\[ \newcommand{R}{\mathbb{R}} \newcommand{N}{\mathbb{N}} \newcommand{defarrow}{\quad \stackrel{\text{def}}{\Longleftrightarrow} \quad} \]

# Bounded linear transformations

Let \(X\) and \(Y\) be normed spaces (both real, or both complex), and \(T \in L(X,Y)\) a linear mapping between them.

## Boundedness

A linear mapping \(T\colon X \to Y\) is called **bounded**, \(T \in B(X,Y)\), if \(T\) maps bounded sets into bounded sets:
\[
T \in B(X,Y) \defarrow \exists C; \quad \|Tx\|_Y \leq C \|x\|_X \quad \text{ for all } x \in X.
\]
Thus, if \(T\) is bounded, the number
\[
\|T\| \stackrel{\text{def.}}{=} \sup_{x \neq 0} \frac{\| Tx \|_{Y}}{\|x\|_X}
\]
is finite; it is the **(operator) norm of** \(T\).

**N.b.** In some sources \(B(X,Y)\) is denoted by \(L(X,Y)\); to us, \(L(X,Y)\) is the space of linear transformations between two (not necessarily normed) vector spaces; if \(X\) and \(Y\) are normed spaces, then \(B(X,Y) \subset L(X,Y)\).

* From now on, we will not always write out the indices for the norms; just recall that * \( x \in X\) *and* \(Lx \in Y\).

**Ex.**

- The integral \(\int_0^t f(s)\,ds\) defines a linear transformation on the space of bounded and continuous functions \(f\colon [0,1] \to \R\), \[T \colon BC([0,1],\R) \to BC([0,1],\R), \qquad (Tf)(t) = \int_0^t f(s)\,ds. \] This transformation is bounded, since \[ \|Tf\|_{BC([0,1],\R)} = \sup_{t \in [0,1]} \Big| \int_0^t f(s)\,ds \Big| \leq \int_0^1 \max_{s \in [0,1]}|f(s)|\,ds = \|f\|_{BC([0,1],\R)},\] so that \(\|T\| \leq 1\). In fact, \(\|T\| =1\) (can you see why?).

- If \(g \in BC([0,1],\R)\), a similar argument yields that \[T \colon BC([0,1],\R) \to BC([0,1],\R), \qquad (Tf)(t) = \int_0^t f(s) g(s)\,ds \] is bounded too, with \(\|T\| \leq \max_{t \in [0,1]} |g(t)| = \|g\|_{BC([0,1],\R)}\) (see if you can make this better).

- By the same argument, with \(t=1\) and \(g \in BC([0,1],\R)\), the definite integral \(\int_0^1 f(s) g(s) \,ds\) defines a
**bounded linear functional**^{1)}\[T \colon BC([0,1],\R) \to \R, \qquad f \mapsto \int_0^1 f(s)g(s)\,ds.\] Remember the form of this functional – it is the inner product for the Hilbert space \(L_2((0,1),\R)\).

- The derivative \(\frac{d}{dx}\) is in general
*not*a bounded operator.^{2)}To see why, consider a sequence of functions like \[ f_n(x) := \sin(nx).\] These functions are uniformly bounded, but not their derivatives. This indicates that, to solve differential equations, it is better to reformulate them as integral operators.

### Equivalence of norm expressions

For \(T \in B(X,Y)\), \[ \|T\| = \sup_{x \neq 0} \frac{\| Tx \|}{\|x\|} = \sup_{\|x\| = 1} \|Tx\| = \sup_{\|x\| \leq 1} \|Tx\|\] all describe the least possible bound on \(C\) such that \(\|Tx\| \leq C\|x\|\) for all \(x \in X\).

### › B(X,Y) is a normed space

\[ B(X,Y) = \{T \in L(X,Y) \colon T \text{ is bounded}\}\] is a normed space when equipped with the operator norm \(\|\cdot\|\).

**Ex.**

- As we shall see, \(B(\R^n,\R^m) = L(\R^n,\R^m)\) (as sets and linear spaces): given bases for \(\R^n, \R^m\) there is a bijective correspondence between matrices \(A \in M_{m \times n}(\R)\) and bounded linear transformations \(T \in B(\R^n, \R^m)\).

