Lecture 10: Linear Operators

Definition — Bounded Linear Operator

Boundedness and Continuity

Theorem

Let $T\colon V\to W$ be a linear operator, and let $B=\{x\in V\mid \|x\|_1\le 1\}$ be the closed unit ball in $V$. The following statements are equivalent.

$T$ is bounded.
$T(B)$ is a bounded subset of $W$.
$T$ is Lipschitz continuous.
$T$ is continuous at every point of $V$.
$T$ is continuous at $0$.

Proof

$(1)\Rightarrow(2).$ Suppose $T$ is bounded with constant $C>0$. For every $x\in B$, we have

$$ \|T(x)\|_2\le C\|x\|_1\le C. $$

Thus $T(B)$ lies in the closed ball of radius $C$ centered at the origin in $W$.

$(2)\Rightarrow(1).$ Suppose $T(B)$ is bounded. Choose $M>0$ such that $\|T(u)\|_2\le M$ for every $u\in B$. For any nonzero $x\in V$, the vector $u=x/\|x\|_1$ belongs to $B$. By linearity,

$$ \|T(x)\|_2 = \|x\|_1 \left\|T\!\left(\frac{x}{\|x\|_1}\right)\right\|_2 \le M\|x\|_1. $$

The same bound holds at $x=0$ because $T(0)=0$. Hence $T$ is bounded.

$(1)\Rightarrow(3).$ Suppose $T$ is bounded with constant $C>0$. For every $x,y\in V$, linearity gives

$$ \|T(x)-T(y)\|_2 = \|T(x-y)\|_2 \le C\|x-y\|_1. $$

Thus $T$ is Lipschitz continuous with Lipschitz constant $C$.

$(3)\Rightarrow(4).$ Let $L>0$ be a Lipschitz constant for $T$. Fix $x_0\in V$. Given $\epsilon>0$, take $\delta=\epsilon/L$. Whenever $\|x-x_0\|_1\lt\delta$, we have

$$ \|T(x)-T(x_0)\|_2 \le L\|x-x_0\|_1 \lt L\delta =\epsilon. $$

Thus $T$ is continuous at every point of $V$.

$(4)\Rightarrow(5).$ Continuity at every point of $V$ implies continuity at the origin.

$(5)\Rightarrow(1).$ Suppose $T$ is continuous at $0$. Since $T(0)=0$, taking $\epsilon=1$ gives a radius $\delta>0$ such that $\|y\|_1\lt\delta$ implies $\|T(y)\|_2\lt 1$. For any nonzero $x\in V$, set

$$ y=\frac{\delta}{2\|x\|_1}\,x. $$

Then $\|y\|_1=\delta/2\lt\delta$, so $\|T(y)\|_2\lt 1$. Scaling back gives

$$ \|T(x)\|_2 = \frac{2\|x\|_1}{\delta}\,\|T(y)\|_2 \lt \frac{2}{\delta}\|x\|_1. $$

Hence $\|T(x)\|_2\le (2/\delta)\|x\|_1$ for every $x\in V$, including $x=0$. This proves boundedness and completes the proof.

Consequence

A linear operator is unbounded if and only if it is discontinuous at every point. Indeed, if $T$ is continuous at any point $x_0$, then $T(h)=T(x_0+h)-T(x_0)$ tends to zero as $h\to 0$. Thus $T$ is continuous at the origin, and the theorem implies that $T$ is bounded.

Linear Operators on Finite-Dimensional Spaces

Theorem

If $V$ is a finite-dimensional normed space and $W$ is any normed space over the same scalar field, then every linear operator $T\colon V\to W$ is bounded.

