# How to prove that $\langle Au_n, u_n-u\rangle_{X^*\times X}\to 0$

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 $$\langle Au_n, u_n-u\rangle_{X^*\times X}\to 0?$$

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 $$u_n\rightharpoonup 0$$).
• 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). – zorro47 Mar 4 '19 at 11:31
• @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. – s.harp Mar 4 '19 at 12:34
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 $$n\in \mathbb{N}$$.
• 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. – Giuseppe Negro Mar 4 '19 at 22:51