Lecture 11: The Space of Bounded Linear Operators

Recall that a linear operator $T\colon V_1\to V_2$ is bounded if there exists a constant $B\ge 0$ such that

$$ \|Tx\|_2\le B\|x\|_1 $$

for every $x\in V_1$. The operator norm gives the smallest such constant.

Definition — Operator Norm

Let $T\colon (V_1,\|\cdot\|_1)\to(V_2,\|\cdot\|_2)$ be a bounded linear operator. Its operator norm is \[ \displaystyle \|T\|=\sup_{\|x\|_1\le 1}\|Tx\|_2. \] When $V_1\ne\{0\}$, this can also be written as \[ \displaystyle \|T\|=\sup_{x\ne 0}\frac{\|Tx\|_2}{\|x\|_1} \]

For each nonzero $x\in V_1$, the ratio $\|Tx\|_2/\|x\|_1$ is at most its supremum. Multiplying by $\|x\|_1$ gives

$$ \|Tx\|_2\le \|T\|\,\|x\|_1. $$

This inequality also holds at $x=0$ because $T0=0$. Any constant $B$ that bounds all the ratios must also bound their supremum, so $\|T\|$ is the smallest possible bound.

Completeness of the Space of Bounded Linear Operators

Theorem

Let $V_1$ and $V_2$ be normed spaces over the same scalar field, and let $\mathcal{B}(V_1,V_2)$ denote the space of bounded linear operators from $V_1$ to $V_2$. If $V_2$ is a Banach space, then $\mathcal{B}(V_1,V_2)$, equipped with the operator norm, is a Banach space.

Proof

1. The bounded linear operators form a vector space.

For $S,T\in\mathcal{B}(V_1,V_2)$ and a scalar $\alpha$, define addition and scalar multiplication pointwise:

$$ (S+T)x=Sx+Tx, \qquad (\alpha T)x=\alpha(Tx). $$

Both operators are linear. To check boundedness, let $x\in V_1$. The triangle inequality gives

$$ \|(S+T)x\|_2 \le \|Sx\|_2+\|Tx\|_2 \le (\|S\|+\|T\|)\|x\|_1. $$

Similarly, $\|(\alpha T)x\|_2\le|\alpha|\,\|T\|\,\|x\|_1$. Thus bounded linear operators are closed under addition and scalar multiplication. They also include the zero operator, so they form a vector space.

2. The operator norm satisfies the norm axioms.

The operator norm is nonnegative. If $\|T\|=0$, then $\|Tx\|_2\le\|T\|\|x\|_1=0$ for every $x\in V_1$. Thus $Tx=0$ for every $x$, so $T$ is the zero operator. Conversely, the zero operator has norm zero.

Absolute homogeneity follows by taking the supremum over the closed unit ball:

$$ \|\alpha T\| = \sup_{\|x\|_1\le 1}|\alpha|\,\|Tx\|_2 = |\alpha|\,\|T\|. $$

Taking the same supremum in the estimate for $S+T$ gives the triangle inequality $\|S+T\|\le\|S\|+\|T\|$.

3. Construct the limit operator.

Let $(T_n)$ be a Cauchy sequence in the operator norm. Fix $x\in V_1$. By the operator norm inequality,

$$ \|T_nx-T_mx\|_2 = \|(T_n-T_m)x\|_2 \le \|T_n-T_m\|\,\|x\|_1. $$

Since $x$ is fixed and $(T_n)$ is Cauchy, the right-hand side tends to zero as $n,m\to\infty$. Thus $(T_nx)$ is Cauchy in $V_2$. This is where we use completeness of $V_2$: for each $x$, the sequence $(T_nx)$ has a limit in $V_2$. Define $T\colon V_1\to V_2$ by

$$ Tx=\lim_{n\to\infty}T_nx. $$

To check linearity, let $x,y\in V_1$ and let $\alpha,\beta$ be scalars. Each $T_n$ is linear, so

$$ \begin{aligned} T(\alpha x+\beta y) &= \lim_{n\to\infty}T_n(\alpha x+\beta y) \\ &= \lim_{n\to\infty}(\alpha T_nx+\beta T_ny) \\ &= \alpha Tx+\beta Ty. \end{aligned} $$

4. Show that the limit operator is bounded.

We need one constant that bounds $\|Tx\|_2$ by a multiple of $\|x\|_1$ for every $x$. Since $(T_n)$ is Cauchy, choose $N_0$ such that $\|T_n-T_{N_0}\|\lt 1$ for every $n\ge N_0$. Then

$$ \|T_n\| \le \|T_n-T_{N_0}\|+\|T_{N_0}\| \lt 1+\|T_{N_0}\|. $$

The finitely many earlier terms also have finite norms. Hence $B=\sup_{n\in\mathbb{N}}\|T_n\|\lt\infty$, and for every $x\in V_1$,

$$ \|T_nx\|_2\le \|T_n\|\,\|x\|_1\le B\|x\|_1. $$

Since $T_nx\to Tx$, the reverse triangle inequality gives

$$ \bigl|\|T_nx\|_2-\|Tx\|_2\bigr| \le \|T_nx-Tx\|_2\to 0. $$

Taking the limit in the bound yields $\|Tx\|_2\le B\|x\|_1$. Thus $T\in\mathcal{B}(V_1,V_2)$.

5. Prove convergence in the operator norm.

So far we know that $T_nx\to Tx$ for each fixed $x$. To prove operator norm convergence, we need a single estimate that holds for every $x$ in the closed unit ball.

Given $\epsilon>0$, choose $N$ such that $\|T_n-T_m\|\lt\epsilon/2$ whenever $n,m\ge N$. For every $x$ with $\|x\|_1\le 1$, we then have

$$ \|T_nx-T_mx\|_2 \le \|T_n-T_m\|\,\|x\|_1 \lt \frac{\epsilon}{2}. $$

Fix $n\ge N$ and let $m\to\infty$. Since $T_mx\to Tx$, we obtain $\|T_nx-Tx\|_2\le\epsilon/2$. The same $N$ works for every $x$ in the closed unit ball, so we can take the supremum:

$$ \|T_n-T\| = \sup_{\|x\|_1\le 1}\|T_nx-Tx\|_2 \le \frac{\epsilon}{2} \lt \epsilon. $$

Thus $T_n\to T$ in the operator norm. Every Cauchy sequence in $\mathcal{B}(V_1,V_2)$ therefore converges to an element of that space, proving that $\mathcal{B}(V_1,V_2)$ is a Banach space.

References

SFSU Fourier Analysis by Professor Chun-Kit Lai
Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi