Bounded Linear Operators

From Linear Maps to Operator Bounds

A linear map between normed spaces is bounded when one constant controls the size of every output in terms of the size of its input. That uniform control is useful in estimates: it turns information about an input into information about its image, regardless of which vector is chosen. In earlier tutorials, the boundedness criterion and the operator norm were established. Here we use them to study concrete operators on spaces of continuous functions.

For normed spaces $V$ and $W$, a linear map $T:V\to W$ is bounded if there is a finite constant $C\geq0$ such that $\|T(x)\|_W\leq C\|x\|_V$ for every $x\in V$. The operator norm $\|T\|$, when $T$ is bounded, is the least such uniform bound. Both norms matter: the same algebraic rule can be bounded for one choice of norms and unbounded for another.

Key idea: To prove an operator is bounded, find a constant independent of the input that bounds the output norm by the input norm. To prove it is unbounded, it is enough to find inputs whose norms stay controlled while their output norms grow without bound.

The examples below use $C[a,b]$, the space of continuous real-valued functions on $[a,b]$, equipped with the supremum norm $\|f\|_\infty=\sup_{x\in[a,b]}|f(x)|$. For nonempty compact intervals, this is a norm and $C[a,b]$ is complete, as established earlier in the course. We will also use the fact that a continuous function on a compact set is bounded and uniformly continuous.

Continuous Kernels Give Bounded Integral Operators

An integral operator takes a function as input and returns a new function. A common form is to integrate the input against a kernel $k(x,t)$, where $t$ is the variable of integration and $x$ is the variable of the output. Continuity of the kernel gives enough control both to ensure that the output is continuous and to bound its size.

Theorem (Boundedness of a Continuous-Kernel Integral Operator): Let $a<b$ and $c<d$, and let $k:[c,d]\times[a,b]\to\mathbb{R}$ be continuous. Define $T:C[a,b]\to C[c,d]$ by $$ (Tf)(x)=\int_a^b k(x,t)f(t)\,dt. $$ Then $T$ is a bounded linear map, and $$ \|T\|\leq (b-a)\sup_{\substack{x\in[c,d]\\t\in[a,b]}}|k(x,t)|. $$

Proof. The kernel $k$ is continuous on the compact rectangle $[c,d]\times[a,b]$, so it is bounded there. Set $K=\sup\{|k(x,t)|:x\in[c,d],\,t\in[a,b]\}$, which is finite. For each $f\in C[a,b]$ and fixed $x\in[c,d]$, the function $t\mapsto k(x,t)f(t)$ is continuous. Its Riemann integral therefore exists.

We first check that $Tf$ is continuous. If $f$ is identically zero, then $Tf$ is identically zero. Otherwise, fix $\varepsilon>0$. By uniform continuity of $k$ on its compact rectangle, there is a $\delta>0$ such that whenever $x,y\in[c,d]$ and $|x-y|<\delta$,

|k(x,t)-k(y,t)|< \frac{\varepsilon}{(b-a)(\|f\|_\infty+1)} \qquad\text{for every }t\in[a,b].

Using the integral triangle inequality and $|f(t)|\leq\|f\|_\infty$, we obtain

\begin{aligned} |(Tf)(x)-(Tf)(y)| &\leq\int_a^b |k(x,t)-k(y,t)|\,|f(t)|\,dt\\ &< (b-a)\frac{\varepsilon}{(b-a)(\|f\|_\infty+1)}\|f\|_\infty\\ &=\varepsilon\frac{\|f\|_\infty}{\|f\|_\infty+1} <\varepsilon. \end{aligned}

Thus $Tf$ is continuous on $[c,d]$, so $T$ maps into $C[c,d]$. Linearity follows from linearity of the Riemann integral: for $f,g\in C[a,b]$ and scalars $\alpha,\beta$, $(T(\alpha f+\beta g))(x)=\alpha(Tf)(x)+\beta(Tg)(x)$ for every $x\in[c,d]$.

Finally, for every $x\in[c,d]$,

\begin{aligned} |(Tf)(x)| &\leq\int_a^b |k(x,t)|\,|f(t)|\,dt\\ &\leq (b-a)K\|f\|_\infty. \end{aligned}

Taking the supremum over $x$ gives $\|Tf\|_\infty\leq(b-a)K\|f\|_\infty$. This is a uniform bound independent of $f$, so $T$ is bounded and its operator norm is at most $(b-a)K$. $\square$

The estimate in the theorem is often enough to prove boundedness, but it need not be the exact operator norm. It uses the largest value of $|k|$ on the whole rectangle and the full interval length; the actual action of integration can give a smaller bound. The following example finds the exact norm by using the positivity of its kernel.

Worked Example: An Exponential-Kernel Operator

Define $T:C[0,1]\to C[0,1]$ by

(Tf)(x)=\int_0^1 e^{xt}f(t)\,dt.

The kernel $e^{xt}$ is continuous on $[0,1]\times[0,1]$, so the theorem shows that $T$ is bounded and maps into $C[0,1]$. We can sharpen its norm estimate. For $x,t\in[0,1]$, we have $xt\leq t$, and hence $0<e^{xt}\leq e^t$. Therefore, for every $f\in C[0,1]$,

|(Tf)(x)| \leq\|f\|_\infty\int_0^1 e^{xt}\,dt \leq\|f\|_\infty\int_0^1 e^t\,dt =(e-1)\|f\|_\infty.

This proves $\|T\|\leq e-1$. Take the constant function $f(t)=1$, which has $\|f\|_\infty=1$. At $x=1$,

(Tf)(1)=\int_0^1 e^t\,dt=e-1.

