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The answer is no (a unit is not necessary within the algebra). In the case the product is continuous (i.e. there exists $M>0$ such that, identically, $||xy||\leq M||x||.||y||$ see this question) and of a unital algebra, the common way to construct a multiplicative norm equivalent to the given one is to consider the left-regular representation i.e. $s\rightarrow \gamma(s)$ where $\gamma(s)\in \mathcal{L}_A$ is defined by $\gamma(s)[x]=sx$ and setting the new norm $||s||':=|||\gamma(s)|||$ (the last norm being that of the bounded convergence within $\mathcal{L}_A$). In this respect, the representation $s\rightarrow \gamma(s)$ must be faithful (it is the case, in particular, when $A$ is unital but not only). If it is not, then $||\ ||_1$ is only a seminorm. Following Yemon's comment, in the case when the given algebra is not unital, a way to circumvent this is to consider the left-regular $s\rightarrow \gamma_1(s)$ representation of $A$ on the Banach space $A\oplus\mathbb{C}$ (endowed, for example, with the norm $||x+\lambda||_1=||x||+|\lambda|$). This representation is given by $\gamma_1(s)[x+\lambda]=sx+\lambda.s$. It is an easy exercise to prove that the new seminorm $||s||'':=|||\gamma_1(s)|||$ is, in any case, a norm equivalent to the given one. Hope that it helps.

The answer is no (a unit is not necessary within the algebra). In the case the product is continuous (i.e. there exists $M>0$ such that, identically, $||xy||\leq M||x||.||y||$ see this question) and of a unital algebra, the common way to construct a multiplicative norm equivalent to the given one is to consider the left-regular representation i.e. $s\rightarrow \gamma(s)$ where $\gamma(s)\in \mathcal{L}_A$ is defined by $\gamma(s)[x]=sx$ and setting the new norm $||s||':=|||\gamma(s)|||$ (the last norm being that of the bounded convergence within $\mathcal{L}_A$). In this respect, the representation $s\rightarrow \gamma(s)$ must be faithful (it is the case, in particular, when $A$ is unital but not only). If it is not, then $||\ ||_1$ is only a seminorm. Following Yemon's comment, in the case when the given algebra is not unital, a way to circumvent this is to consider the left-regular $s\rightarrow \gamma_1(s)$ representation of $A$ on the Banach space $A\oplus\mathbb{C}$ (endowed, for example, with the norm $||x+\lambda||_1=||x||+|\lambda|$). This representation is given by $\gamma_1(s)[x+\lambda]=sx+\lambda.s$. It is an easy exercise to prove that the new seminorm $||s||'':=|||\gamma_1(s)|||$ is, in any case, a norm equivalent to the given one. Hope that it helps.

The answer is no (a unit is not necessary within the algebra). In the case the product is continuous (i.e. there exists $M>0$ such that, identically, $||xy||\leq M||x||.||y||$ see A question concerning separate and joint continuity of bilinear maps) and of a unital algebra, the common way to construct a multiplicative norm equivalent to the given one is to consider the left-regular representation i.e. $s\rightarrow \gamma(s)$ where $\gamma(s)\in \mathcal{L}_A$ is defined by $\gamma(s)[x]=sx$ and setting the new norm $||s||':=|||\gamma(s)|||$ (the last norm being that of the bounded convergence within $\mathcal{L}_A$). In this respect, the representation $s\rightarrow \gamma(s)$ must be faithful (it is the case, in particular, when $A$ is unital but not only). If it is not, then $||\ ||_1$ is only a seminorm. Following Yemon's comment, in the case when the given algebra is not unital, a way to circumvent this is to consider the left-regular $s\rightarrow \gamma_1(s)$ representation of $A$ on the Banach space $A\oplus\mathbb{C}$ (endowed, for example, with the norm $||x+\lambda||_1=||x||+|\lambda|$). This representation is given by $\gamma_1(s)[x+\lambda]=sx+\lambda.s$. It is an easy exercise to prove that the new seminorm $||s||'':=|||\gamma_1(s)|||$ is, in any case, a norm equivalent to the given one. Hope that it helps.

