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Section 4.2 Bounded Operators

Given two vector spaces \(X,Y\text{,}\) it is natural to study the set of all linear maps \(L:X\to Y\text{,}\) denoted by \(\cL(X,Y)\text{.}\) Unlike what happens in finite dimension, though, in infinite dimension linear maps are not necessarily continuous.
Let \(\Omega \) be a bounded domain and let \(k\in L^2(\Omega\times\Omega)\text{.}\)

The integral operator \(K:L^2(\Omega)\to L^2(\Omega) \) given by
\begin{equation*} Ku(x)=\int_\Omega k(x,y)u(y)dy \end{equation*}
is bounded on \(L^2(\Omega)\)

Indeed, by the Cauchy Schwarz inequality,
\begin{align*} \|Ku\|_{L^2(\Omega)}^2\amp=\int_\Omega \left|\int_\Omega k(x,y)u(y) dy\right|^2dx\\ \\ \amp\leq\int_\Omega\left(\int_\Omega|k(x,y)|^2dy\right)\left(\int_\Omega|u(y)|^2dy\right)dx\\ \\ \amp= \|k\|^2_{{L^2(\Omega\times\Omega)}}\cdot \|u\|_{L^2(\Omega)}^2. \end{align*}
Hence \(\|K\|_{op} \leq \|k\|_{{L^2(\Omega\times\Omega)}} \text{.}\)
Let \(B\) be a Banach space with mutually linearly independent vectors \(e_1,e_2,\dots\text{.}\) We can assume without loss of generality that \(\|e_n\|=1\) for all \(n=1,2,\dots\text{.}\) Then
\begin{equation*} e'_=\frac{e_n}{\sqrt{n}}\to0. \end{equation*}
Let now \(A:B\to B\) be the linear map defined by \(Ae_n=ne_n\text{.}\) If \(A\) were continuous, then we would have that \(Ae'_n\to0\text{.}\) On the contrary, \(A(e'_n)=\sqrt{n}\cdot e_n\) does not converge to any vector since
\begin{equation*} \|\sqrt{n}\cdot e_n\|=\sqrt{n}\to\infty. \end{equation*}
Hence, \(A\) is not continuous.

We have precisely this kind of behavior in case of the derivative operator. Consider indeed the derivative as an operator
\begin{equation*} \frac{d}{dx}:C^1([0,1])\subset C^0([0,1])\to C^0([0,1]). \end{equation*}
Notice that this operator is not defined over the whole space but rather only on a dense subset.

Now, while
\begin{equation*} f_n(x)=\frac{\sin(nx)}{\sqrt{n}}\to0 \end{equation*}
in \(C^0\) as \(n\to\infty\text{,}\) on the contrary
\begin{equation*} \|f'_n\|=\|\sqrt{n}\sin(nx)\|\to\infty. \end{equation*}
Hence, the first derivative is not continuous under the \(C^0\) topology.

Notice also that, since \(C^0([0,1])\) embeds continuously into each space \(L^p([0,1])\text{,}\) then \(d/dx\) is not continuous also under the \(L^p\) topologies.

The example above shows the origin of the problem. In finite dimension, there is a finite number of possible eigendirections for a linear map and so a finite bound for how much a vector can be stretched by the map. In infinite dimension, loosely speaking, there are infinitely many possible eigendirections and so the stretching factor may diverge.

We discuss first the case of Banach spaces, that is the most interesting for us. At the end of this section, we dedicate a subsection to the case of Fréchet spaces.

