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Let $X$ be a reflexive Banach space and consider a linear and continuous operator $A\colon X\to X^*$. Suppose that $(u_n)$ is a sequance of elements from $X$ and assume that $u_n\to u$ weakly in $X$. How to deduce that $$\langle Au_n, u_n-u\rangle_{X^*\times X}\to 0\,?$$

Let $x\in X$. Consider $$\lim_{n\to \infty}\langle Au_n-Au,x\rangle_{X^*\times X}=\langle\lim_{\ n\to \infty}A(u_n-u),x\rangle_{X^*\times X}=0.$$ This means $Au_n\to Au$ weakly$^*$ in $X^*$. As $X$ is reflexive, so is its dual. Since in reflexive Banach spaces weak and weak$^*$ topologies coincide, we deduce that $Au_n\to Au$ weakly in $X^*$. How to use that and linearity of $A$ to deduce $$\langle Au_n, u_n-u\rangle_{X^*\times X}\to 0?$$

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This is false. Take $X=L^2(\mathbb R)$, and $A=I$, the identity operator. Fix $f\in L^2$, $f\ne 0$ and define $$ u_n(x):=f(x-n).$$ Then $u_n \rightharpoonup 0$, so $u=0$ in the notation of the original question, but $$\langle u_n | u_n\rangle = \int_{-\infty}^\infty f(x-n)^2\, dx = \lVert f\rVert_{L^2}^2\ne 0.$$

(See here for a proof that $u_n\rightharpoonup 0$).

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  • $\begingroup$ Thank you! So this ($\langle Au_n, u_n-u\rangle_{X^*\times X}\to 0$) would be true if $u_n\to u$ strongly in $X$ only (I mean without any additional assumptions). $\endgroup$ – zorro47 Mar 4 '19 at 11:31
  • $\begingroup$ Yes, that's easy; just use Cauchy-Schwarz and the fact that Au_n is bounded. $\endgroup$ – Giuseppe Negro Mar 4 '19 at 11:46
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    $\begingroup$ @zorro47 the proof works with any sequence weakly converging to $0$ but not doing so in norm. The specific example does converge weakly to zero, you need to check that $$\int_{\Bbb R}f(x-n)g(x)\,dx$$ converges to zero for any $f,g\in L^2$. This is a measure theory exercise. $\endgroup$ – s.harp Mar 4 '19 at 12:34
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I have another counter example. Let $X$ be a Hilbert space and $A\colon X\to X^*$ be defined by: $$\langle Au,v\rangle=\langle-u,v\rangle_{X}.$$ Let $(u_n)$ be an orthonormal set of vecotrs from $X$. Thus, for every $v\in X$, by the Bessel inequality $$\sum_{n=1}^{\infty}|\langle u_n,v\rangle|^2\le \|v\|^2_X$$ we deduce that $\lim\langle u_n,v\rangle=0$. Since $X$ is Hilbert, we have that $$u_n\to0$$ weakly in $X$. Then $$\langle Au_n,u_n-u\rangle=\langle Au_n,u_n\rangle=-\langle u_n, u_n\rangle_{X}=-\|u_n\|^2=-1\ne0$$ for all $n\in \mathbb{N}$.

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    $\begingroup$ Yes; this is another case of a sequence $u_n$ that converges weakly and does not converge strongly. The operator $A=I$ furnishes a counterexample in this case, too. $\endgroup$ – Giuseppe Negro Mar 4 '19 at 22:51

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