[proofplan]
We prove submultiplicativity by estimating $\|(T \circ S)x\|_Z$ for unit vectors $x$ using the operator norm definitions of $T$ and $S$ separately. The composition $T \circ S$ is linear (as a composition of linear maps) and bounded (by the product of norms), establishing $T \circ S \in \mathcal{L}(X, Z)$. For the Banach algebra statement, we combine submultiplicativity with the [Completeness of the Operator Space](/theorems/1053) and verify the identity element.
[/proofplan]
[step:Establish that $T \circ S$ is a bounded linear operator from $X$ to $Z$]
The composition $T \circ S: X \to Z$ is linear: for all $x_1, x_2 \in X$ and $\lambda \in \mathbb{R}$ (or $\mathbb{C}$),
\begin{align*}
(T \circ S)(\lambda x_1 + x_2) &= T(S(\lambda x_1 + x_2)) \\
&= T(\lambda S(x_1) + S(x_2)) \\
&= \lambda T(S(x_1)) + T(S(x_2)) \\
&= \lambda (T \circ S)(x_1) + (T \circ S)(x_2),
\end{align*}
where the second equality uses the linearity of $S$ and the third uses the linearity of $T$.
To show boundedness, let $x \in X$ with $\|x\|_X \le 1$. Since $S \in \mathcal{L}(X, Y)$, we have $\|Sx\|_Y \le \|S\|_{\mathcal{L}(X, Y)} \|x\|_X \le \|S\|_{\mathcal{L}(X, Y)}$. Since $T \in \mathcal{L}(Y, Z)$, applying $T$ to the element $Sx \in Y$ gives
\begin{align*}
\|(T \circ S)x\|_Z = \|T(Sx)\|_Z \le \|T\|_{\mathcal{L}(Y, Z)} \|Sx\|_Y \le \|T\|_{\mathcal{L}(Y, Z)} \cdot \|S\|_{\mathcal{L}(X, Y)}.
\end{align*}
The right-hand side is finite and independent of $x$, so $T \circ S$ is bounded. Hence $T \circ S \in \mathcal{L}(X, Z)$.
[guided]
We must first verify that $T \circ S$ is a well-defined element of $\mathcal{L}(X, Z)$, meaning it is both linear and bounded as a map from $X$ to $Z$.
**Linearity.** The composition of two linear maps is linear. We write this out explicitly: for $x_1, x_2 \in X$ and scalar $\lambda$,
\begin{align*}
(T \circ S)(\lambda x_1 + x_2) &= T(S(\lambda x_1 + x_2)) \\
&= T(\lambda S(x_1) + S(x_2)) \quad \text{(linearity of $S$)} \\
&= \lambda T(S(x_1)) + T(S(x_2)) \quad \text{(linearity of $T$)} \\
&= \lambda (T \circ S)(x_1) + (T \circ S)(x_2).
\end{align*}
**Boundedness.** We need to show $\sup_{\|x\|_X \le 1} \|(T \circ S)x\|_Z < \infty$. The strategy is to apply the definition of the operator norm twice — once for $S$ and once for $T$ — in a chain:
\begin{align*}
\|(T \circ S)x\|_Z = \|T(Sx)\|_Z \le \|T\|_{\mathcal{L}(Y,Z)} \cdot \|Sx\|_Y \le \|T\|_{\mathcal{L}(Y,Z)} \cdot \|S\|_{\mathcal{L}(X,Y)} \cdot \|x\|_X.
\end{align*}
The first inequality uses the defining property of the operator norm $\|T\|_{\mathcal{L}(Y,Z)} = \sup_{\|y\|_Y \le 1} \|Ty\|_Z$, which implies $\|Ty\|_Z \le \|T\|_{\mathcal{L}(Y,Z)} \|y\|_Y$ for all $y \in Y$, applied with $y = Sx$. The second inequality uses the analogous property for $S$. Since $\|T\|_{\mathcal{L}(Y,Z)} \cdot \|S\|_{\mathcal{L}(X,Y)} < \infty$ (both $T$ and $S$ are bounded), the map $T \circ S$ is bounded.
[/guided]
[/step]
[step:Take the supremum over unit vectors to obtain the submultiplicative bound]
From the estimate in the previous step, for every $x \in X$ with $\|x\|_X \le 1$,
\begin{align*}
\|(T \circ S)x\|_Z \le \|T\|_{\mathcal{L}(Y, Z)} \cdot \|S\|_{\mathcal{L}(X, Y)}.