Definition 4.2.3. Bounded operators between Banach spaces.
Let \((X ,\|\cdot \|_X )\) and \((Y ,\|\cdot \|_Y )\) be two Banach spaces and let \(A\in\cL(X,Y)\text{,}\) namely
\begin{equation*} A ( \alpha x + \beta y ) = \alpha A x + \beta A y\text{ for each } x,y\in X\text{ and }\alpha,\beta\in \RR. \end{equation*}

We say that \(A\) is bounded if \(A\) takes bounded sets to bounded sets, namely if there is some \(M>0\) such that
\begin{equation*} \| A x \|_Y \leq M \|x\|_X\text{ for all } x \in X. \end{equation*}
We denote the set of all linear and bounded operators by
\begin{equation*} \mathcal{B}(X,Y) := \{A : X \rightarrow Y, A \text{ is linear and bounded } \}. \end{equation*}
This space can be equipped with the operator norm
\begin{align*} \|A\|_{op} \amp : = \inf \{M: \| A x \|_Y \leq M \|x\|_Y \quad \mbox{for all}\quad x \in X \}\\ \amp = \sup_{x \neq 0} \frac{\| A x \|_Y } {\|x\|_X }\\ \amp = \sup_{\|x\|=1}\| A x \|_Y. \end{align*}

We justify the definition above by pointing out that, due to linearity, a map \(L\in\cL(X,Y)\) takes bounded sets to bounded sets if and only if it takes the unit ball into a bounded set, namely if and only if \(\sup_{\|x\|=1}\| A x \|_Y\) is finite. We leave to the reader as an exercise the verification that the operator norm is, indeed, a norm.

In order to get an intuition on the meaning of the operator norm, let us see what is its meaning in finite dimension. In this case, it is easier to consider the case when the scalar field is \(\bC\text{.}\) Let \(M\in M_n(\bC)\) and assume \(M\) to be normal, namely that \(MM^*=M^*M\text{.}\) Then we know from linear algebra (spectral theorem) that \(M\) is diagonalizable, namely there is a basis \(e_1,\dots,e_n\) of \(\bC^n\) and scalars \(\lambda_1,\dots,\lambda_n\) such that \(Me_i=\lambda_ie_i\) for every \(i=1,\dots,n\text{.}\)

We can assume without loss of generality that \(|\lambda_1|\leq\dots\leq|\lambda_n|\text{.}\) A direct check shows that
\begin{equation*} \|M\|_{op} = |\lambda_n|. \end{equation*}
When \(M\) is not normal, one can show that
\begin{equation*} \|M\|_{op} = \sqrt{\|MM^*\|_{op}} = \sqrt{\|M^*M\|_{op}}. \end{equation*}

"\(\Rightarrow\)" If \(A\) is bounded, then for all \(x_n, x\in X\)
\begin{equation*} \| A(x_n-x)\|_Y \leq \|A\|_{op} \|x_n-x\|_X, \end{equation*}
hence \(A\) is continuous.

"\(\Leftarrow\)" Now let \(A\) be continuous and assume \(A\) is not bounded. Then for each \(n\in\mathbb{N}\) there exists \(y_n \in X\) with
\begin{equation*} \|A y_n\|_Y \geq n^2\|y_n\|_X. \end{equation*}
Define \(x_n:=\frac{y_n}{n\|y_n\|}\text{.}\) Then \(\|x_n\|_X\rightarrow 0\) for \(n\rightarrow \infty\text{.}\) But \(\|A x_n\|\geq n\rightarrow\infty\) so \(A\) is not continuous, which is a contradiction.

Let \(\{A_n\}\) be a Cauchy sequence in \(\mathcal{L} (X ,Y )\text{.}\) Given \(\epsilon>0\) there exists an \(N\) such that
\begin{equation*} \| A_m - A_n \|_{ op }\leq \epsilon \qquad \mbox{for all }\quad m , n \geq N. \end{equation*}
Now for fixed \(x \in X\) we have
\begin{equation*} \|A_m x - A_n x\|\leq \|(A_m - A_n) x \| \leq \|A_m - A_n\|_{op} \|x\| \leq \epsilon \|x\|. \end{equation*}
Hence \(\{A_n x\}\) is a Cauchy sequence in \(Y\) and since \(Y\) is complete, we have a limit
\begin{equation*} A_n x \rightarrow y \text{ in } Y. \end{equation*}

As we do this for each \(x\in X\) we define a mapping \(A:X \rightarrow Y\) given by \(Ax=y\text{.}\) \(A\) is the candidate for the limit of \(\{A_n\}\text{.}\) We show below that \(A\in\cB(X,Y)\text{.}\)

1. \(A\) is linear. Let \(\lim A_nx=z=Ax\) and \(\lim A_ny=w=Ay\text{.}\) Then, by definition, \(A(x+y)=\lim(A_nx+A_ny)=\lim A_n(x+y)\) and, by continuity of the sum operator, \(\lim A_n(x+y)=Ax+Ay\text{.}\) Similarly happens for the multiplication by a scalar.