Proof

If $V=\{0\}$, the result follows from $T(0)=0$. Otherwise, choose a basis $\{e_1,\ldots,e_n\}$ of $V$. Write any $x\in V$ as $x=\sum_{i=1}^{n}c_i e_i$. By linearity and the triangle inequality,

$$ \begin{aligned} \|T(x)\|_2 &= \left\|\sum_{i=1}^{n}c_iT(e_i)\right\|_2 \\ &\le \sum_{i=1}^{n}|c_i|\,\|T(e_i)\|_2 \\ &\le \left(\max_{1\le i\le n}\|T(e_i)\|_2\right) \sum_{i=1}^{n}|c_i|. \end{aligned} $$

$\max_i\|T(e_i)\|_2$ is a constant but what about the sum? Recall that we need a bound of the form

$$ \begin{aligned} \|T(x)\|_2\le C\|x\|_1 \end{aligned} $$

So take the sum $\sum_{i=1}^{n}|c_i|$. This sum defines a norm on $V$ in the chosen basis. Since all norms on a finite-dimensional space are equivalent, there exists $K>0$ such that

$$ \sum_{i=1}^{n}|c_i|\le K\|x\|_1. $$

for every $x\in V$. Substituting this bound into the estimate gives

$$ \|T(x)\|_2 \le K\left(\max_{1\le i\le n}\|T(e_i)\|_2\right)\|x\|_1. $$

The factor multiplying $\|x\|_1$ is independent of $x$, so $T$ is bounded.

Example 1: The Fourier Coefficient Operator

For a function $f\in L^1[0,2\pi]$, its Fourier coefficient at frequency $n\in\mathbb{Z}$ is

$$ \widehat{f}(n) = \frac{1}{2\pi}\int_0^{2\pi} f(x)e^{-inx}\,dx. $$

Now define the Fourier operator by

$$ \mathcal{F}f=\bigl(\widehat{f}(n)\bigr)_{n\in\mathbb{Z}}. $$

It takes a function and returns its sequence of Fourier coefficients. This operator is linear by linearity of the integral. In the next four examples, we studying different spaces and norms for this operator. To see whether it’s bounded. This depends on the space and the norm equipped. Several choices are discussed in here.

Example 2: Convolution with a Fixed Function

Fix $g\in L^1[0,2\pi]$. Using periodic convolution, define

$$ \begin{align*} T_g\colon L^2[0,2\pi] &\to L^2[0,2\pi] \\ (T_gf)(x) &= (g*f)(x) \\ &= \frac{1}{2\pi}\int_0^{2\pi}g(t)f(x-t)\,dt, \end{align*} $$

where $f$ is extended periodically. Since $g$ is fixed, $T_g$ is linear in $f$.

Young’s Convolution Inequality

If $g\in L^1[0,2\pi]$ and $f\in L^2[0,2\pi]$, then

$g*f\in L^2[0,2\pi]$ and
$\|g*f\|_{L^2}\le\|g\|_{L^1}\|f\|_{L^2}$.

Thus $T_g\colon L^2[0,2\pi]\to L^2[0,2\pi]$ is bounded, since the factor $\|g\|_{L^1}$ is independent of $f$. For example, take $g=F_N$, the Fejér kernel. It is nonnegative and has normalized integral $1$, so $\|F_N\|_{L^1}=1$. Young’s inequality gives

$$ \|F_N*f\|_{L^2}\le\|f\|_{L^2}. $$

Example 3: Differentiation Is Unbounded in the Supremum Norm

Give both the domain and the codomain $C^\infty[0,2\pi]$ the supremum norm. The differentiation operator $Df=f'$ is linear. To test whether it is bounded, take $f_n(x)=\sin(nx)$ for $n\ge 1$. Then

$$ \begin{aligned} \|f_n\|_\infty &= \sup_{x\in[0,2\pi]}|\sin(nx)|=1, \\ \|Df_n\|_\infty &= \sup_{x\in[0,2\pi]}|n\cos(nx)|=n. \end{aligned} $$

If $D$ were bounded with constant $C$, we would have $n=\|Df_n\|_\infty\le C\|f_n\|_\infty=C$ for every $n\ge 1$. Since $n$ can be arbitrarily large, $D$ is unbounded.

Example 4 — Integration Is Bounded in the Supremum Norm

On the space of complex-valued continuous functions on $[0,1]$, define the linear functional

$$ T\colon C[0,1]\to\mathbb{C}, \qquad T(f)=\int_0^1 f(x)\,dx. $$

For every $f\in C[0,1]$, we have

$$ |T(f)| = \left|\int_0^1 f(x)\,dx\right| \le \int_0^1|f(x)|\,dx \le \int_0^1\|f\|_\infty\,dx = \|f\|_\infty. $$

Thus $T$ is bounded with constant $1$.

References

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