# If $T(B)$ is a bounded subset of $K$, then, is the linear functional $T$ bounded (norm of $T$ is finite)? [duplicate]

This question is an exact duplicate of:

Let $H$ be a infinite dimensional Hilbert space on $K$ and let $B$ be a basis of $H$ that $H=\overline{span(B)}$

moreover, let $T : H \rightarrow K$ be a linear functional (i don't know if it is bounded)

If $T(B)$ is a bounded subset of $K$, then, is the linear functional $T$ bounded (norm of $T$ is finite)?

## marked as duplicate by colormegone, JonMark Perry, user91500, Watson, Joel Reyes NocheMay 26 '16 at 10:16

This question was marked as an exact duplicate of an existing question.

• Didn't you just ask this question? math.stackexchange.com/questions/1799257/… – s.harp May 25 '16 at 11:52
• yes it is, but this time i explained the question, because if you see the previous answers they ask me some clarifications, sorry i didn't know how i do otherwise – Matey Math May 25 '16 at 13:20

If $X$ is a Banach space and $B=\{ e_j \}_{j\in J}$ its Hamel basis and $$P_j (x) =P_j \left( \sum a_j (x) e_j \right) =a_j (x)$$ be a projection on th i-th axis then only for finite number of $j$ the function $P_j$ is continuous.
Proof. Suppose contrary that there is an infinite sequence $\{ v_k \} \subset J$ such that $P_{v_k }$ are continuous. Then for $x=\sum_{k=1}^{\infty} \frac{1}{2^k ||e_{v_k} ||} e_{v_k }$ we have $$P_{v_k } (x) \neq 0$$ for $k=1,2, ...$ but this is imposible since $B$ is he basis. Contradiction.
Now take any $j$ such that $P_j$ is not continuous. Then $P(B) =\{1\}$ so the answer to your question is no.