1 Tuesday, January 11

This course: solving \(Lf=g\) for \(L\) a linear operator, in analogy to solving \(Ax=b\) in matrices. References:

The issue when passing to infinite-dimensional vector spaces: the topology matters. E.g. the closure of the unit ball is closed and bounded and thus compact in finite dimensions, but this may no longer be true in \({\mathbb{R}}^\infty\) or \({\mathbb{C}}^\infty\). Recall that a Banach space is a complete normed space, and is further a Hilbert space if the norm is induced by an inner product. See the textbook for a review of vector spaces, metric spaces, norms, and inner products.

Our first example of infinite dimensional vector spaces: sequence spaces \(\ell\) with elements \(f \mathrel{\vcenter{:}}=(f_1, f_2, \cdots )\) with each \(f_i\in {\mathbb{R}}\).

Linear subspaces are subspaces that contain zero, as opposed to affine subspaces. An example is \(C_0([0, L]; {\mathbb{R}}) \leq C([0, L]; {\mathbb{R}})\), the subspace of bounded continuous functionals on \([0, L]\) which vanish at the endpoints. For any subset \(S \subseteq V\), write \([S]\) or \(\mathop{\mathrm{span}}S\) for the linear span of \(S\): all finite linear combinations of elements in \(S\).

Let \(V = C([-1, 1])\) and \(x_1\neq x_2\in [-1, 1]\), and set \(M_i \mathrel{\vcenter{:}}=\left\{{f\in V {~\mathrel{\Big\vert}~}f(x_i) = 0}\right\}\). Then \(M_i \leq V\) is a linear subspace, and in fact \(V = M_1 + M_2\) but \(V\neq M_1 \oplus M_2\) since the zero function is in both subspaces.

Limits of finite operators are compact. The classical example: set \((A_N)_{i, i} = {1\over i}\), so \(A_N = \operatorname{diag}\qty{{1}, {1\over 2}, {1\over 3}, \cdots, {1\over N}}\). Then \(\operatorname{Spec}A_N = \left\{{1\over n}\right\}_{n\leq N}\), but \(A\mathrel{\vcenter{:}}=\lim_N A_N\) is an operator with \(0\in \mkern 1.5mu\overline{\mkern-1.5mu\operatorname{Spec}(A)\mkern-1.5mu}\mkern 1.5mu\) as an accumulation point. Exercise: what is \(\ker A\)? Is it nontrivial?

A subset \(S \subseteq V\) is convex iff \begin{align*} tf + (1-t)g \in S \qquad \forall f, g\in S,\quad \forall t\in (0, 1] .\end{align*} Equivalently, \begin{align*} {af+bg\over a+b}\in S \qquad \forall f, g\in S,\quad \forall a,b\geq 0 \end{align*} where not both of \(a\) and \(b\) are zero. The convex hull of \(S\) is the smallest convex set containing \(S\).

Recall Holder’s inequality: \begin{align*} {\left\lVert {fg} \right\rVert}_1 \leq {\left\lVert {f} \right\rVert}_p \cdot {\left\lVert {g} \right\rVert}_q ,\end{align*} Schwarz’s inequality \begin{align*} {\left\lvert {{\left\langle {f},~{g} \right\rangle}} \right\rvert} \leq {\left\lVert {f} \right\rVert} {\left\lVert {g} \right\rVert} \qquad {\left\lVert {f} \right\rVert} \mathrel{\vcenter{:}}=\sqrt{{\left\langle {f},~{f} \right\rangle}} ,\end{align*} and Minkowski’s inequality \begin{align*} {\left\lVert {f+g} \right\rVert}_p \leq {\left\lVert {f} \right\rVert}_p + {\left\lVert {g} \right\rVert}_p .\end{align*}

A nice proof of Cauchy-Schwarz:

2 Thursday, January 13

My notes:

Last time: any norm yields a metric: \(d(f, g) \mathrel{\vcenter{:}}={\left\lVert {f-g} \right\rVert}\).

3 Tuesday, January 18

Last time:


Some theorems that hold in Hilbert spaces but not necessarily Banach spaces:

Absolute continuity:


\(\ker L = 0\) may not be sufficient to guarantee bijectivity in infinite dimensions:


Bounded iff continuous:

4 More Banach Spaces (Thursday, January 20)

Last time: complete iff absolutely convergent implies convergent. Today: wrapping up some results on Banach and Hilbert spaces, skipping a review of \(L^p\) spaces, and starting on operators next week.

