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Let $V$ be a Banach separable space over $\mathbb{C}$

Let $B$ a subset of $V$ such that is linearly independent, infinite and $\forall b \in B: \left\|b\right\|=1$

Let $U=\operatorname{span}(B)$

Let $\{v_m\} \in U$ be a sequence of $U$ with $$ v_m = \sum_j a_{m,j}b_{m,j} $$ and $a_{m,j} \in \mathbb{C}$ and $b_{m,j} \in B$ and the sum is finite and such that $$ \lim_{m \to \infty} v_m = b $$ with $b \in B$

My question is if $|a_{m,j}|$ is limited, that is $\exists M>0 : \forall m,j |a_{m,j}| < M$

Thanks.

Matey Math
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1 Answers1

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The answer is no:

Let $V = (L^1[0,1], \|\cdot\|_1)$ be our separable Banach space and $B = \{1, 2x, 3x^2, 4x^3\ldots\} \subseteq V$. Notice that $B$ is linearly independent, and $\left\|nx^{n-1}\right\|_1 = 1$ for $n\in\mathbb{N}$.

Consider the sequence $(f_n)_{n=1}^\infty \subseteq \mathrm{span}\,B$ defined as $f_n(x) = 1 + (1-x)^n$, $x \in [0,1]$.

We have $f_n \xrightarrow{n\to\infty} 1 \in B$:

$$\|f_n - 1\|_1 = \int_0^1 (1-x)^n\,dx = \frac{1}{n+1} \xrightarrow{n\to\infty} 0$$

However, we have

$$1 + (1-x)^n = 1 + \sum_{k=0}^n {n\choose k}(-1)^kx^k$$

so the coefficient of the element $2x$ is $-\frac{n}2$, which is unbounded as $n\to\infty$.

mechanodroid
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  • thanks @mechanodroid again and again and again , but does it exit any theorem such that coefficients are bounded? please read tha comment below my previous question https://math.stackexchange.com/questions/2454115/in-a-sequence-in-a-normed-space-are-the-coefficientes-bounded – Matey Math Oct 02 '17 at 12:50
  • @MateyMath I think it has something to do with the fact that coefficient functionals of Hamel bases can be unbounded: look here. – mechanodroid Oct 02 '17 at 12:54
  • but if $v_m \to 0$ and $\left| b \right| = 1$, are coefficients bounded? – Matey Math Oct 02 '17 at 13:04
  • @MateyMath Again, no: take $f_n(x) = (1-x)^n \xrightarrow{n\to\infty} 0$, but the coefficient of $2x$ is still $-\frac{n}2$. – mechanodroid Oct 02 '17 at 13:09
  • thanks @mechanodroid for your helping so much – Matey Math Oct 02 '17 at 14:04