## June 10th, 2021

### On normed vector spaces

Originally published at 狗和留美者不得入内. You can comment here or there.

Here, the underlying field of any vector space shall be either or . Moreover, subspace will always denote the subspace of a vector space.

Definition 1 A seminorm on a vector space over is a function that satisfies the following properties.

1. Absolute homogeneity: for all , , .
2. Triangle inequality: for all , .

Proposition 1 For any seminorm , .

Proof: Follows directly from absolute homogeneity.

Definition 2 A norm on a vector space is a seminorm such that iff .

Definition 3 A vector space equipped with a norm is called a normed vector space. The topology it defines is called the norm topology on

Definition 4 A sequence of vectors in vector space converges with respect to norm iff .

Definition 5 A normed vector space that is complete with respect to the norm metric is called a Banach space.

Definition 6 A series converges absolutely iff .

Theorem 1 A normed vector space is complete iff every series in it that converges absolutely also converges with respect to the norm topology.

Proof: We assume the space is complete. This means that for any Cauchy sequence , for some . Now take any such that , which of course means that . To show that it converges, it suffices to show that is Cauchy. We have that for all

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<p><small>Originally published at <a href="https://gmachine1729.wpcomstaging.com/2021/06/10/on-normed-vector-spaces/">狗和留美者不得入内</a>. You can comment here or <a href="https://gmachine1729.wpcomstaging.com/2021/06/10/on-normed-vector-spaces/#comments">there</a>.</small></p><div class="RichText ztext Post-RichText"> <p>Here, the underlying field <img src="https://www.zhihu.com/equation?tex=K" alt="[公式]" eeimg="1" data-formula="K"> of any vector space shall be either <img src="https://www.zhihu.com/equation?tex=%5Cmathbb%7BR%7D" alt="[公式]" eeimg="1" data-formula="\mathbb{R}"> or <img src="https://www.zhihu.com/equation?tex=%5Cmathbb%7BC%7D" alt="[公式]" eeimg="1" data-formula="\mathbb{C}"> . Moreover, subspace will always denote the subspace of a vector space.</p> <p><b>Definition 1</b> A <i>seminorm</i> on a vector space <img src="https://www.zhihu.com/equation?tex=X" alt="[公式]" eeimg="1" data-formula="X"> over <img src="https://www.zhihu.com/equation?tex=K" alt="[公式]" eeimg="1" data-formula="K"> is a function <img src="https://www.zhihu.com/equation?tex=%5Cparallel+%5Ccdot+%5Cparallel+%3A+X+%5Cto+%5B0%2C+%5Cinfty%29" alt="[公式]" eeimg="1" data-formula="\parallel \cdot \parallel : X \to [0, \infty)"> that satisfies the following properties.</p> <ol> <li>Absolute homogeneity: for all <img src="https://www.zhihu.com/equation?tex=x+%5Cin+X" alt="[公式]" eeimg="1" data-formula="x \in X"> , <img src="https://www.zhihu.com/equation?tex=%5Clambda+%5Cin+K" alt="[公式]" eeimg="1" data-formula="\lambda \in K"> , <img src="https://www.zhihu.com/equation?tex=%5Cparallel+%5Clambda+x%5Cparallel+%3D+%7C%5Clambda%7C+%5Cparallel+x%5Cparallel" alt="[公式]" eeimg="1" data-formula="\parallel \lambda x\parallel = |\lambda| \parallel x\parallel"> .