Note that \(S \mathrel{\vcenter{:}}=(0, 1]\) is open and not complete, but \({ \operatorname{cl}} _{\mathbb{R}}(S) = [0, 1]\) is both closed and complete. Generalizing:

A subset \(S \subseteq B\) of a Banach space is complete iff \(S\) is closed in \(B\).

\(\impliedby\): If \(S\) is closed, a Cauchy sequence \((f_n)\) in \(S\) converges to some \(f\in B\). Since \(S\) is closed in \(B\), in fact \(f\in S\).

\(\implies\): Suppose that for \(f\in { \operatorname{cl}} _B(S)\), there is a Cauchy sequence \(f_n\to f\) with \(f_n \in S\) and \(f\in B\). Since \(S\) is complete, \(f\in S\), so \({ \operatorname{cl}} _B(S) \subseteq S\) making \(S\) closed.

For \(\Omega \subseteq {\mathbb{R}}^n\), the space \(X = (C(\Omega; {\mathbb{C}}), {\left\lVert {{-}} \right\rVert}_\infty)\) is a Banach space.

This is not complete with respect to any other \(L^p\) norm with \(p<\infty\)!

Use the lemma – write \(B\) for the space of all bounded (not necessarily continuous) functions on \(\Omega\), which is clearly a normed vector space, so it suffice to show

Step 1: we’ll show \({ \operatorname{cl}} _B(X) \subseteq X\). Take \(f\) to be a limit point of \(X\), then for every \({\varepsilon}>0\) there is a \(g\in X\) with \({\left\lVert {f-g} \right\rVert} < {\varepsilon}\). Apply the triangle inequality: \begin{align*} {\left\lVert {f} \right\rVert} \leq {\left\lVert {f-g+g} \right\rVert}\leq {\left\lVert {f-g} \right\rVert} + {\left\lVert {g} \right\rVert} = {\varepsilon}+ C < \infty ,\end{align*} so \(f\in { \operatorname{cl}} _B(X) \subseteq B\) since it is bounded. It remains to show \(f\) is continuous. Use that \({\left\lVert {f-g} \right\rVert}\infty <{\varepsilon}\) and continuity of \(g\) to get \({\left\lvert {g(x) - g(x_0)} \right\rvert} < {\varepsilon}\) for \({\left\lvert {x-x_0} \right\rvert}<{\varepsilon}\). Now \begin{align*} {\left\lvert {f(x) - f(x_0)} \right\rvert} &= {\left\lvert {f(x) - g(x) + g(x) -g(x_0) + g(x_0) - f(x_0)} \right\rvert} \\ &\leq {\left\lvert {f(x) - g(x)} \right\rvert} + {\left\lvert {g(x) - g(x_0)} \right\rvert} + {\left\lvert {f(x_0) - g(x_0)} \right\rvert} \\ &\leq 3{\varepsilon} .\end{align*} So \(X\) is closed in \(B\).

Step 2: Let \(f_n\) be Cauchy in \(B\), and note that we have a pointwise bound \({\left\lvert {f_n(x) - f_n(x_0)} \right\rvert} \leq {\left\lVert {f_n - f_m} \right\rVert} \to 0\). So pointwise, \(f_n(x)\) is a Cauchy sequence in \({\mathbb{C}}\) which is complete, so \(f_n(x) \to f(x)\) for some \(f: \Omega\to {\mathbb{C}}\). We now want to show \(f_n\to f\) in \(X\). Using that \(f_n\) is Cauchy in \(X\), produce an \(N_0\) such that \(n, m\geq N_0 \implies {\left\lVert {f_n - f_m} \right\rVert}< {\varepsilon}\). Now \begin{align*} {\left\lVert {f - f_n} \right\rVert} &= \sup_{x\in \Omega} {\left\lvert {f(x) - f_n(x)} \right\rvert} \\ &\leq \sup_{x\in \Omega} \sup_{m\geq N_0} {\left\lvert {f_m(x) - f_n(x)} \right\rvert} \\ &= \sup_{m\geq N_0} \sup_{x\in \Omega} {\left\lvert {f_m(x) - f_n(x)} \right\rvert} \\ &= \sup_{m\geq N_0} {\left\lVert {f_m - f_n} \right\rVert} \\ &\leq {\varepsilon} .\end{align*} Now use the reverse triangle inequality to show \(f_n\) is bounded \begin{align*} {\left\lVert {f} \right\rVert} - {\left\lVert {f_n} \right\rVert} \leq {\left\lVert {f-f_n} \right\rVert} < {\varepsilon}\implies {\left\lVert {f} \right\rVert} < \infty .\end{align*}

Now by problem 1.13, every Cauchy sequence is bounded, so \(f_n \to f\in B\).

