u(t))$. Aug 27, 2018 at 20:44
• I think you mean that $(u_n(t))_n$ is a bounded sequence in $X$, but that still does not follow from boundedness of $(u_n)$ in $L^2(0,T;X)$ I think. Aug 28, 2018 at 11:14
• Please notice $u_n(t)$ are continuous functions. Aug 28, 2018 at 14:01
• Consider $L^2(0,1)$ and an enumeration $(q_n)$ of $[0,1] \cap \mathbb{Q}$, set $f_n(t) = n^{1/4}$ if $|t-q_n| \leq 1/n$ and $|t-q_n|^{-1/4}$ otherwise for $t \in (0,1)$. The $f_n$ are continuous and $\|f_n\|_{L^2(0,1)}^2 \leq 2\sqrt{2}$. Take $t^* \in [0,1] \setminus \mathbb{Q}$ and consider a sequence $(q_k) \subset \mathbb{Q}$ such that $q_k \to t^*$. For $k$ large enough, $f_k(t^*) = k^{1/4}$, so $f_k(t^*) \to \infty$. This is the case for every $t^* \in [0,1] \setminus \mathbb{Q}$ and the latter is not a zero set. Aug 28, 2018 at 15:12
• Uniform continuity is not a property that applies to collections of functions (not anywhere that I have seen, at least). Uniform continuity is a property which a single function may or may not possess. Did you mean uniform equicontinuity? And it sounds as though you want uniform equicontinuity and uniform boundedness in the supremum norm.
– user114263
Aug 28, 2018 at 16:47