</li> <li>Triangle inequality: for all <img src="https://www.zhihu.com/equation?tex=x%2Cy%5Cin+X" alt="[公式]" eeimg="1" data-formula="x,y\in X"> , <img src="https://www.zhihu.com/equation?tex=%5Cparallel+x%2By%5Cparallel+%5Cleq+%5Cparallel+x+%5Cparallel+%2B+%5Cparallel+y+%5Cparallel" alt="[公式]" eeimg="1" data-formula="\parallel x+y\parallel \leq \parallel x \parallel + \parallel y \parallel"> .</li> </ol> <p><b>Proposition 1</b> For any seminorm <img src="https://www.zhihu.com/equation?tex=%5Cparallel+%5Ccdot+%5Cparallel" alt="[公式]" eeimg="1" data-formula="\parallel \cdot \parallel"> , <img src="https://www.zhihu.com/equation?tex=%5Cparallel+0%5Cparallel+%3D+0" alt="[公式]" eeimg="1" data-formula="\parallel 0\parallel = 0"> .</p> <p><i>Proof</i>: Follows directly from absolute homogeneity. <img src="https://www.zhihu.com/equation?tex=%5Csquare" alt="[公式]" eeimg="1" data-formula="\square"> </p> <p><b>Definition 2</b> A <i>norm</i> <img src="https://www.zhihu.com/equation?tex=%5Cparallel+%5Ccdot+%5Cparallel" alt="[公式]" eeimg="1" data-formula="\parallel \cdot \parallel"> on a vector space is a <i>seminorm</i> such that <img src="https://www.zhihu.com/equation?tex=%5Cparallel+x+%5Cparallel+%3D+0" alt="[公式]" eeimg="1" data-formula="\parallel x \parallel = 0"> iff <img src="https://www.zhihu.com/equation?tex=x+%3D+0" alt="[公式]" eeimg="1" data-formula="x = 0"> .</p> <p><b>Definition 3</b> A vector space equipped with a norm is called a <i>normed vector space</i>. The topology it defines is called the <i>norm topology</i> on <img src="https://www.zhihu.com/equation?tex=X" alt="[公式]" eeimg="1" data-formula="X"> </p> <p><b>Definition 4</b> A sequence of vectors <img src="https://www.zhihu.com/equation?tex=%5C%7Bx_n%5C%7D" alt="[公式]" eeimg="1" data-formula="\{x_n\}"> in vector space <img src="https://www.zhihu.com/equation?tex=X" alt="[公式]" eeimg="1" data-formula="X"> converges with respect to norm <img src="https://www.zhihu.com/equation?tex=%5Cparallel+%5Ccdot+%5Cparallel" alt="[公式]" eeimg="1" data-formula="\parallel \cdot \parallel"> iff <img src="https://www.zhihu.com/equation?tex=%5Cparallel+x_n+-+x%5Cparallel+%5Cto+0" alt="[公式]" eeimg="1" data-formula="\parallel x_n - x\parallel \to 0"> .</p> <p><b>Definition 5</b> A normed vector space that is complete with respect to the norm metric is called a <i>Banach space</i>.</p> <p><b>Definition 6</b> A series <img src="https://www.zhihu.com/equation?tex=%5Csum_%7Bi%3D1%7D%5E%5Cinfty+x_i" alt="[公式]" eeimg="1" data-formula="\sum_{i=1}^\infty x_i"> <i>converges absolutely</i> iff <img src="https://www.zhihu.com/equation?tex=%5Csum_%7Bi%3D1%7D%5E%5Cinfty+%5Cparallel+x_i%5Cparallel+%3C+%5Cinfty" alt="[公式]" eeimg="1" data-formula="\sum_{i=1}^\infty \parallel x_i\parallel < \infty"> .</p> <p><b>Theorem 1</b> A normed vector space <img src="https://www.zhihu.com/equation?tex=X" alt="[公式]" eeimg="1" data-formula="X"> is complete iff every series in it that converges absolutely also converges with respect to the norm topology.