- Let \(X\) be a real normed space. The space \(X' := B(X,\R)\) is called the
**dual of**\(X\); its elements are**bounded linear functionals**on \(X\). If \(X\) is complex, \(B(X,\mathbb{C})\) is its dual.^{3)}

- The dual of \(\R\) is \(\R\): each bounded linear functional \( T \in B(\R,\R) \) is realized by multiplication with a real constant: \[ T \in B(\R,\R) \quad\Longleftrightarrow\quad Tx = \lambda x, \quad \lambda \in \R. \]

**Riesz representation theorem**: Let \(L_2(I,\R)\) be the space of real-valued square-integrable functions on an open interval \(I \subset \R\), with norm \[ \|f\|_{L_2(I,\R)} = \big(\int_I |f(t)|^2\,dt\big)^{1/2}. \] The Riesz representation theorem asserts that each bounded linear functional \(T\) on \(L_2(I,\R)\) can be identified with an element \(g \in L_2(I,\R)\), via \[ T f = \int_I f(t)g(t)\,dt.\] Thus \( L_2(I,\R) \cong B(L_2(I,\R),\R)\) is its own dual.^{4)}

### › B(X,Y) is Banach for Y Banach

If \(Y\) is complete, so is \(B(X,Y)\).

**N.b.** Note that \(X\) has no role in the completeness of \(B(X,Y)\). ^{5)}

## Boundedness and continuity

### Continuity

A mapping \(f \colon X \to Y\) between to metric spaces is said to be **continuous at** \(x_0\) if
\[
f(x_n) \to f(x_0) \:\text{ in } Y \qquad\text{ as }\qquad x_n \to x_0 \:\text{ in } X.
\]
Since continuous and sequential limits agree, this is the same as
\[
\forall\, \varepsilon > 0 \quad \exists\, \delta > 0; \qquad d_Y(f(x),f(x_0)) < \varepsilon \quad\text{ for }\quad d_X(x,x_0) < \delta.
\]
A mapping that is continuous at all points in \(X\) is called **continuous**.

**Ex.**

- In a normed space, \((X,\|\cdot\|)\), the norm is a continuous function \(X \to \R\): if \(x_n \to x_0\) in \(X\), then \[ d_\R(\|x_n\|,\|x_0\|) = \big| \|x_n\| - \|x_0\| \big| \leq \|x_n - x_0\| = d_X(x_n,x_0) \to 0, \] by the reverse triangle inequality.

### › For linear operators, continuity means boundedness

Let \(T \in L(X,Y)\). Then the following statements are equivalent:

- \(T\) is everywhere continuous.
- \(T\) is continuous at \(x = 0\).
- \(T\) is bounded.

**Ex.**

- Any linear operator \(T \in L(X,Y)\) defined on a
*finite-dimensional*normed space \(X\) is continuous. Reason: identify \(X \cong \R^n\) and note that \[\mathrm{ran}(T) = \mathrm{span}\{Te_1, \ldots, Te_n\} \cong \R^m \quad\text{ for some } m \leq n,\] where \(\{e_1, \ldots, e_n\}\) is a basis for \(\R^n\). Hence, \[ T \colon X \cong \R^n \to \R^m \cong \tilde Y \subset Y\] is a linear transformation onto a finite-dimensional subspace \(\tilde Y\) of \(Y\), and, as such, has a matrix representation \[ T \colon x \mapsto Ax = \big(\sum_{j=1}^n a_{ij} x_j\big)_{i=1}^m. \] All norms on a finite-dimensional vector space are equivalent, so whatever the norms of \(X\) and \(Y\), we can consider any suitable norms for \(\R^n \cong X\) and \(\R^m \cong \tilde Y\). Choose, for example, the \(l_\infty\)-norm: then \[ \|Ax\|_{l_\infty} = \max_{1 \leq i \leq m} \big| \sum_{j=1}^n a_{ij} x_j \big| \leq n \max_{i,j} |a_{ij}| \max_j |x_j| = n \max_{i,j} |a_{ij}| \|x\|_{l_\infty}. \] This means that \(T\) is bounded with \(\|T\| \leq n \max_{i,j} |a_{ij}|\), and therefore also continuous.

**N.b.** Equivalent norms yield the same open and closed sets, the same convergence, but *not the same constants* in the estimates – in particular, the exact value of \(\|T\|\) depends on the norms for \(X\) and \(Y\).

### › The kernel of a bounded operator is closed

Let \(T \in B(X,Y)\). Then \(\mathrm{ker}(T)\) is a closed subspace of \(X\). In particular, if \(X\) is a Banach space, so is \(\mathrm{ker}(T)\).

**Ex.**

- The null space of a matrix \(A \in M_{m\times n}(\R)\) is a closed subspace of \(\R^n\).

- In \(L_2((-\pi,\pi),\R)\), the kernel of the bounded linear functional \[T \colon f \mapsto \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \sin(t)\,dt \] is a closed subspace; it consists of all functions with zero Fourier coefficient before \(\sin(t)\) in its Fourier expansion.

^{1)}

^{2)}

^{3)}

**continuous dual**; it is possible to define more general duals.

^{4)}

^{5)}