Is the set of bounded operators a linear space?

Question

Let $H_1$ and $H_2$ two Hilbert Spaces. Is the set of bounded operators from $H_1$ to $H_2$ a linear space? My first feeling is that it is true, but I don’t know how to prove it. Could anyone help or give me some references?

Thank you!

Answer 1

Let $X$ and $Y$ be vector spaces over the same field $F$, and let $\mbox{ Hom }(X, Y)$ denote the set of all the linear transformations $T \colon X \rightarrow Y$, that is, let $$ \mbox{ Hom } (X, Y) \colon= \big\{ \, T \colon X \rightarrow Y \, \colon \, T \mbox{ is linear } \, \big\}. $$ Then $\mbox{ Hom } (X, Y)$ is also a vector space over the field $F$ under the operations $+_{ \mbox{ Hom } (X, Y) }$ and scalar multiplication defined as follows:

Let $S, T \in \mbox{ Hom } (X, Y)$. Then $S \colon X \rightarrow Y$ and $T \colon X \rightarrow Y$ are both linear transformations. Now let $S +_{ \mbox{ Hom } (X, Y) } T \colon X \rightarrow Y$ be the mapping defined by $$ \left( S +_{ \mbox{ Hom } (X, Y) } T \right) (x) \colon= S(x) +_Y T(x) \ \mbox{ for } x \in X. \tag{Formula for Sum} $$ We can easily show that the mapping $S +_{ \mbox{ Hom } (X, Y) } T \colon X \rightarrow Y$ is a linear transformation.

Now let $T \in \mbox{ Hom }(X, Y)$ and let $\alpha \in F$. Then the mapping $\alpha T \colon X \rightarrow Y$ defined by the formula $$ \big( \alpha T \big)(x) \colon= \alpha T(x) \ \mbox{ for } x \in X. \tag{Formula for Scalar Multiple} $$ The scalar multiplication on the right-hand-side of this formula of course takes place in our vector space $Y$.

Under the above operations of vector addition and scalar multiplication $\mbox{ Hom }(X, Y)$ is a vector space over $F$.

[ If you want a detailed proof, then please contact me directly.]

Now let $X$ and $Y$ be normed spaces, either both real or both complex. Then a linear transformation $T \colon X \rightarrow Y$ is said to be a bounded linear operator if there exists a real number $\lambda_T$ such that $$ \big\lVert T(x) \big\rVert_Y \leq \lambda_T \lVert x \rVert_X \ \mbox{ for all } x \in X. \tag{Definition of Boundedness} $$ Such a real number $\lambda_T$, if it exists, is of course non-negative. And, if $X \neq \left\{ \mathbf{0}_X \right\}$, then in this case the set $$ \left\{ \, \frac{ \big\lVert T(x) \big\rVert_Y }{ \lVert x \rVert_X } \, \colon \, x \in X, x \neq \mathbf{0}_X \, \right\} \tag{A} $$ is a non-empty bounded subset of $\mathbb{R}$; in fact $\lambda_T$ is an upper bound of this set.

Let $B(X, Y)$ denote the set of all the bounded linear operators $T \colon X \rightarrow Y$, where $X$ and $Y$ are normed spaces, either both real or both complex. Then $B(X, Y)$ is a subspace of $\mbox{ Hom }(X, Y)$.

The zero mapping $\mathbf{0}_{ \mbox{ Hom }(X, Y) } \colon X \rightarrow Y$ defined by $$ \mathbf{0}_{ \mbox{ Hom }(X, Y) } (x) \colon= \mathbf{0}_Y \ \mbox{ for all } x \in X $$ is a bounded linear operator as per (Definition of Boundedness) above.

