Does weak convergence imply norm convergence in finite dimensional subspace?

Question

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).

Answer 1

The answer is yes.

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 $X$.