Let $X$ be a closed subspace of $\ell^2$ such that $X$ is contained in $\ell^1$. It is easy to show that the inclusion operator $J \colon X \hookrightarrow \ell^1$ is closed, hence, by the closed graph theorem $J$ is bounded.
Is it true that $X$ is automatically finite dimensional?
I would really appreciate any hints.
On
The norms $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent on $X.$ Assume $X$ is infinite dimensional. Let $\{v_n\}_{n=1}^\infty$ be an orthonormal basis in $X.$ Any element $\varphi\in \ell^\infty$ determines a bounded linear functional on $X.$ Thus it corresponds to an element $u\in X$ such that $$\varphi(v_n) =\langle v_n,u\rangle$$ Hence $\varphi(v_n)\to 0$ by the Bessel inequality. This means the sequence $v_n$ tends weakly (in $\ell^1$) to $0.$ By the Schur theorem every weakly convergent sequence in $\ell^1$ is norm convergent. This gives a contradiction as $\|v_n\|_2=1,$ hence $\|v_n\|_1\not\to 0.$
This is true. Indeed, composing the orthogonal projection $\ell^2 \to X$ with $J$, then restricting to $\ell^1$ (recall that $\ell^1 \subset \ell^2$ and $\|\cdot\|_1 \geq \|\cdot\|_2$) shows that there exists a bounded projection $\ell^1 \to J(X)$. Thus, $J(X)$ is a closed and complemented subspace of $\ell^1$. All such subspaces are either finite-dimensional or isomorphic to $\ell^1$ (see https://mathworld.wolfram.com/ComplementarySubspaceProblem.html). The latter case is impossible since $J$ is an isomorphism between $X$ and $J(X)$, but $\ell^1$ is not isomorphic to a Hilbert space, whilst $X$ is a Hilbert space. Hence, $J(X)$ and thus $X$ must be finite-dimensional.