is $\ell_{2}$ space Banach algebra? Is $\ell_{2}$ space Banach algebra ? By $\ell_{2}$, we mean the space of all sequences of complex
numbers $( x_{n})$ for which $\sum|x_{k}|^{2}&lt;\infty$.
The space l₂ is closed under multiplication. For let x, y be in this space. Then there is N such that ∣x_{i}∣,∣y_{i}∣&lt;1,i≥N. Then ∑(x_{i}y_{i})²=∑_{i≥N}(x_{i}y_{i})²+∑_{i&lt;N}(x_{i}y_{i})²
&lt;∑_{i&lt;N}(x_{i}y_{i})²+∑_{i≥N}∣x_{i}y_{i}∣. For positive a,b we have a^{1/2}b^{1/2}≤(a+b)/2. Thus 
((∣x_{i}y_{i}∣)/(∥x∥∥y∥))≤((∣x_{i}∣²/∥x∥²+∣y_{i}∣²/∥y∥²)/2). A little work shows that ∑_{i&lt;N}(x_{i}y_{i})²+∑_{i≥N}∣x_{i}y_{i}∣&lt;∞. Minkowwski's inequality holds. Thus the space is an algebra. But it does not contain 1 and so it is not a Banach algebra.
The space l₂ is closed under multiplication. For let x, y be in this space. Then there is N such that ∣x_{i}∣,∣y_{i}∣&lt;1,i≥N. Then ∑(x_{i}y_{i})²=∑_{i≥N}(x_{i}y_{i})²+∑_{i&lt;N}(x_{i}y_{i})²
&lt;∑_{i&lt;N}(x_{i}y_{i})²+∑_{i≥N}∣x_{i}y_{i}∣. For positive a,b we have a^{1/2}b^{1/2}≤(a+b)/2. Thus 
((∣x_{i}y_{i}∣)/(∥x∥∥y∥))≤((∣x_{i}∣²/∥x∥²+∣y_{i}∣²/∥y∥²)/2). A little work shows that ∑_{i&lt;N}(x_{i}y_{i})²+∑_{i≥N}∣x_{i}y_{i}∣&lt;∞. Minkowwski's inequality holds. Thus the space is an algebra. But it does not contain 1 and so it is not a Banach algebra.