</p> <p><i>Proof</i>: We assume the space is complete. This means that for any Cauchy sequence <img src="https://www.zhihu.com/equation?tex=%5C%7Bx_n%5C%7D" alt="[公式]" eeimg="1" data-formula="\{x_n\}"> , <img src="https://www.zhihu.com/equation?tex=x_n+%5Cto+x" alt="[公式]" eeimg="1" data-formula="x_n \to x"> for some <img src="https://www.zhihu.com/equation?tex=x+%5Cin+X" alt="[公式]" eeimg="1" data-formula="x \in X"> . Now take any <img src="https://www.zhihu.com/equation?tex=%5C%7By_n%5C%7D" alt="[公式]" eeimg="1" data-formula="\{y_n\}"> such that <img src="https://www.zhihu.com/equation?tex=%5Csum_%7Bi%3D1%7D%5E%5Cinfty+%5Cparallel+y_i%5Cparallel+%3C+%5Cinfty" alt="[公式]" eeimg="1" data-formula="\sum_{i=1}^\infty \parallel y_i\parallel < \infty"> , which of course means that <img src="https://www.zhihu.com/equation?tex=%5Cparallel+y_n+%5Cparallel+%5Cto+0" alt="[公式]" eeimg="1" data-formula="\parallel y_n \parallel \to 0"> . To show that it converges, it suffices to show that <img src="https://www.zhihu.com/equation?tex=S_n+%3D+%5Csum_%7Bi%3D1%7D%5En+y_i" alt="[公式]" eeimg="1" data-formula="S_n = \sum_{i=1}^n y_i"> is Cauchy. We have that for all <img src="https://www.zhihu.com/equation?tex=%5Cepsilon%3E0" alt="[公式]" eeimg="1" data-formula="\epsilon>0&#8243;> , there exists <img src="https://www.zhihu.com/equation?tex=N" alt="[公式]" eeimg="1" data-formula="N"> such that <img src="https://www.zhihu.com/equation?tex=n+%5Cgeq+m%5Cgeq+N" alt="[公式]" eeimg="1" data-formula="n \geq m\geq N"> implies <img src="https://www.zhihu.com/equation?tex=%5Csum_%7Bi%3Dm%7D%5En+%5Cparallel+y_i%5Cparallel+%3C+%5Cepsilon" alt="[公式]" eeimg="1" data-formula="\sum_{i=m}^n \parallel y_i\parallel < \epsilon"> . The triangle inequality implies that <img src="https://www.zhihu.com/equation?tex=%5Cleft%5ClVert+%5Csum_%7Bi%3Dm%7D%5En+y_i+%5Cright%5CrVert+%3C+%5Cepsilon" alt="[公式]" eeimg="1" data-formula="\left\lVert \sum_{i=m}^n y_i \right\rVert < \epsilon"> , which shows that <img src="https://www.zhihu.com/equation?tex=S_n" alt="[公式]" eeimg="1" data-formula="S_n"> is Cauchy.</p> <p>We now assume that absolute convergence implies convergence with respect to norm topology. Take any Cauchy sequence <img src="https://www.zhihu.com/equation?tex=%5C%7By_n%5C%7D" alt="[公式]" eeimg="1" data-formula="\{y_n\}"> . For all <img src="https://www.zhihu.com/equation?tex=k+%5Cin+%5Cmathbb%7BN%7D" alt="[公式]" eeimg="1" data-formula="k \in \mathbb{N}"> , there exists a minimum <img src="https://www.zhihu.com/equation?tex=n_k" alt="[公式]" eeimg="1" data-formula="n_k"> such that <img src="https://www.zhihu.com/equation?tex=n+%5Cgeq+m+%5Cgeq+n_k" alt="[公式]" eeimg="1" data-formula="n \geq m \geq n_k"> implies <img src="https://www.zhihu.com/equation?tex=%5ClVert+y_n+-+y_%7Bm%7D+%5CrVert+%3C+%5Cfrac%7B1%7D%7B2%5Ek%7D" alt="[公式]" eeimg="1" data-formula="\lVert y_n - y_{m} \rVert < \frac{1}{2^k}"> . From this we derive a subsequence <img src="https://www.zhihu.com/equation?tex=%5C%7By_%7Bn_k%7D%5C%7D" alt="[公式]" eeimg="1" data-formula="\{y_{n_k}\}"> . Let <img src="https://www.zhihu.com/equation?tex=%5C%7Bx_k+%3D+y_%7Bn_%7Bk%2B1%7D%7D+-+y_%7Bn_%7Bk%7D%7D%5C%7D" alt="[公式]" eeimg="1" data-formula="\{x_k = y_{n_{k+1}} - y_{n_{k}}\}"> . We have that <img src="https://www.zhihu.com/equation?tex=y_%7Bn_%7Bk%2B1%7D%7D+%3D+y_%7Bn_1%7D%2B%5Csum_%7Bi%3D1%7D%5Ek+x_i" alt="[公式]" eeimg="1" data-formula="y_{n_{k+1}} = y_{n_1}+\sum_{i=1}^k x_i"> . That upper bound by <img src="https://www.zhihu.com/equation?tex=%5Cfrac%7B1%7D%7B2%5Ek%7D" alt="[公式]" eeimg="1" data-formula="\frac{1}{2^k}"> tells