Example of an algebra that is a Banach space but not a Banach algebra

Question

I'm looking for an example of a space $\mathbb{A} $ such,

• $\mathbb{A} $ is an algebra;
• $\mathbb{A}$ is equipped with a norm that makes it a Banach space;
• $\mathbb{A}$ is not a Banach algebra, i.e., the norm is not submultiplicative.

I have not been able to think of any examples for this case, but I believe there must be a space satisfying this.

And if I assume that $\mathbb{A}$ has unity, is it still possible to find any examples?

Edit: The example given by José made me question whether it is possible to construct an example in which the norm is not equivalent to another norm that turns the space into a Banach algebra.

Answer 1

We are going to construct an algebra, which is a Banach space, but multiplication is not continuous. Therefore it cannot be a Banach algebra with respect to any norm equivalent to the original norm. The plan is to start with a Banach algebra $\ell^1(\mathbb{N}_0)$ and introduce a new norm, such that the space is complete with this norm and the convolution is not continuous with respect to the new norm.

Let $X=\mathcal{F}$ denote the space of sequences with finitely many nonzero terms, equipped with two norms $$\|x\|_1=\sum_{n=0}^\infty |x_n|,\qquad \|x\|=\|x\|_1+\sum_{n=1}^\infty n\,|x_{2n}|$$ We complete the space with respect to both norms obtaining separable Banach spaces $X_1=\ell^1$ and $X_2.$ We have $$X_1=X\oplus Y_1,\qquad X_2=X\oplus Y_2$$ for subspaces $Y_1\subset X_1$ and $Y_2\subset X_2.$ The Hamel bases of $Y_1$ and $Y_2$ have the same cardinality (continuum). Therefore there exists an algebraic isomorphism $\varphi:Y_1\to Y_2.$ Then the mapping $\Phi:X_1\to X_2$ defined by $$\Phi(x+y)=x+\varphi(y),\quad x\in X,\ y\in Y_1$$ is an algebraic isomorphism from $X_1$ to $X_2.$ Define a new norm on $X_1$ by $\|x\|_2=\|\Phi(x)\|.$ Then $X_1$ is complete with respect to this norm, as $(X_1,\|\cdot\|_2)$ is isometrically isomorphic with $(X_2,\|\cdot\|).$ The space $(X_1,\|\cdot\|_1)$ is a Banach algebra with convolution of sequences. But $(X_1,\|\cdot\|_2)$ is not a Banach algebra with the same operation. Indeed, by definition $\Phi(x)=x $ for $x\in X.$ Let $\{\delta_n\}_{n=1}^\infty$ be the standard basis sequence in $X.$ Then $$\displaylines{ \|\delta_{2n+1}\|_2=\|\delta_{2n+1}\|=1\\ \|\delta_{2n}\|_2=\|\delta_{2n}\|=n+1}$$ We have $\delta_1*\delta_{2n-1}=\delta_{2n}.$ Therefore convolution with $\delta_1$ is unbounded with respect to $\|\cdot\|_2$ norm.

Answer 2

Let $\Bbb A$ be the algebra of all $2\times2$ real matrices. Consider the norm$$\left\|\begin{bmatrix}a&b\\c&d\end{bmatrix}\right\|=\max\{|a|,|b|,|c|,|d|\}.$$Then $(\Bbb A,\|\cdot\|)$ is a Banach space, but it's not a Banach algebra. For instance, $A=\left[\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right]$ has $\|A\|=1$, but$$\left\|A^2\right\|=\left\|\begin{bmatrix}1&2\\0&1\end{bmatrix}\right\|=2.$$Note that if you consider instead the norm$$\left\|A\right\|_{\text{op}}=\max\{\|A.v\|\mid\|v\|=1\}$$(where in the RHS $\|\cdot\|$ is any norm defined on $\Bbb R^2$), then $\|\cdot\|_{\text{op}}$ is equivalent to $\|\cdot\|$, but $(\Bbb A,\|\cdot\|_{\text{op}})$ is a Banach algebra.