If $S, T \in B(X, Y)$. Then there exist real numbers $\lamba_S$ and $\lambda_T$ such that $$ \big\lVert T(x) \big\rVert_Y \leq \lambda_T \lVert x \rVert_X \ \mbox{ for all } x \in X, $$ and $$ \big\lVert T(x) \big\rVert_Y \leq \lambda_T \lVert x \rVert_X \ \mbox{ for all } x \in X. $$ Therefore for all $x \in X$, we obtain $$ \begin{align} \left\lVert \left( S +_{ \mbox{ Hom }(X, Y) } T \right)(x) \right\rVert_Y &= \left\lVert S(x) +_Y T(x) \right\rVert_Y \\ &\leq \big\lVert S(x) \big\rVert_Y + \big\lVert T(x) \big\rVert_Y \\ &\leq \lambda_S \lVert x \rVert_X + \lambda_T \lVert x \rVert_X \\ &= \left( \lambda_S + \lambda_T \right) \lVert x \rVert_X. \end{align} $$ Thus we obtain $$ \left\lVert \left( S +_{ \mbox{ Hom }(X, Y) } T \right)(x) \right\rVert_Y \leq \left( \lambda_S + \lambda_T \right) \lVert x \rVert_X \ \mbox{ for all } x \in X, \tag{1} $$ which shows that $S +_{ \mbox{ Hom }(X, Y) } T \colon X \rightarrow Y$ is a bounded linear operator, that is, $S +_{ \mbox{ Hom }(X, Y) } T \in B(X, Y)$.

And, if $T \in B(X, Y)$ and $\alpha$ is a scalar, then from (Definition of Boundedness) we obtain for any $x \in X$, $$ \begin{align} \big\lVert \big( \alpha T \big)(x) \big\rVert_Y &= \big\lVert \alpha T(x) \big\rVert \\ &= \lvert \alpha \rvert \, \lVert T(x) \rVert_Y \\ &\leq \lvert \alpha \rvert \, \lambda_T \lVert x \rVert_X. \end{align} $$ Thus we have a real number $\lvert \alpha \rvert \times \lambda_T$ such that $$ \big\lVert \big( \alpha T \big)(x) \big\rVert_Y \leq \big( \lvert \alpha \rvert \, \lambda_T \big) \times \lVert x \rVert_X \ \mbox{ for all } x \in X. \tag{2} $$ Thus $\alpha T \colon X \rightarrow Y$ is also a bounded linear operator, that is, $\alpha T \in B(X, Y)$.

Thus the set $B(X, Y)$ is a subset of $\mbox{ Hom }(X, Y)$, and this subset contains the zero vector of the vector space $\mbox{ Hom }(X, Y)$ and is also closed under the operations of vector addition and scalar multiplication for $\mbox{ Hom }(X, Y)$. Therefore $B(X, Y)$ is a (vector) subspace of $\mbox{ Hom }(X, Y)$.

For any $T \in B(X, Y)$, let us define $$ \big\lVert T \big\rVert_{B(X, Y)} \colon= \sup \left\{ \, \frac{ \big\lVert T(x) \big\rVert_Y }{ \lVert x \rVert_X } \, \colon \, x \in X, x \neq \mathbf{0}_X \, \right\} \tag{Definition of Norm} $$ Then we have the inequality $$ \big\lVert T(x) \big\rVert_Y \leq \big\lVert T \big\rVert_{B(X, Y)} \, \lVert x \rVert_X \ \mbox{ for all } x \in X. \tag{Norm Inequality} $$ And, it can also be shown that $$ \big\lVert T \big\rVert_{B(X, Y)} = \sup \left\{ \, \big\lVert T(x) \big\rVert_Y \, \colon \, x \in X, \lVert x \rVert_X = 1 \, \right\}. \tag{Formula II for Norm} $$

If $S, T \in B(X, Y)$, then using (Norm Inequality) and the same procedure as used for deriving (1) above, we obtain $$ \big\lVert S +_{\mbox{ Hom }(X, Y)} T \big\rVert_{B(X, Y)} \leq \big\lVert S \big\rVert_{B(X, Y)} + \big\lVert T \big\rVert_{B(X, Y)}. $$

And, if $T \in B(X, Y)$ and $\alpha$ is a scalar, then we can use the procedure adopted for deriving (2) above along with (Norm Inequality) above to obtain $$ \big\lVert \alpha T \big\rVert_{B(X, Y)} = \lvert \alpha \rvert \, \big\lVert T \big\rVert_{B(X, Y)}. $$

Thus if $X$ and $Y$ are any normed spaces, either both real or both complex, then the set $B(X, Y)$ of all the bounded linear operaotrs $T \colon X \rightarrow Y$ is also a normed space with the norm defined by (Formula for Norm) or (Formula II for Norm) above.

Please also refer to Sec. 2.7 in the book Introductory Functional Analysis With Applications by Erwine Kreyszig.