us that <img src="https://www.zhihu.com/equation?tex=%5Csum_%7Bi%3D1%7D%5E%5Cinfty+%5ClVert+x_k+%5CrVert+%3C+%5Cinfty" alt="[公式]" eeimg="1" data-formula="\sum_{i=1}^\infty \lVert x_k \rVert < \infty"> , which by our hypothesis implies that <img src="https://www.zhihu.com/equation?tex=%5Csum_%7Bi%3D1%7D%5E%5Cinfty+x_k" alt="[公式]" eeimg="1" data-formula="\sum_{i=1}^\infty x_k"> is convergent, which tells us that <img src="https://www.zhihu.com/equation?tex=%5C%7By_%7Bn_k%7D%5C%7D" alt="[公式]" eeimg="1" data-formula="\{y_{n_k}\}"> is convergent to the same value. Its being a subsequence of Cauchy sequence <img src="https://www.zhihu.com/equation?tex=%5C%7By_n%5C%7D" alt="[公式]" eeimg="1" data-formula="\{y_n\}"> means that <img src="https://www.zhihu.com/equation?tex=%5C%7By_n%5C%7D" alt="[公式]" eeimg="1" data-formula="\{y_n\}"> converges to the same value too. This completes our proof. <img src="https://www.zhihu.com/equation?tex=%5Csquare" alt="[公式]" eeimg="1" data-formula="\square"> </p> <p>This proposition felt initially elusive or not very intuitive to me. However, once one realize that one can telescope a Cauchy sequence to express it as a sequence of partial sums, it is natural to realize that the condition regarding absolute convergence would imply completeness.</p> <p><b>Definition 7</b> Let <img src="https://www.zhihu.com/equation?tex=X%2C+Y" alt="[公式]" eeimg="1" data-formula="X, Y"> be two normed vector spaces. Let <img src="https://www.zhihu.com/equation?tex=A%3A+X+%5Cto+Y" alt="[公式]" eeimg="1" data-formula="A: X \to Y"> be a linear operator. Moreover, the following are equivalent.</p> <ol> <li><img src="https://www.zhihu.com/equation?tex=A" alt="[公式]" eeimg="1" data-formula="A"> is <i>bounded</i>.</li> <li>There exists <img src="https://www.zhihu.com/equation?tex=C+%5Cin+%5Cmathbb%7BR%7D" alt="[公式]" eeimg="1" data-formula="C \in \mathbb{R}"> such that for all <img src="https://www.zhihu.com/equation?tex=x+%5Cin+X" alt="[公式]" eeimg="1" data-formula="x \in X"> ,<img src="https://www.zhihu.com/equation?tex=%5ClVert+Ax%5CrVert+%5Cleq+C%5ClVert+x%5CrVert" alt="[公式]" eeimg="1" data-formula="\lVert Ax\rVert \leq C\lVert x\rVert"> .</li> <li>For some <img src="https://www.zhihu.com/equation?tex=%5Cdelta+%3E+0" alt="[公式]" eeimg="1" data-formula="\delta > 0&#8243;> , there exists <img src="https://www.zhihu.com/equation?tex=C+%5Cin+%5Cmathbb%7BR%7D" alt="[公式]" eeimg="1" data-formula="C \in \mathbb{R}"> such that for all <img src="https://www.zhihu.com/equation?tex=x+%5Cin+X" alt="[公式]" eeimg="1" data-formula="x \in X"> such that <img src="https://www.zhihu.com/equation?tex=%5ClVert+x%5CrVert+%3D+%5Cdelta" alt="[公式]" eeimg="1" data-formula="\lVert x\rVert = \delta"> ,<img src="https://www.zhihu.com/equation?tex=%5ClVert+Ax%5CrVert+%5Cleq+C%5ClVert+x%5CrVert" alt="[公式]" eeimg="1" data-formula="\lVert Ax\rVert \leq C\lVert x\rVert"> .</li> </ol> <p><b>Proposition 2</b> Let <img src="https://www.zhihu.com/equation?tex=X%2C+Y" alt="[公式]" eeimg="1" data-formula="X, Y"> be two normed vector spaces. Let <img src="https://www.zhihu.com/equation?tex=A%3A+X+%5Cto+Y" alt="[公式]" eeimg="1" data-formula="A: X \to Y"> be a linear operator. The following are equivalent.