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Let $H$ be a hilbert space and let $u_n\in L^2(0,T;H)$ be a sequence. Then $u_n$ converges weakly to $u\in L^2(0,T;H)$ iff

$$\int_0^T (u_n(t),v(t))_H\,dt \to \int_0^T (u(t),v(t))_H\,dt$$ for every $v\in L^2(0,T;H)$. Does the following hold:

$$u_n \rightharpoondown u\quad \text{in } L^2(0,T;H) \quad\Leftrightarrow\quad u_n \text{ is bounded in } L^2(0,T;H)\text{ and } u_n(t)\rightharpoondown u(t)\quad \text{for almost all } t\in (0,T).$$

$\Leftarrow:$ Let $v\in L^2(0,T;H)$. By assumption, we have $$(u_n(t),v(t))_H \to (u(t),v(t))_H\quad\text{for almost all } t\in (0,T).$$

Since $$\lvert(u_n(t),v(t))_H\rvert\leq \lvert\lvert u_n(t)\rvert\rvert_H \lvert\lvert v(t)\rvert\rvert_H,$$ we can use the hölder inequality and the boundedness of $u_n$ to bound the righthandside by an integrable function. Lebesgues dominated convergence then implies

$$\int_0^T (u_n(t),v(t))_H\,dt \to \int_0^T (u(t),v(t))_H\,dt$$

and thus $u_n \rightharpoondown u$ in $L^2(0,T;H)$. Is my reasoning correct?

I am also stuck on the other direction.

Mandelbrot
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1 Answers1

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Take $H=\mathbb{R}$ (or any finite-dimensional space), then weak convergence in $H$ is equivalent to strong convergence. But weak convergence in $L^2((0,T),\mathbb{R})$ does not imply pointwise almost everywhere convergence. For example, $u_n(t) = \sin(nt)$ converges weakly to zero in $L^2((0,T),\mathbb{R})$ as $n\to\infty$, but it does not converge a.e. to zero since every $u_n$ is nonzero except at a finite number of points. In fact, you can’t even extract an a.e. convergent subsequence. (See Pointwise almost everywhere convergent subsequence of $\{\sin(nx)\}$)

kieransquared
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