# Does weak convergence imply norm convergence in finite dimensional subspace?

Let $X$ be a reflexive Banach space and suppose that $$x_k\to x_0\quad \mbox{weakly in}\quad X.$$ Suppose there exists a finite dimensional subspace $Y\subset X$ such that $x_k, x_0 \in Y$. Does it imply that $$x_k\to x_0\quad (\mbox{strongly})?\qquad (1)$$

I thought that we may endow $Y$ with topology induced by $X$ topology. Since $Y$ is finite dimensional, then weak and strong topologies coincide and we get (1).

• Your idea will work. The only thing that's missing is checking that $x_n \to x_0$ in the weak topology on $Y$ (that is the one induced by $Y^*$ where we have given $Y$ the topology arising from the norm on $X$). This follows from an application of the Hahn-Banach theorem. Commented Jun 2, 2018 at 9:59
• Exactly $x_n\to x_0$ weakly in $Y$ is missing. I was trying to use adjoints operator, but I think it's not good enough. Commented Jun 2, 2018 at 10:01
• $x_n \to x_0$ weakly in $Y$ if and only if for every $\phi \in Y^*$, $\phi(x_n) \to \phi(x_0)$. Try using Hahn-Banach to extend $\phi$ to an element of $X^*$ and apply the weak convergence there. Commented Jun 2, 2018 at 10:03
• Take $\varphi \in Y^*$. From H-B theorem, there exists $\overline{\varphi}\in X^*$ such that $\overline{\varphi}|_Y=\varphi$. Now, apply weak convergence in X. Commented Jun 2, 2018 at 10:09

Lemma

Let $X$ be a topological space and $Y$ its subspace. Suppose that $(x_n)_n$ is a sequence in $Y$ and $x \in Y$ such that $x_n \to x$ in $X$. Then $x_n \to x$ in $Y$.

Proof.

Let $V$ be an open neighbourhood of $x$ in $Y$. By definition of subspace topology, there exists an open set $U$ in $X$ such that $U \cap Y = V$. $U$ is then an open neighbourhood of $x$ in $X$ so there exists $n_0 \in \mathbb{N}$ such that $x_n \in U, \forall n \ge n_0$. But then since $x_n \in Y$, we have $x_n \in U \cap Y = V, \forall n \ge n_0$.

Applied to this situation: the weak topology of $Y$ is the subspace topology w.r.t to the weak topology of $X$. Since $x_n \to x$ weakly in $X$, by the lemma we conclude $x_n \to x$ weakly in $Y$.

Now since $Y$ is finite-dimensional, weak and strong topology on $Y$ coincide, so $x_n \to x$ strongly in $Y$. Then of course $x_n \to x$ strongly in $X$.

• I must admit that in my reasoning I wanted to avoid purely topological insight, but now I see, that it also works giving even more general result. I'm very greatful for your answer! Commented Jun 2, 2018 at 12:54
• It might be worth noting that you still need to check that giving $Y$ the topology arising from the norm of $X$ and then taking the corresponding weak topology in fact gives you the same topology as the subspace topology on $Y$ induced by the weak topology on $X$. (this was the point of my comments essentially) Commented Jun 3, 2018 at 14:17
• @RhysSteele True, I've thought about that, a basis for the weak topology on $Y$ is given by $\{f^{-1}(U) : f \in Y^*, U\subseteq \mathbb{C} \text{ open}\}$ and a basis for the weak subspace topology on $Y$ is $\{f^{-1}(U) \cap Y : f \in X^*, U\subseteq \mathbb{C} \text{ open} \}$. The equality of these basis follows from $f^{-1}(U) \cap Y = (f|_Y)^{-1}(U)$ for $f \in X^*$, because $f|_Y \in Y^*$ and every $g \in Y^*$ can be extended by Hahn-Banach to some $f \in X^*$ such that $f|_Y = g$. Commented Jun 3, 2018 at 14:52
• I think we should assume that $X$ is Hausforff. Commented Jun 3, 2018 at 17:26
• @zorro47 $X$ is assumed to be a normed space in the above comment. $Y$ is a vector subspace of $X$. Commented Jun 3, 2018 at 18:06