2. \(A\) is bounded. Since \(A_n\) is Cauchy, for every \(\eps>0\) there is a \(N>0\) such that \(\|A_n-A_m\|_{op}<\eps\) for all \(n,m>N\text{.}\) Then
\begin{equation} \|A_nx-A_mx\|_{op} <\eps\|x\|\tag{4.2.1} \end{equation}
and, passing to the limit for \(n\to\infty\text{,}\)
\begin{equation} \|Ax-A_mx\|\leq\eps\|x\|,\tag{4.2.2} \end{equation}
for all \(m>N\text{.}\) Hence
\begin{equation} \|Ax\|_{op}\leq \|A-A_m\|_{op}+\|A_m\|_{op}\leq(\eps+\|A_m\|\eps)\|x\|,\tag{4.2.3} \end{equation}
so \(A\) is bounded.

The range of an operator \(A:D(A)\rightarrow Y\) is defined as
\begin{equation} R(A)=\{g\in Y: g=Af, f\in D(A)\}\tag{4.2.4} \end{equation}
and the kernel or nullspace is
\begin{equation*} \text{Ker}(A)=\{f\in D(A):Af=0\}. \end{equation*}

For each \(y\in R(A)\) find a unique solution \(x\in X\) with \(Ax=y\text{.}\) If \(\text{Ker}(A)\neq\{0\}\) then there exists a \(\varphi \neq0\text{,}\) \(\varphi\in \text{Ker}(A) \) and \(A\varphi=0\text{.}\) But also \(A(0)=0\text{,}\) which means \(A\) is not invertible. Now assume \(\text{Ker}(A)=\{0\}\text{.}\) \(A\) is surjective by assumption. To show injectivity assume \(Ax_1=y\text{,}\) \(Ax_2=y\text{.}\) Then \(A(x_1-x_2)=y-y=0\text{,}\) which implies \(x_1-x_2\in \text{Ker}(A)\text{,}\) i.e. \(x_1=x_2\text{.}\)

Subsection 4.2.1 Bounded operators in Fréchet spaces

As we saw previously, many functional spaces are not normable. This includes the spaces of smooth functions on open sets and on their closure and also the spaces of \(C^k\) functions on open sets and of \(L^p_{loc}\) spaces on unbounded sets. Hence, it is important to discuss the case of continuous operators on this kind of space.
Definition 4.2.7. Fréchet spaces.
We say that a sequence \(p_1,p_2,\dots\) of seminorms on a vector space \(X\)separates points if, for any \(x\in X\text{,}\) there is a \(k\geq1\) such that \(p_k(x)>0\text{.}\)

A space \(X\) is Fréchet if there is a countable family \(p_k\) of seminorms on \(X\) such that:
  1. the \(p_k\) separate points;
  2. \(X\) is complete under the distance
    \begin{equation*} d(x,y) = \sum_{i=1}^\infty\frac{1}{2^k}\frac{p_k(x-y)}{1+p_k(x-y)}. \end{equation*}
We say that a subset \(B\subset X\) is bounded if every seminorm \(p_k\) is bounded on \(B\text{.}\)
Remark 4.2.8.
With this distance, \(d(x,y)<1\) for every \(x,y\in X\text{.}\)
Monotone families of seminorms. We can always assume that, in a Fréchet space \(X\) with seminorms \(p_1,p_2,\dots\text{,}\) the seminorms are such that
\begin{equation*} p_1 \leq p_2 \leq \dots. \end{equation*}
If it is not so, indeed, we can replace \(p_k\) with \(p'_k(x)=\max\{p_1(x),\dots,p_k(x)\}\text{.}\) One can prove that the family \(p'_1,p'_2,\dots\) induces on \(X\) the same topology as the original family.

