\[ \newcommand{R}{\mathbb{R}} \newcommand{N}{\mathbb{N}} \newcommand{defarrow}{\quad \stackrel{\text{def}}{\Longleftrightarrow} \quad} \]
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 \(Lx \in Y\).
- 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 \(\|\cdot\|\).
- 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.
- 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.
- 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 \(\mathrm{ker}(T)\).
- 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.