Extending to vector-valued functions: for \(\Omega \subseteq {\mathbb{R}}^n\), take \(\mathbf{x} = {\left[ {x_1, \cdots, x_n} \right]}\) and \(F = {\left[ {f_1, \cdots, f_m} \right]}: {\mathbb{C}}^n\to {\mathbb{C}}^m\). Then there is a Banach space \begin{align*} X = C^(\Omega, {\mathbb{C}}^m), \qquad {\left\lVert {f} \right\rVert}_{C_1(\Omega)} \mathrel{\vcenter{:}}=\sum_{i\leq m} \sup_{x\in \Omega} {\left\lvert {f(x)} \right\rvert} + \sum_{i\leq m, j\leq n} \sup_{x\in \Omega} {\left\lvert {{\frac{\partial f_i}{\partial x_j}\,}(x) } \right\rvert} .\end{align*} Similarly define \(L^p(\Omega)\), noting that \({\left\lVert {f} \right\rVert}_{L^\infty(\Omega)}\) is the essential supremum.

For \(p\in (1, \infty)\), the sequence space \(X = (\ell^p, {\left\lVert {{-}} \right\rVert}_{\ell^p})\) is a Banach space.

A closed subspace of a Banach space is a linear subspace \(M \leq B\) which is norm-closed in \(B\).

\begin{align*} M \mathrel{\vcenter{:}}=\ker { \nabla\cdot }= \left\{{f\in C(\Omega) {~\mathrel{\Big\vert}~}{ \nabla\cdot }f = 0}\right\} \leq C(\Omega) \end{align*} is closed, where \({ \nabla\cdot }f\) is the divergence of a function \(f\).

For any \(S \subseteq B\), one can also take the corresponding closed subspace \(\mkern 1.5mu\overline{\mkern-1.5mu[S]\mkern-1.5mu}\mkern 1.5mu \mathrel{\vcenter{:}}={ \operatorname{cl}} _B \mathop{\mathrm{span}}_{\mathbb{C}}\left\{{s\in S}\right\}\), i.e. all linear combinations of elements in \(S\) and their limits. This is called the closed linear span of \(S\).

Let \(B = ( C[a, b], {\left\lVert {{-}} \right\rVert}_{L^\infty}\) and for \(x_0 \in [a, b]\) define \begin{align*} M \mathrel{\vcenter{:}}=\left\{{f\in C[a, b] {~\mathrel{\Big\vert}~}f(x_0) = 0}\right\}, \qquad N \mathrel{\vcenter{:}}=\left\{{f\in C[a, b] {~\mathrel{\Big\vert}~}f(x_0) \leq c}\right\} .\end{align*} Show that these are closed subspaces with no nontrivial open subsets of \(B\), since any \(f\in M\) can be perturbed to be nonzero at \(x_0\) with an arbitrarily small norm difference.

Recall that for \(S_1 \subseteq S_2 \subseteq B\), \(S_1\) is dense in \(S_2\) iff \({ \operatorname{cl}} _{S_2}(S_1) = S_2\). Recall Weierstrass’ theorem: for \(\Omega \subseteq {\mathbb{R}}^n\) is closed and bounded and write \({\mathcal{O}}\mathrel{\vcenter{:}}={\mathbb{R}}[x_1, \cdots, x_n]\) for the polynomials in \(n\) variables. Then \({\mathcal{O}}\subseteq C(\Omega)\), and \({ \operatorname{cl}} _{C(\Omega)} {\mathcal{O}}= C(\Omega)\), i.e. \({\mathcal{O}}\) is a dense subspace in the \(L^\infty\) norms. In fact, piecewise linear functions are dense.

Norms are equivalent iff \(c_1 {\left\lVert {f} \right\rVert}_a \leq {\left\lVert {f} \right\rVert}_b\leq c_2 {\left\lVert {f} \right\rVert}_a\) for some constants \(c_i\). All norms on \({\mathbb{R}}^n\) (resp. \({\mathbb{C}}^n\)) are equivalent.