</p> <ol> <li><img src="https://www.zhihu.com/equation?tex=A" alt="[公式]" eeimg="1" data-formula="A"> is continuous.</li> <li><img src="https://www.zhihu.com/equation?tex=A" alt="[公式]" eeimg="1" data-formula="A"> is continuous at <img src="https://www.zhihu.com/equation?tex=0" alt="[公式]" eeimg="1" data-formula="0">. </li> <li><img src="https://www.zhihu.com/equation?tex=A" alt="[公式]" eeimg="1" data-formula="A"> is bounded.</li> </ol> <p><i>Proof</i>: That (1) implies (2) is immediate. Assume (2), which means that every neighborhood <img src="https://www.zhihu.com/equation?tex=N_Y" alt="[公式]" eeimg="1" data-formula="N_Y"> of <img src="https://www.zhihu.com/equation?tex=0+%5Cin+Y" alt="[公式]" eeimg="1" data-formula="0 \in Y"> , there exists an open ball of radius <img src="https://www.zhihu.com/equation?tex=%5Cdelta" alt="[公式]" eeimg="1" data-formula="\delta"> centered at <img src="https://www.zhihu.com/equation?tex=0+%5Cin+X" alt="[公式]" eeimg="1" data-formula="0 \in X"> , which we denote via <img src="https://www.zhihu.com/equation?tex=B%280%2C+%5Cdelta%29" alt="[公式]" eeimg="1" data-formula="B(0, \delta)"> , such that <img src="https://www.zhihu.com/equation?tex=A%28B%280%2C+%5Cdelta%29%29+%5Csubset+N_Y" alt="[公式]" eeimg="1" data-formula="A(B(0, \delta)) \subset N_Y"> . Let <img src="https://www.zhihu.com/equation?tex=N_Y" alt="[公式]" eeimg="1" data-formula="N_Y"> be bounded above in norm by <img src="https://www.zhihu.com/equation?tex=M+%3E+0" alt="[公式]" eeimg="1" data-formula="M > 0&#8243;> . Then, (3) of Definition 7 is satisfied, or equivalently, <img src="https://www.zhihu.com/equation?tex=A" alt="[公式]" eeimg="1" data-formula="A"> is bounded. Now we show that (3) implies (1). Assume (3), namely that there exists <img src="https://www.zhihu.com/equation?tex=C%3E0" alt="[公式]" eeimg="1" data-formula="C>0&#8243;> such that <img src="https://www.zhihu.com/equation?tex=%5ClVert+Ax%5CrVert+%5Cleq+C+%5ClVert+x%5CrVert" alt="[公式]" eeimg="1" data-formula="\lVert Ax\rVert \leq C \lVert x\rVert"> for all <img src="https://www.zhihu.com/equation?tex=x+%5Cin+X" alt="[公式]" eeimg="1" data-formula="x \in X"> . In that case, if <img src="https://www.zhihu.com/equation?tex=%5ClVert+x_1+-+x_2+%5CrVert+%3C+C%5E%7B-1%7D%5Cepsilon" alt="[公式]" eeimg="1" data-formula="\lVert x_1 - x_2 \rVert < C^{-1}\epsilon"> , then <img src="https://www.zhihu.com/equation?tex=%5ClVert+Ax_1+-+Ax_2%5CrVert+%3D+%5ClVert+A%28x_1+-+x_2%29+%5CrVert+%3C+%5Cepsilon" alt="[公式]" eeimg="1" data-formula="\lVert Ax_1 - Ax_2\rVert = \lVert A(x_1 - x_2) \rVert < \epsilon"> . This implies that if <img src="https://www.zhihu.com/equation?tex=%5C%7Bx_n%5C%7D" alt="[公式]" eeimg="1" data-formula="\{x_n\}"> is Cauchy than <img src="https://www.zhihu.com/equation?tex=%5C%7BAx_n%5C%7D" alt="[公式]" eeimg="1" data-formula="\{Ax_n\}"> is also Cauchy, which is a definition of continuity in a metric space. <img src="https://www.zhihu.com/equation?tex=%5Csquare" alt="[公式]" eeimg="1" data-formula="\square"> </p> <p><b>Definition 8</b> We define a function <img src="https://www.zhihu.com/equation?tex=T+%5Cmapsto+%5ClVert+T%5CrVert" alt="[公式]" eeimg="1" data-formula="T \mapsto \lVert T\rVert"> by</p> <p><img