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

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.

• If $A$ is a Banach space with a multiplication structure that satisfies $x(y+z)=xy+xz$ and $(y+z)x=yx+zx$ and $\|xy\|\leq\|x\|\|y\|$, $A$ can be embedded in a Banach algebra with unit. This is a well known result. See pp. 228 of FA by Rudin. Further, even if $\|\;\|$ is not submultiplicative, if is left and right continuous ($x\mapsto\|yx\|$, $x\mapsto\|yx\|$ continuous for each $y$) then you there is an equivalent norm that makes the space a Banach algebra. Same reference. Commented Apr 22, 2022 at 18:57
• Yes, theorem 10.2 in Rudin functional analysis book.
– Raju
Commented Dec 22, 2022 at 7:29

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.

• (+1) This is an Interesting and not trivial example where things cannot be fixed in contrast to the rather easy cases of finite dimensional spaces. Commented Apr 22, 2022 at 20:12
• Thanks. The second norm could be made slightly simpler. But this particular norm has an additional feature described in the Remark. Commented Apr 22, 2022 at 20:37
• @RyszardSzwarc Thanks for the example! But isn't convolution continuous? Because I know that the space $\ell(\mathbb{Z})$ is a Banach algebra, so the operation is continuous on this space... But in $X_1$ isn't it? Commented Apr 22, 2022 at 20:41
• It is indeed continuous on $\ell^1(\mathbb{N}_0)$ but with respect to the standard norm there. I have introduced another inequivalent norm on the same space. The convolution turns out to be not continuous with respect to that norm. Commented Apr 22, 2022 at 20:55
• @Mrcrg: interesting indeed. it is the first time that I see a Banach space with a product structure where the product is not continuous, and so, no obvious changed of (equivalent) of norm (an operator norm-like) makes the space a Banach algebra. Like I said before, norms in finite dimensional examples (matrices), cute as they might be, are easy to modify to get Banach algebras for the product continuous in one norm will be continuous in all norms. Commented Apr 22, 2022 at 21:17

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.

• It might be worth mentioning that the failure is not that grave in this example (there exists an equivalent norm that is submultiplicative), similar as picking $\mathbb{A}=\mathbb{R}$ with $\Vert x\Vert =2^{-1} \vert x\vert$. Nice answer (+1) Commented Apr 22, 2022 at 18:22
• @SeverinSchraven Nice suggestion. I've edited my answer. Commented Apr 22, 2022 at 18:29
• @JoséCarlosSantos Thanks for the example, I thought it would be a weirder space. But with Severin's comment, I wonder, is there a space that the norm is not equivalent to one that turns the space into a banach algebra? Commented Apr 22, 2022 at 18:33
• @Mrcrg You need to ensure that multiplication is not jointly continuous (otherwise we find an equivalent norm, see here math.stackexchange.com/questions/1746342/…). I am pretty sure that such a thing exists. Commented Apr 22, 2022 at 18:48
• @JoséCarlosSantos: In finite dimensional algebras, the norm can be modified to make the algebra a Banach algebra. For all practical purposes all finite dimensional normed algebras as Banach algebras. More interesting would be to find a Banach space $(X,\|\;\|)$ (or normed space) with an associative product $\cdot:X\times X\rightarrow X$ that distributes with $+$ form left and right, and which is not left or right continuous for then, $(X,\|\;\|)$ is not a Banach algebra, and there is may not be an equivalent norm $\||\;\||$ that makes $(X,\||\;\||)$ a Banach algebra. Commented Apr 22, 2022 at 19:52

Very interesting construction by Pr. Szwarc indeed, and nice example by José Carlos Santos. This is the only such construction that exists, to my knowledge.

I came here by looking for a counterexample to the absolute summability of a product family in a complete algebra with a non submultiplicative norm. If you have two absolutely summable sequences $$u$$ and $$v$$, you do need a continuous algebraic product for $$(u_p v_q)_{p,q\in \mathbb{N}^2}$$ to be absolutely summable as well, even in ambient completeness.

You might be interested in the counterexample I came up with, now that the heavy lifting of constructing our wishy-washy algebra has been done by Pr. Szwarc :

First I need to slightly modify the new proposed norm (power of $$n$$) : $$\|x\|=\|x\|_1+\sum_{n=1}^\infty n^4\,|x_{2n}|$$

This changes absolutely nothing for the proposed construction, but makes $$\|\delta_{2n}\|_2=\|\delta_{2n}\|=1+n^4$$.

Now consider the sequence of $$\ell^1(\mathbb{N})$$ : $$u := (\delta_1,\delta_3/3^2, \delta_5/5^2, ..)$$ and $$v:=u$$.

$$\|u_n\|_2 = \|\delta_{2n+1}\|_1 / (2n+1)^2 = 1 / (2n+1)^2$$, so $$u$$ is absolutely summable (and $$\|u\|_2$$ sums to $$\pi^2/8$$).

The term of the product family is : $$u_p \star v_q = \delta_{2p+1} \star \delta_{2q+1} / (2p+1)^2(2q+1)^2 = \delta_{2(p+q+1)} / (2p+1)^2(2q+1)^2$$

If the product family $$(u_p \star v_q)_{(p,q) \in \mathbb{N}^2}$$ was absolutely summable, so would be the sub-family indexed by $$\{(p, p-1), p \in \mathbb{N}^*\}$$, but $$\|u_p \star v_{p-1}\|_2 = \|\delta_{4p}\|_2 / (2p+1)^2(2p-1)^2 = (1+2^4 p^4)/(4p^2-1)^2$$ doesn't go to zero with $$p \to \infty$$.

I would be very interested in any similar counterexample, but, again, I never saw an other example of a complete algebra with non submultiplicative norm.