The answer is no (a unit is not necessary within the algebra). In the case the product is continuous (i.e. there exists $M>0$ such that, identically, $||xy||\leq M||x||.||y||$ see this question) and of a unital algebra, the common way to construct a multiplicative norm equivalent to the given one is to consider the left-regular representation i.e. $s\rightarrow \gamma(s)$ where $\gamma(s)\in \mathcal{L}_A$ is defined by $\gamma(s)[x]=sx$ and setting the new norm $||s||':=|||\gamma(s)|||$ (the last norm being that of the bounded convergence within $\mathcal{L}_A$). In this respect, the representation $s\rightarrow \gamma(s)$ must be faithful (it is the case, in particular, when $A$ is unital but not only). If it is not, then $||\ ||_1$ is only a seminorm. Following Yemon's comment, in the case when the given algebra is not unital, a way to circumvent this is to consider the left-regular $s\rightarrow \gamma_1(s)$ representation of $A$ on the Banach space $A\oplus\mathbb{C}$ (endowed, for example, with the norm $||x+\lambda||_1=||x||+|\lambda|$). This representation is given by $\gamma_1(s)[x+\lambda]=sx+\lambda.s$. It is an easy exercise to prove that the new seminorm $||s||'':=|||\gamma_1(s)|||$ is, in any case, a norm equivalent to the given one. Hope that it helps.

The answer is no (a unit is not necessary within the algebra). In the case the product is continuous (i.e. there exists $M>0$ such that, identically, $||xy||\leq M||x||.||y||$ see A question concerning separate and joint continuity of bilinear maps) and of a unital algebra, the common way to construct a multiplicative norm equivalent to the given one is to consider the left-regular representation i.e. $s\rightarrow \gamma(s)$ where $\gamma(s)\in \mathcal{L}_A$ is defined by $\gamma(s)[x]=sx$ and setting the new norm $||s||':=|||\gamma(s)|||$ (the last norm being that of the bounded convergence within $\mathcal{L}_A$). In this respect, the representation $s\rightarrow \gamma(s)$ must be faithful (it is the case, in particular, when $A$ is unital but not only). If it is not, then $||\ ||_1$ is only a seminorm. Following Yemon's comment, in the case when the given algebra is not unital, a way to circumvent this is to consider the left-regular $s\rightarrow \gamma_1(s)$ representation of $A$ on the Banach space $A+\mathbb{C}$ (endowed, for example, with the norm $||x+\lambda||_1=||x||+|\lambda|$). This representation is given by $\gamma_1(s)[x+\lambda]=sx+\lambda.s$. It is an easy exercise to prove that the new seminorm $||s||'':=|||\gamma_1(s)|||$ is, in any case, a norm equivalent to the given one. Hope that it helps.

The answer is no (a unit is not necessary within the algebra). In the case the product is continuous (i.e. there exists $M>0$ such that, identically, $||xy||\leq M||x||.||y||$ see A question concerning separate and joint continuity of bilinear maps) and of a unital algebra, the common way to construct a multiplicative norm equivalent to the given one is to consider the left-regular representation i.e. $s\rightarrow \gamma(s)$ where $\gamma(s)\in \mathcal{L}_A$ is defined by $\gamma(s)[x]=sx$ and setting the new norm $||s||':=|||\gamma(s)|||$ (the last norm being that of the bounded convergence within $\mathcal{L}_A$). In this respect, the representation $s\rightarrow \gamma(s)$ must be faithful (it is the case, in particular, when $A$ is unital but not only). If it is not, then $||\ ||_1$ is only a seminorm. Following Yemon's comment, in the case when the given algebra is not unital, a way to circumvent this is to consider the left-regular $s\rightarrow \gamma_1(s)$ representation of $A$ on the Banach space $A\oplus\mathbb{C}$ (endowed, for example, with the norm $||x+\lambda||_1=||x||+|\lambda|$). This representation is given by $\gamma_1(s)[x+\lambda]=sx+\lambda.s$. It is an easy exercise to prove that the new seminorm $||s||'':=|||\gamma_1(s)|||$ is, in any case, a norm equivalent to the given one. Hope that it helps.