src="https://www.zhihu.com/equation?tex=%5Cbegin%7Beqnarray%7D+%5ClVert+T%5CrVert+%26%3D%26+%5Csup%5C%7B%5ClVert+Tx%5CrVert%3A+%5ClVert+x%5CrVert+%3D+1%5C%7D%5C%5C+%26%3D%26%5Csup%5Cleft%5C%7B%5Cfrac%7B%5ClVert+Tx%5CrVert%7D%7B%5ClVert+x%5CrVert%7D%3A+x+%5Cneq+0%5Cright%5C%7D%5C%5C+%26%3D%26+%5Cinf+%5C%7BC+%3A+%5ClVert+Tx%5CrVert+%5Cleq+C%5ClVert+x%5CrVert%2C+%5Cforall+x%5C%7D+%5Cend%7Beqnarray%7D%5C%5C" alt="[公式]" eeimg="1" data-formula="\begin{eqnarray} \lVert T\rVert &amp;=&amp; \sup\{\lVert Tx\rVert: \lVert x\rVert = 1\}\\ &amp;=&amp;\sup\left\{\frac{\lVert Tx\rVert}{\lVert x\rVert}: x \neq 0\right\}\\ &amp;=&amp; \inf \{C : \lVert Tx\rVert \leq C\lVert x\rVert, \forall x\} \end{eqnarray}\\"> </p> <p>on the vector space <img src="https://www.zhihu.com/equation?tex=%5Cmathcal%7BL%7D%28X%2CY%29" alt="[公式]" eeimg="1" data-formula="\mathcal{L}(X,Y)"> of linear transformations from <img src="https://www.zhihu.com/equation?tex=X" alt="[公式]" eeimg="1" data-formula="X"> to <img src="https://www.zhihu.com/equation?tex=Y" alt="[公式]" eeimg="1" data-formula="Y"> , which is called the <i>operator norm</i>. We leave to the reader to verify that it is a norm.</p> <p><b>Proposition 3</b> If <img src="https://www.zhihu.com/equation?tex=Y" alt="[公式]" eeimg="1" data-formula="Y"> is complete, so is <img src="https://www.zhihu.com/equation?tex=%5Cmathcal%7BL%7D%28X%2CY%29" alt="[公式]" eeimg="1" data-formula="\mathcal{L}(X,Y)"> .</p> <p><i>Proof</i>: For any Cauchy sequence <img src="https://www.zhihu.com/equation?tex=%5C%7BA_n%5C%7D" alt="[公式]" eeimg="1" data-formula="\{A_n\}"> in <img src="https://www.zhihu.com/equation?tex=%5Cmathcal%7BL%7D%28X%2CY%29" alt="[公式]" eeimg="1" data-formula="\mathcal{L}(X,Y)"> , <img src="https://www.zhihu.com/equation?tex=%5C%7BA_n+x%5C%7D" alt="[公式]" eeimg="1" data-formula="\{A_n x\}"> is also Cauchy. Thus, we can define <img src="https://www.zhihu.com/equation?tex=A%3A+X+%5Cto+Y" alt="[公式]" eeimg="1" data-formula="A: X \to Y"> by <img src="https://www.zhihu.com/equation?tex=Ax+%3D+%5Clim+A_n+x" alt="[公式]" eeimg="1" data-formula="Ax = \lim A_n x"> . We leave to the reader to verify that <img src="https://www.zhihu.com/equation?tex=A+%5Cin+%5Cmathcal%7BL%7D%28X%2CY%29" alt="[公式]" eeimg="1" data-formula="A \in \mathcal{L}(X,Y)"> and that it is indeed in limit with respect to the operator norm. <img src="https://www.zhihu.com/equation?tex=%5Csquare" alt="[公式]" eeimg="1" data-formula="\square"> </p> <p><b>Proposition 4</b> If <img src="https://www.zhihu.com/equation?tex=B+%5Cin+%5Cmathcal%7BL%7D%28X%2CY%29%2C+A+%5Cin+%5Cmathcal%7BL%7D%28Y%2CZ%29" alt="[公式]" eeimg="1" data-formula="B \in \mathcal{L}(X,Y), A \in \mathcal{L}(Y,Z)"> , then</p> <p><img src="https://www.zhihu.com/equation?tex=%5ClVert+AB%5CrVert+%5Cleq+%5ClVert+A%5ClVert+%5ClVert+B%5ClVert.%5C%5C+" alt="[公式]" eeimg="1" data-formula="\lVert AB\rVert \leq \lVert A\lVert \lVert B\lVert.\\ "> </p> <p><i>Proof</i>: Very mechanical and left to the reader. <img src="https://www.zhihu.com/equation?tex=%5Csquare" alt="[公式]" eeimg="1" data-formula="\square"> </p> <p><b>References</b></p> <ul> <li>[1] Gerald B. Folland. <i>Real Analysis &#8211; Modern Techniques and their Applications</i>. John Wiley &amp; Sons, Inc., 1999.</li> </ul> </div>