\end{align*}
Taking the supremum over all such $x$:
\begin{align*}
\|T \circ S\|_{\mathcal{L}(X, Z)} = \sup_{\substack{x \in X \\ \|x\|_X \le 1}} \|(T \circ S)x\|_Z \le \|T\|_{\mathcal{L}(Y, Z)} \cdot \|S\|_{\mathcal{L}(X, Y)}.
\end{align*}
This establishes the submultiplicativity inequality.
[/step]
[step:Verify that $\mathcal{L}(X)$ is a unital Banach algebra when $X$ is a Banach space]
When $X = Y = Z$, the space $\mathcal{L}(X) := \mathcal{L}(X, X)$ is a normed vector space (with the operator norm) that is complete by the [Completeness of the Operator Space](/theorems/1053) (since $X$ is a Banach space, it serves as both the domain and the complete codomain). The submultiplicativity $\|T \circ S\|_{\mathcal{L}(X)} \le \|T\|_{\mathcal{L}(X)} \cdot \|S\|_{\mathcal{L}(X)}$ gives $\mathcal{L}(X)$ the structure of a Banach algebra under composition.
The identity element is $\mathrm{Id}_X: X \to X$, $x \mapsto x$. We verify $\|\mathrm{Id}_X\|_{\mathcal{L}(X)} = 1$ when $X \neq \{0\}$:
\begin{align*}
\|\mathrm{Id}_X\|_{\mathcal{L}(X)} = \sup_{\substack{x \in X \\ \|x\|_X \le 1}} \|\mathrm{Id}_X(x)\|_X = \sup_{\substack{x \in X \\ \|x\|_X \le 1}} \|x\|_X = 1,
\end{align*}
where the supremum equals $1$ because $X \neq \{0\}$ guarantees the existence of $x \in X$ with $\|x\|_X = 1$ (take any nonzero $x_0 \in X$ and set $x = x_0 / \|x_0\|_X$). Thus $\mathcal{L}(X)$ is a unital Banach algebra with identity $\mathrm{Id}_X$.
[guided]
A **Banach algebra** is a Banach space $\mathcal{A}$ equipped with an associative, bilinear multiplication satisfying $\|ab\| \le \|a\| \cdot \|b\|$ for all $a, b \in \mathcal{A}$. A **unital** Banach algebra additionally has a multiplicative identity $e$ with $\|e\| = 1$.
We verify that $\mathcal{L}(X)$ satisfies each requirement:
1. **Complete normed space.** By the [Completeness of the Operator Space](/theorems/1053), $\mathcal{L}(X, Y)$ is a Banach space whenever $Y$ is a Banach space. Taking $Y = X$ (which is Banach by hypothesis), $\mathcal{L}(X) = \mathcal{L}(X, X)$ is a Banach space.
2. **Associative bilinear multiplication.** Composition of linear maps is associative: $(T \circ S) \circ R = T \circ (S \circ R)$ for all $T, S, R \in \mathcal{L}(X)$. Bilinearity follows from the linearity of each factor: $T \circ (\lambda S_1 + S_2) = \lambda(T \circ S_1) + (T \circ S_2)$ and $(\lambda T_1 + T_2) \circ S = \lambda(T_1 \circ S) + (T_2 \circ S)$.
3. **Submultiplicativity.** This is exactly the inequality established above: $\|T \circ S\|_{\mathcal{L}(X)} \le \|T\|_{\mathcal{L}(X)} \cdot \|S\|_{\mathcal{L}(X)}$.
4. **Identity element.** The identity map $\mathrm{Id}_X: X \to X$, $x \mapsto x$ satisfies $T \circ \mathrm{Id}_X = \mathrm{Id}_X \circ T = T$ for all $T \in \mathcal{L}(X)$. To compute its norm when $X \neq \{0\}$:
\begin{align*}
\|\mathrm{Id}_X\|_{\mathcal{L}(X)} = \sup_{\substack{x \in X \\ \|x\|_X \le 1}} \|x\|_X.
\end{align*}
The supremum is at most $1$ (since $\|x\|_X \le 1$ in the constraint). It equals $1$ because $X \neq \{0\}$: pick any nonzero $x_0 \in X$ and set $x = x_0/\|x_0\|_X$, which satisfies $\|x\|_X = 1$, attaining the supremum.
[/guided]
[/step]
