In a Hilbert space $x_n \overset{\text{w}}{\to} x$ and $y_n \to y$. Prove $\langle x_n, y_n \rangle \to \langle x, y \rangle$

Question

Proposition

In a Hilbert space, suppose $x_n \overset{\text{w}}{\longrightarrow} x$ and $y_n \longrightarrow y$. Then, $\langle x_n, y_n \rangle \longrightarrow \langle x, y \rangle$.

Proof

By definition of weak convergence, we have for any bounded linear functional $f \in H'$

\begin{align} \lim f(x_n) = f(x) &\implies \lim\langle x_n, y\rangle = \langle x, y\rangle \\ &\implies \lim \langle x_n- x, y\rangle = 0 \end{align} From there we observe

\begin{align} 0 \leq |\langle x_n, y_n\rangle - \langle x, y\rangle| &= |\langle x_n, y_n\rangle -\langle x, y_n\rangle + \langle x, y_n\rangle - \langle x, y\rangle| \\ &\leq |\langle x_n, y_n\rangle -\langle x, y_n\rangle | + |\langle x, y_n\rangle - \langle x, y\rangle| \\ &= |\langle x_n - x, y_n\rangle| + |\langle x, y_n - y\rangle| \longrightarrow 0 \end{align}

as $n \longrightarrow \infty$.

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Seemed a bit too simple so I'm afraid. So I'm requesting to verification or where I went wrong. Thanks.

Answer 1

Note: As noted by Evan William Chandra below your question, the last step is invalid, because $y_n$ does not remain fixed as $n$ grows. This means that you cannot use the weak convergence there. However, your idea can be easily modified.

We have that

$$\langle x_n,y_n\rangle-\langle x,y\rangle= \langle x_n,y_n-y\rangle+\langle x_n-x,y\rangle.\ \ \ (1)$$

By weak convergence $\langle x_n-x,y\rangle\rightarrow 0$. Furthermore, a weakly convergent sequence is bounded. Consequently, $\exists M>0$ such that $||x_n||\leq M,\ \forall n\in \mathbb{N}$ and Cauchy-Schwarz yields that

$$\langle x_n,y_n-y\rangle\leq ||x_n||\cdot||y_n-y||\leq M\cdot||y_n-y||\rightarrow 0,$$

as $n$ approaches infinity (because $y_n\rightarrow y$). Finally, $\langle x_n-x,y\rangle\rightarrow 0$ and $\langle x_n,y_n-y\rangle\rightarrow 0$. So, $(1)$ gives the desired result.