For \(a>0, f\in C[0, 1]\), define \begin{align*} {\left\lVert {f} \right\rVert} \mathrel{\vcenter{:}}=\sup_{x\in I} {\left\lvert {e^{-ax} f(x)} \right\rvert} ,\end{align*} which can be thought of as a weighting on the uniform norm which de-emphasizes the tails of functions near the endpoints. This is equivalent to \({\left\lVert {{-}} \right\rVert}_\infty\), since \begin{align*} e^a \sup {\left\lvert {f} \right\rvert} \leq \sup {\left\lvert {e^{-ax} f} \right\rvert} \leq 1\cdot \sup {\left\lvert {f} \right\rvert} .\end{align*}

Note that a basis for a norm can be used as a basis with respect to an equivalent norm in finite dimensions. In infinite dimensions this may not hold – e.g. for Fourier series, \(\left\{{e_k(x)}\right\}_{k\in {\mathbb{Z}}}\) is not a basis for \(C[0, 2\pi]\) with the sup norm.

An \(B\in \mathcal{B}\) is separable iff \(X\) contains a countable dense subset \(S = \left\{{f_k}\right\}_{k\geq 0}\) such that for each \(f\in B\) and \({\varepsilon}>0\), there is an \(f_k\in S\) with \({\left\lVert {f-f_k} \right\rVert} < {\varepsilon}\).

Show that

4.1 Random Notes

5 Tuesday, January 25

\begin{align*} {\left\lvert {{\left\langle {f},~{g} \right\rangle}} \right\rvert}\leq {\left\lVert {f} \right\rVert} {\left\lVert {g} \right\rVert} .\end{align*}

Use \({\left\langle {f},~{g} \right\rangle} = {\left\lvert {f} \right\rvert} {\left\lvert {g} \right\rvert} \cos \theta_{fg}\) where \({\left\lvert { \cos \theta } \right\rvert} \leq 1\). - Assume \(g\neq 0\), then STS \({\left\langle {f},~{g\over {\left\lVert {g} \right\rVert}} \right\rangle} \leq {\left\lVert {f} \right\rVert}\). - Use \begin{align*} 0 &\leq {\left\lVert {f - {\left\langle {f},~{g} \right\rangle} g} \right\rVert}^2 \\ &= {\left\langle {f - {\left\langle {f},~{g} \right\rangle} g},~{f - {\left\langle {f},~{g} \right\rangle} g} \right\rangle} \\ &= {\left\langle {f},~{f} \right\rangle} - {\left\langle {f},~{g} \right\rangle} {\left\langle {g},~{f} \right\rangle} - \mkern 1.5mu\overline{\mkern-1.5mu{\left\langle {f},~{g} \right\rangle}\mkern-1.5mu}\mkern 1.5mu {\left\langle {f},~{g} \right\rangle} + {\left\langle {f},~{g} \right\rangle} \mkern 1.5mu\overline{\mkern-1.5mu{\left\langle {f},~{g} \right\rangle} \mkern-1.5mu}\mkern 1.5mu \\ &= {\left\lVert {f} \right\rVert}^2 - {\left\lvert { {\left\langle {f},~{g} \right\rangle}} \right\rvert}^2 .\end{align*}

Let \(f\in M\leq {\mathcal{H}}\) be a closed (and thus complete) subspace of a Hilbert space \({\mathcal{H}}\). Then there is a unique element \(g\) in \({\mathcal{H}}\) closest to \(M\) in the norm.

Let \(d\mathrel{\vcenter{:}}=\operatorname{dist}(f, M)\), choose a sequence \(g_n \in M\) such that \({\left\lVert {f-g_n} \right\rVert}\to d\), which is possible since \(d = \inf_{g\in M}{\left\lVert {f-g} \right\rVert}\). Apply the parallelogram law to write \begin{align*} {\left\lVert {g_n - g_m} \right\rVert}^2 &= {\left\lVert {(g_n - f) - (g_m - f) } \right\rVert}^2 \\ &= 2{\left\lVert {g_n - f} \right\rVert}^2 + 2{\left\lVert {g_m - f} \right\rVert}^2 - 4{\left\lVert { {1\over 2} (g_n + g_m) - f} \right\rVert}^2 \\ &\leq 2{\left\lVert {g_n - f} \right\rVert}^2 + 2{\left\lVert {g_m - f} \right\rVert}^2 - 4d^2 \\ &\leq 2d^2 + 2d^2 - 4d^2 \\ &= 0 ,\end{align*} so the \(g_n\) are Cauchy. Here we’ve used that \({1\over 2}(g_n + g_m) = m \in M\) since \(M\) is a subspace, and \({\left\lVert {m-f} \right\rVert} \geq d\). Since \(M\) is complete, \(g_n\to g\in M\) and moreover \({\left\lVert {f-g} \right\rVert} = d\). For uniqueness, if \({\left\lVert {f-g'} \right\rVert} = d\) then \begin{align*} {\left\lVert { f - {1\over 2} (g+g')} \right\rVert}^2 = d^2 - {\left\lVert {g-g'} \right\rVert}^2 < d^2 \qquad \contradiction .\end{align*}