Topology of a Fréchet space. On a topological vector space \(X\text{,}\) it is enough to discuss the case of neighborhoods of the origin. As a metric space with distance \(d\text{,}\) the basic neighborhoods of the origin are given by the \(\eps\)-balls
\begin{equation*} B_{0,\eps}=\{x\in X\,:\,d(x,0)<\eps\}. \end{equation*}
It is critical to notice the following facts.
Definition 4.2.9. Balls and seminorm tubes.
Given a Fréchet space \(X\text{,}\) we call ball any set
\begin{equation*} B_{\eps} = \{x:X\,:\,d(0,x)<\eps\} \end{equation*}
and we call seminorm tube any set
\begin{equation*} T_{k,\eps} = \{p_k(x)<\delta\} \end{equation*}
Notice that \(p_k/(1+p_k)\leq1\text{.}\) Hence, for every \(\eps>0\text{,}\) there is a \(N\geq1\) such that
\begin{equation*} \sum_{i=N+1}^\infty \frac{1}{2^i}\frac{p_i(x)}{1+p_i(x)}<\frac{\eps}{2}\text{ for all }x\in X. \end{equation*}
Now, consider the seminorm tube \(T_{N,\delta}\text{.}\) Since norms are non-decreasing and \(t/(1+t)\) is monotonically increasing,
\begin{equation*} \sum_{i=1}^N \frac{1}{2^i}\frac{p_i(x)}{1+p_i(x)}\leq \delta\sum_{i=1}^N \frac{1}{2^i}. \end{equation*}
By taking \(\delta\) small enough, this quantity can be made smaller than \(\eps/2\) and so \(d(x,0)<\eps\text{,}\) namely \(x\in B_\eps\text{,}\) for every \(x\in T_{N,\delta}\text{.}\) Notice that this means that, in a Fréchet space, each open sets is determined by only finitely many seminorms!

Let now \(\eps>0\) and suppose that, for every \(k\geq1\) and \(\delta>0\text{,}\) there is a \(x\in B_\delta\) such that \(p_k(x)\geq\eps\text{.}\) Then
\begin{equation*} d(x,0) \geq \sum_{n=1}^\infty 2^{-n}\frac{\eps}{1+\eps} = \frac{\eps}{1+\eps}, \end{equation*}
which is incompatible with \(x\in B_\delta\) for \(\delta\) small enough. Hence, for every \(\eps>0\text{,}\) there must be a \(\delta>0\) and a \(k\geq1\) such that \(p_k(x)<\eps\) for all \(x\in B_\delta\text{.}\) In other words, \(B_\delta\subset T_{k,\eps}\text{.}\)
Recall first of all that, in case of linear maps, continuity it equivalent with "continuity at 0". The continuity of \(L\) at 0 means that, for every \(\eps>0\text{,}\) there is a \(\delta>0\) such that
\begin{equation*} x\in B_{\delta}\,\implies\, Lx\in B_{\eps}. \end{equation*}
By Proposition 4.2.10, there are \(k\geq1\) and \(\delta'>0\) such that \(T_{k,\delta'}\subset B_{\delta}\) and \(k'\geq1\) and \(\eps'>0\) such that \(B_{\eps}\subset T_{k',\eps'}\text{.}\) Hence,
\begin{equation*} x\in T_{k,\delta'}\,\implies\, Lx\in T_{k',\eps'}. \end{equation*}
Similarly one can prove the inverse direction.

