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If $X$ is a Banach space admitting a Schauder basis, then can we choose a set $\{e_1,e_2,e_3 \cdots \}$ as the basis such that there are bounded linear functional $f_i$ such that $f_i(e_j)=\delta_{ij}$?

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Yes, you can; and in fact more is true.

For any basis $(x_n)$ of a Banach space $X$, the coefficient functionals $(x_n^*)$ defined by $x_n^*\bigl(\sum\limits_{i=1}^\infty\alpha_i x_i\bigr)=\alpha_n$ are automatically continuous. Note that $x_n^*(x_m)=\delta_{nm}$. See, e.g., Joseph Diestel, Sequences and Series in Banach Spaces, pg. 33. (More can be said: the sequence $(x_n^*)$ is a basis of its closed linear span.)

The above remark also follows from the fact that for a basis $(x_n)$ of $X$, the projection maps $P_n:X\rightarrow X$ defined by $P_n\bigl(\sum\limits_{i=1}^\infty \alpha_i x_i\bigr)=\sum\limits_{i=1}^n \alpha_i x_i$ are bounded linear operators and $\sup_n\Vert P_n\Vert<\infty$. To see why the remark follows from this fact, note that $x_n^*=P_n-P_{n-1}$. The quantity $\sup_n\Vert P_n\Vert$ is called the basis constant of $(x_n)$.

An outline of one proof of this basic fact concerning the projection maps for bases of Banach spaces is:

Consider $X$ as a sequence space, identifying $x=\sum\limits_{i=1}^\infty \alpha_i x_i$ with its basis representation $(\alpha_i)$ equipped with the norm $\vert\Vert x\Vert\vert = \sup_n\Vert\sum\limits_{i=1}^n \alpha_i x_i \Vert$. Show that this is indeed a well-defined norm on $X$, and that $\Vert x\Vert_X\le \vert\Vert x\Vert\vert$ for each $x\in X$. Then show that $X$ is complete under this norm. The Open Mapping Theorem will show that the norms $\Vert\cdot\Vert$ and $\vert\Vert\cdot\Vert\vert$ are equivalent.

See Diestel, e.g., for details.

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