Let \(M\leq {\mathcal{H}}\) be a closed subspace of a Hilbert space. Then \(M^\perp \leq {\mathcal{H}}\) is closed, and \({\mathcal{H}}= M \oplus M^\perp\). In the decomposition \(f= g+h\), in fact \(g\in M\) is the closest approximation to \(f\) in \(M\), making this decomposition unique.

STS \({\mathcal{H}}= M + M^\perp\) by the exercises. If \(f\in M\), take \(f=g+h\) where \(g=f\) and \(h=0\), so suppose \(f\not\in M\). Let \(g = \mathrm{argmin} \operatorname{dist}(f, M) \in M\), and we claim \(f-g\in M^\perp\), so \({\left\langle {f-g},~{h} \right\rangle} = 0\) for any \(h\in M\). For all \(h\in M\) and \(\alpha > 0\), we have \(g + \alpha h\in M\), so \begin{align*} {\left\lVert {f - g} \right\rVert}^2 &\leq {\left\lVert {f - (g+\alpha h)} \right\rVert}^2 \\ &= {\left\lVert {f-g} \right\rVert}^2 -2\Re\alpha {\left\langle {h},~{f-g} \right\rangle} + \alpha^2 {\left\lVert {h} \right\rVert}^2 ,\end{align*} so \begin{align*} 2\Re \alpha {\left\langle {h},~{f-g} \right\rangle} \leq \alpha^2 {\left\lVert {h} \right\rVert}^2 \implies 2\Re {\left\langle {h},~{f-g} \right\rangle} \leq \alpha {\left\lVert {h} \right\rVert}^2 \overset{\alpha\to 0}\longrightarrow 0 ,\end{align*} so \(\Re{\left\langle {h},~{f-g} \right\rangle} = 0\). Similarly \(\Im{\left\langle {h},~{f-g} \right\rangle} = 0\).

Show \(S^\perp\) is closed for any \(S\in {\mathcal{H}}\), and in fact \(S^\perp = \mathop{\mathrm{span}}_{\mathbb{C}}({ \operatorname{cl}} _{\mathcal{H}}(S))^\perp\), and if \(f\in S \cap S^\perp\) then \(f=0\).

Prove the parallelogram law \begin{align*} {\left\lVert {f-g} \right\rVert}^2 + {\left\lVert {f+g} \right\rVert}^2 = 2{\left\lVert {f} \right\rVert}^2 + 2{\left\lVert {g} \right\rVert}^2 .\end{align*}

6 Thursday, January 27


If \({\mathcal{H}}\) is a separable Hilbert spaces and \(K = \left\{{\phi_n}\right\}\) is an orthonormal set, then TFAE

Note \({\left\langle {f},~{\phi_n} \right\rangle}\) is the \(n\)th Fourier coefficient \(\widehat{f}(\xi) = \sum {\left\langle {f},~{\phi_n} \right\rangle}\phi_n(\xi)\), and Parseval says \({\left\lVert {f} \right\rVert}^2 = {\left\lVert {\widehat{f}} \right\rVert}^2\).

\(1\implies 2\): Let \(f\in {\mathcal{H}}\setminus{ \operatorname{cl}} \mathop{\mathrm{span}}(K)\) and project, so \(f= g+h\) with \(g,h\neq 0\). But \(g\in { \operatorname{cl}} \mathop{\mathrm{span}}(K)\) and \(h\in { \operatorname{cl}} \mathop{\mathrm{span}}(K)^\perp = 0\), forcing \(K^\perp = { \operatorname{cl}} (K)^\perp \neq 0\). \(\contradiction\)

\(2\implies 3\): Follows directly from previous lemma that \(f = P_Kf + (I-P_K)f\).