Bounded operators
Definition 4.2.12. Bounded maps between Fréchet spaces.
Let \(X,Y\) be a Fréchet spaces. We say that a map \(A\in\cL(X,Y)\) is bounded if \(A\) takes bounded sets to bounded sets.
The proof below uses an argument about continuity that we show first here in case of Banach spaces. By definition, a continuous linear map \(A:X\to Y\) satisfies the following property: for every \(\eps>0\) there is a \(\delta>0\) such that
\begin{equation*} d_X(x,y)<\delta\;\implies\;d_Y(Lx,Ly)<\eps, \end{equation*}
namely
\begin{equation*} \|x-y\|_X < \delta\;\implies\;\|Lx-Ly\|_Y<\eps. \end{equation*}
In particular, there is a \(\delta>0\) such that
\begin{equation*} \|x-y\|_X < \delta\;\implies\;\|Lx-Ly\|_Y<1. \end{equation*}
The vector \(\frac{\delta}{2}\frac{x-y}{\|x-y\|_X}\) has norm \(\delta/2<\delta\text{,}\) so
\begin{equation*} \left\|L\frac{\delta}{2}\frac{x-y}{\|x-y\|_X}\right\|_Y<1. \end{equation*}
Hence, by linearity,
\begin{equation*} \|L(x-y)\|_X<\frac{2}{\delta}\|x-y\|_X. \end{equation*}
This is an alternate proof of the fact that, in Banach spaces, continuity implies boundedness. In case of Fréchet spaces, repeating the same argument, one can show that, if \(L\) is continuous then, for every \(k\geq 1\text{,}\) there is a \(m\geq1\) and a \(C>0\) such that
\begin{equation*} q_k(L(x-y))<C p_m(x-y). \end{equation*}

"\(\Rightarrow\)" Assume that \(A\) is bounded. We need to prove that, for every \(k\geq 1\text{,}\) there is a \(m\geq1\) and a \(C>0\) such that
\begin{equation*} q_k(L(x-y))<C p_m(x-y). \end{equation*}
Now, for every \(i\geq 1\) and \(\eps>0\text{,}\) the "seminorm tube"
\begin{equation*} T_{i,\eps} = \{x\in X\,|\,p_i(x)\leq\eps\} \end{equation*}
is bounded in \(X\text{.}\) Indeed, every seminorm is a continuous map and so, for every \(p_j\) and \(\eps>0\text{,}\) there is a \(\delta_j\) such that
\begin{equation*} p_i(x)<\delta_j\;\implies\;p_j(x)<1 \end{equation*}
Now, if \(\eps\leq\delta_j\text{,}\) then \(p_j\) is clearly bounded on \(T_{i,\eps}\text{.}\) If \(\eps>\delta_j\text{,}\) then
\begin{equation*} p_i\left(\delta\frac{x}{p_i(x)}\right) = \delta \end{equation*}
and so
\begin{equation*} p_j\left(\delta\frac{x}{p_i(x)}\right) = 1, \end{equation*}
namely
\begin{equation*} p_j(x) = \frac{p_i(x)}{\delta}\leq\frac{\eps}{\delta}\text{ for all }x\in T_{i,\eps}. \end{equation*}
Since \(L\) is bounded, then all sets \(L(T_{i,\eps})\) are bounded in \(Y\text{.}\) Set
\begin{equation*} M_{m,i,\eps} = \sup_{x\in T_{i,\eps}}q_m(Lx). \end{equation*}
Then, if \(p_i(x)\leq\eps\text{,}\)
\begin{equation*} q_m\left(\eps\frac{x}{p_i(x)}\right) \end{equation*}
and let \(x_n, x\in X\) with \(x_n\to x\text{.}\) Since \(A\) is bounded, for every \(k\) there is a seminorm \(r_k\) on \(X\) and a constant \(c_k\) such that
\begin{equation*} q_k(A(x_n-x)) \leq c_k r_k(x_n-x). \end{equation*}
Since \(x_n\to x\text{,}\) then \(r_k(x_n-x)\to0\) and so \(q_k(A(x_n-x))\) for every \(k\text{,}\) which implies that \(Ax_n\to Ax\) in \(Y\text{.}\) Hence, \(A\) is continuous.

"\(\Leftarrow\)" Now let \(A\) be continuous and assume \(A\) is not bounded. Since \(A\) is not bounded, there is at least a bounded set \(B\subset X\) such that \(A(B)\) is not bounded, namely there is a \(k\geq1\) such that \(\sup_{x\in B} q_k(Ax)=\infty\text{.}\) Hence, there is a bounded sequence \(x_n\in X\) such that \(q_k(Ax_n)>n^2\text{.}\)

Since \(x_n\) is bounded, we have that \(x_n/n\to0\) in \(X\text{.}\) On the other side, \(q_k(Ax_n)\to\infty\) and so \(Ax_n\) does not converge in \(Y\text{,}\) against the hypothesis that \(A\) is continuous.