Thus $\|Tf\|_\infty\geq e-1$, giving $\|T\|\geq e-1$. Combining the two inequalities yields $\|T\|=e-1$. Notice that this exact norm is smaller than the general theorem's estimate $e$: the integral of the kernel provides a sharper bound than its maximum value multiplied by the interval length.

Multiplication Operators and Exact Norms

Another basic operator multiplies each function by a fixed function. This gives a useful case where the operator norm can be determined exactly, even if the fixed function does not attain its supremum.

Theorem (Norm of a Multiplication Operator): Let $E$ be a nonempty topological space, and let $g\in C_b(E)$, the space of bounded continuous real-valued functions on $E$ with the supremum norm. Define $M_g:C_b(E)\to C_b(E)$ by $(M_gf)(x)=g(x)f(x)$. Then $M_g$ is bounded and $$ \|M_g\|=\|g\|_\infty. $$

Proof. Since $g$ and $f$ are bounded and continuous, their pointwise product $gf$ is bounded and continuous. For each $f\in C_b(E)$,

\|M_gf\|_\infty =\sup_{x\in E}|g(x)f(x)| \leq\|g\|_\infty\|f\|_\infty.

Thus $M_g$ is bounded and $\|M_g\|\leq\|g\|_\infty$. For the reverse inequality, take the constant function $\mathbf{1}(x)=1$ on $E$. It belongs to $C_b(E)$ and has norm one. Since $M_g\mathbf{1}=g$, the unit-ball formula for the operator norm gives $\|M_g\|\geq\|M_g\mathbf{1}\|_\infty=\|g\|_\infty$. Both inequalities together prove the asserted equality. $\square$

Worked Example: A Multiplication Operator Whose Norm Is Not Attained at a Point

On $E=\mathbb{R}$, let $g(x)=x/(1+|x|)$. This function is continuous, and $|g(x)|=|x|/(1+|x|)<1$ for every real $x$. On the other hand, for positive integers $n$,

g(n)=\frac{n}{1+n}=1-\frac{1}{n+1}\longrightarrow1.

Consequently $\|g\|_\infty=1$, although $|g(x)|$ never equals one at any point of $\mathbb{R}$. The theorem shows that $M_gf=gf$ has operator norm one. Indeed, $\|M_gf\|_\infty\leq\|f\|_\infty$ for every bounded continuous $f$, while $M_g\mathbf{1}=g$ and $\|g\|_\infty=1$. This example distinguishes the supremum of a function from a maximum: an operator norm can be exact even when the supremum of the multiplier is not attained.

When Differentiation Is Not Bounded

Not every natural linear operation on continuous functions is bounded for the supremum norms. Differentiation illustrates why the choice of norms is essential. To use differentiation as a map, its domain must consist of differentiable functions; the question is whether the derivative can be controlled by the size of the function itself.

Worked Example: Differentiation Is Unbounded in the Supremum Norm

Let $V=C^1[0,1]$, the space of continuously differentiable functions on $[0,1]$, and give it the norm inherited from $C[0,1]$: $\|f\|_\infty=\sup_{x\in[0,1]}|f(x)|$. Consider the linear map $D:V\to C[0,1]$ defined by $Df=f'$, with the supremum norm on the target as well.

For each positive integer $n$, set $f_n(x)=x^n$. Since $0\leq x^n\leq1$ on $[0,1]$ and $f_n(1)=1$, we have $\|f_n\|_\infty=1$. But

(Df_n)(x)=nx^{n-1}, \qquad \|Df_n\|_\infty=n,

because $0\leq x^{n-1}\leq1$ and $nx^{n-1}=n$ at $x=1$. If $D$ were bounded, some finite $C$ would satisfy $\|Df_n\|_\infty\leq C\|f_n\|_\infty=C$ for every $n$. That would require $n\leq C$ for every positive integer $n$, which is impossible. Hence differentiation is unbounded with these norms.

This conclusion depends on the norm assigned to the domain. If instead $C^1[0,1]$ is given the norm $\|f\|_{C^1}=\|f\|_\infty+\|f'\|_\infty$, then $\|Df\|_\infty=\|f'\|_\infty\leq\|f\|_{C^1}$, so differentiation is bounded for that norm. The operation has not changed; the measurement of input size has.

Using Bounds Carefully

The examples suggest a practical approach to operator questions. First identify the domain and target spaces, including their norms. Then check linearity and look for an estimate that holds for every input. For an integral operator, bounding the kernel gives a direct estimate; for multiplication, testing the constant function one proves a matching lower bound. To show unboundedness, choose a sequence with bounded input norms and output norms that grow without limit.

A common pitfall is to show only that every individual output is finite. Boundedness requires one constant that works simultaneously for all inputs. The functions $x^n$ in the differentiation example each have a continuous derivative, but that does not provide a common bound on the derivative norms. Conversely, when proving continuity of an integral output, control of the kernel's variation must be chosen with the size of the fixed input in mind. The estimate in the theorem does this explicitly, leaving a strict margin below the requested $\varepsilon$.

Takeaway: Bounded linear operators provide uniform norm estimates. Continuous kernels and bounded multipliers give useful examples, while differentiation on $C^1[0,1]$ with the supremum norm shows that a natural linear map may fail to be bounded. Always specify the norms before deciding.

Check Your Understanding

Use the estimates and examples in this tutorial to answer the following questions.

Where is compactness used to show that a continuous-kernel integral operator is well defined and maps into continuous functions?
For the kernel $e^{xt}$ on $[0,1]\times[0,1]$, why does the constant input $f=1$ prove that the norm is at least $e-1$?
Why does the multiplication-operator norm formula remain valid when the multiplier's supremum is not attained?
Which sequence shows that differentiation is unbounded when both function spaces use the supremum norm?
How does changing the norm on $C^1[0,1]$ make the differentiation estimate possible?

What You'll Learn