\(3\implies 4\): Write \(f\in {\mathcal{H}}\) as \(f = \sum {\left\langle {f},~{\phi_n} \right\rangle} \phi_n\) by sending \(m\to \infty\) in the previous lemma.

\(4\implies 1\): Toward a contradiction, suppose \(f\neq 0\in K^\perp\). Then \({\left\lVert {f} \right\rVert} \neq 0\) but \({\left\langle {f},~{\phi_n} \right\rangle}=0\) for all \(n\), contradicting Parseval. \(\contradiction\)

7 Tuesday, February 01


8 Tuesday, February 08

Motivating question: when is an operator equation solvable? Today: relation between boundedness and continuity for linear operators. Nonlinear operators next week.

Show that the following are equivalent conditions for continuity of \(A: V\to W\) at \(f_0\in D(A)\):

9 Tuesday, February 15

Last time:

If \(L: B\to C\) with dense image (so \({ \operatorname{cl}} _B(D(L)) = B\)), if \(L\) is continuous on \(D(L)\) then it has a unique extension \(\tilde L\) to all of \(B\), so \(D(\tilde L) = B\), with \({\left\lVert {L} \right\rVert} = {\left\lVert {\tilde L} \right\rVert}\).

In steps:

Let \({\mathcal{L}}\in L({\mathbb{C}}^n, {\mathbb{C}}^n)\) be defined in coordinates by \(({\mathcal{L}}f)_i \mathrel{\vcenter{:}}=\sum_{1\leq j\leq n} \alpha_{ij} f_j\) for \(1\leq i\leq n\). Take \({\left\lVert {{-}} \right\rVert}_{\ell^\infty}\) and check \begin{align*} {\left\lVert {Lf} \right\rVert}_\infty &\mathrel{\vcenter{:}}=\sup_i {\left\lvert {\sum \alpha_{ij} f_j} \right\rvert} \\ &\leq \qty{ \sup_i \sum {\left\lvert {\alpha_{ij} } \right\rvert} } \sup_j {\left\lvert {f_j} \right\rvert} \\ &\mathrel{\vcenter{:}}= m {\left\lVert {f} \right\rVert}_{\ell^\infty} .\end{align*} So \({\left\lVert {L} \right\rVert} \leq m\), where \(m\) is the largest row sum. Is there an \(f\) for which equality holds? In this case, we’d need \begin{align*} {\left\lVert {Lf} \right\rVert}_{\ell^\infty} \geq m {\left\lVert {f} \right\rVert}_{\ell^\infty} .\end{align*} Identify the row \(k\) so that \(m = \sum_{1\leq j\leq n} {\left\lvert {\alpha_{kj}} \right\rvert}\). Set \(f\) to be a unit vector with coefficients \((f)_j = \mkern 1.5mu\overline{\mkern-1.5mu\alpha_{kj}\mkern-1.5mu}\mkern 1.5mu / {\left\lvert {\alpha_{kj}} \right\rvert}\). Then \begin{align*} {\left\lVert {Lf} \right\rVert}_\infty &= \sup_i {\left\lvert { \sum_j \alpha_{ij} f_j } \right\rvert} \\ &\geq {\left\lvert {\sum_j \alpha_{kj} f_j } \right\rvert} \\ &= {\left\lvert {\sum_j \alpha_{kj} {\mkern 1.5mu\overline{\mkern-1.5mu\alpha\mkern-1.5mu}\mkern 1.5mu_{kj} / {\left\lvert {\alpha_{k j}} \right\rvert} } } \right\rvert} \\ &= \sum_j {\left\lvert {\alpha_{kj}} \right\rvert} \\ &= m {\left\lVert {f} \right\rVert}_{\ell^\infty} .\end{align*}