"\(\Rightarrow\)" Assume that \(A\) is bounded. Then \(A\) is continuous and so, for each \(k\text{,}\) the map \(F_k:x\to q_k(Ax)\) is continuous. Continuity at 0 of \(F_k\) means that, for every \(\eps>0\text{,}\) there are \(k'\geq1\) and \(\delta>0\text{,}\) depending on \(k\text{,}\) such that
\begin{equation*} p_{k'}(x)<\delta\;\implies\;q_k(Ax)<\eps. \end{equation*}
In particular, there is a \(\delta>0\) such that
\begin{equation*} p_{k'}(x)<\delta\;\implies\;q_k(Ax)<1. \end{equation*}
Now, if \(p_{k'}(x)=0\text{,}\) then \(p_{k'}(\lambda x)=0\) for all \(\lambda\) and so \(\lambda q_k(Ax)<1\) for all \(\lambda\text{,}\) so that \(q_k(Ax)=0\) as well. If \(p_{k'}(x)\neq 0\text{,}\) then
\begin{equation*} p_{k'}\left(\frac{\delta}{2}\frac{x}{p_{k'}}\right)=\frac{\delta}{2} \end{equation*}
and so
\begin{equation*} \frac{\delta}{2p_{k'}(x)} q_{k}(Ax) = q_{k}\left(A\frac{\delta}{2}\frac{x}{p_{k'}(x)}\right)<1, \end{equation*}
namely
\begin{equation*} q_{k}(Ax) \leq C p_{k'}(x) \end{equation*}
for \(C=\frac{\delta}{2}\text{.}\)

"\(\Leftarrow\)" This directon is a tautology.

Notice that, unlike in case of Banach spaces, \(\cB(X,Y)\) is not, in general, a Fréchet space. In fact, it is not even metrizable unless \(X\) is normable.

More can be said when \(Y\) is a Banach space. For example, this happens in case of maps from a Fréchet space to a finite-dimensional space, which includes the case of linear functionals.
Definition 4.2.16.
Let \(X\) be a Fréchet space and \(Y\) a Banach space. The topology of uniform convergence on bounded set on \(\cB(X,Y)\) is the topology generated by the family of seminorms
\begin{equation*} p_B(L)=\sup_{x\in B}\|L(x)\|_Y, \end{equation*}
where \(B\subset X\) is bounded.
Remark 4.2.17.
Unless \(X\) is finite-dimensional, this family cannot be reduced to a countable subset and so this space is not metrizable, hence not Fréchet.
In topological spaces that do not have a countable basis for their topology, as is the case of \(\cB(X,Y)\text{,}\) sequences are not that useful: they have countably many elements and so a single sequence "cannot probe" all elements of a basis for the topology. In this setting, sequences need to be replaced by a more subtle tool called "net".
Definition 4.2.18.
A net in a set \(X\) is a set \(\{x_\alpha|\alpha\in A\}\subset X\) where the set of indices \(A\) is non-empty and indices are ordered by some relation, denoted by \(\leq\text{,}\) so that, for every \(a,b\in A\text{,}\) there is a \(c\in A\) such that \(c\leq a\) and \(c\leq b\text{.}\)

We say that a net \(x_\alpha\) is eventually in \(S\) if there is an \(a\in A\) such that, for every \(b\leq a\text{,}\) \(x_b\in S\text{.}\) We say that \(x_\alpha\to\bar x\) if, for every neighborhhod \(U\) of \(\bar x\text{,}\) \(x_\alpha\) is eventually in \(U\text{.}\)
Remark 4.2.19.
The advantage of nets over sequences is that nets can have uncountably many elements, as long as an ordering with the properties above is given to them.
We like metric spaces because the topology on a metric space can be completely characterized by sequences. The proposition below, that we will not prove here, shows that nets do the same job over general topological spaces.