So the answer is yes in this case. Does this also work for \({\left\lVert {{-}} \right\rVert}_{\ell^p}\) with \(p\in (1, \infty)\)? Recall Holder’s inequality: \begin{align*} {\left\lvert {\sum \alpha_{ij} f_j } \right\rvert} &\leq \qty{ \sum {\left\lvert {\alpha_{ij}} \right\rvert}^q }^{1\over q} \qty{\sum {\left\lvert {f_j} \right\rvert}^p }^{1\over p} \\ &= \qty{ \sum {\left\lvert {\alpha_{ij}} \right\rvert}^q }^{1\over q} {\left\lVert {f} \right\rVert}_{\ell^p} .\end{align*} Check that \begin{align*} {\left\lVert {Lf} \right\rVert}_{\ell^p}^p &= \sum_i {\left\lvert {(Lf)_i } \right\rvert}^p \\ &= \sum_i {\left\lvert {\sum_j \alpha_{ij} f_j } \right\rvert}^p \\ &\leq \sum_i \qty{ \sum_j {\left\lvert {\alpha_{ij} } \right\rvert} }^{p\over q} {\left\lVert {f} \right\rVert}_{\ell^p}^p ,\end{align*} where we’ve applied Holder in the last line. Thus \begin{align*} {\left\lVert {L} \right\rVert} \leq \qty{ \sum_i \qty{ \sum_j {\left\lvert { \alpha_{ij}} \right\rvert} }^{p\over q} }^{1\over p} .\end{align*}

Is there an \(f\) that attains this bound in the \(\ell^p\) case?

For \({\mathcal{L}}\in L({\mathbb{C}}^\infty, {\mathbb{C}}^\infty)\) defined by \((Lf)_i = \sum_{j\geq 11} \alpha_{ij} f_j\) for \(j\geq 1\), one needs a notion of convergence of the coordinates \(\alpha_{ij}\) in order for \({\mathcal{L}}\) to be bounded. A sufficient condition is \(m\mathrel{\vcenter{:}}=\sup_i \sum_{j\geq 1} {\left\lvert {\alpha_{ij}} \right\rvert} < \infty\).

Some notation:

\begin{align*} {\left\lVert \left\lVert {\alpha} \right\rVert\right\rVert}_1 &\mathrel{\vcenter{:}}=\sup_j \sum_i {\left\lvert { \alpha_{ij}} \right\rvert} \\ {\left\lVert \left\lVert {\alpha} \right\rVert\right\rVert}_p &\mathrel{\vcenter{:}}=\qty{ \sum_i \qty{ \sum_j {\left\lvert {\alpha_{ij} } \right\rvert}^{q} }^{p\over q} }^{1\over p} \\ {\left\lVert \left\lVert {\alpha} \right\rVert\right\rVert}_\infty &\mathrel{\vcenter{:}}=\sup_i \sum_j {\left\lvert { \alpha_{ij}} \right\rvert} .\end{align*}

Note that if \({\mathcal{L}}: \ell^p\to \ell^p\), then \({\left\lVert {L} \right\rVert} \leq {\left\lVert \left\lVert {\alpha} \right\rVert\right\rVert}_p\) for \(p\in [1, \infty)\) and for \(p=\infty\) this is an equality.

Consider \(C[a, b]\) with \({\left\lVert {{-}} \right\rVert}_\infty\) and \(k\in C^0( [a,b]{ {}^{ \scriptscriptstyle\times^{2} } }, {\mathbb{C}})\). Define \begin{align*} K: C[a, b] &\to C[a, b] \\ f &\mapsto \int_a^b k(x, y) f(y) \,dy .\end{align*} What is \({\left\lVert {K} \right\rVert}\)? Estimate \begin{align*} {\left\lVert {Kf} \right\rVert} &\leq \sup_{y\in [a, b]}{\left\lvert {f(y)} \right\rvert} \sup_{x\in [a, b]} \int_a^b {\left\lvert {k(x, y)} \right\rvert} \,dy\\ &\leq {\left\lVert {f} \right\rVert}_\infty {\left\lVert {k} \right\rVert}_\infty ,\end{align*} so \({\left\lVert {K} \right\rVert} \leq {\left\lVert {k} \right\rVert}_\infty\).

Define \begin{align*} {\left\lVert \left\lVert {k} \right\rVert\right\rVert}_1 &\mathrel{\vcenter{:}}=\sup_y \int {\left\lvert {k} \right\rvert} \,dx\\ {\left\lVert \left\lVert {k} \right\rVert\right\rVert}_p &\mathrel{\vcenter{:}}=\qty{\int \qty{ \int {\left\lvert {k} \right\rvert}^q \,dy}^{p\over q} \,dx}^{1\over p} \\ {\left\lVert \left\lVert {k} \right\rVert\right\rVert}_\infty &\mathrel{\vcenter{:}}=\sup_x \int {\left\lvert {k} \right\rvert} \,dy .\end{align*}

10 Bibliography