### How to interpret the Hahn-Banach theorem

Originally published at 狗和留美者不得入内. You can comment here or there.

The motivation behind the Hahn-Banach theorem can come across to a functional analysis newbie as somewhat elusive. I shall here try to explain this to the extent that I understand it.

Suppose that a seminorm on a vector space is such that iff , for some subspace of . Let be the dual space of . If is bounded with respect to this seminorm, then . We wish to induce via this seminorm a norm on the quotient space . Since a norm induces a metric (and a seminorm induces a pseudometric), it is natural then to define the norm on to be the distance corresponding to the seminorm between and , which is formally . It is easy to verify that this is well defined and a norm.

Similarly, induces an element of . Since is arbitrary, is also an arbitrary functional on its domain. We wish to show that given the constraint that for some subspace of , , we can for any , define functional on such that .

We are interested in extending a functional defined on a subspace to the full space with agreement of values on the subspace and a certain degree of boundnesses, more specifically an upper bound by the norm. Since this is trivially obtained by simply mapping to the elements outside the subspace, we are interested in an extension that is as non-zero or as large in absolute value as possible. In attempt to achieve this, we can try extending the functional with the requirement that its value on any input , which can be negative, is bounded above by the value of the application of another function on that has the reals, including negative ones, as its codomain. With this along with the properties of norms and seminorms in mind, we define the following.

Definition 1 Let be a real vector space. A sublinear functional on is a map such that for all and ,

1. Triangle inequality: .
2. Non-negative homogeneity: .

We immediately notice that the constraints defining a sublinear functional are a subset of the constraints defining a seminorm or norm, which means that any seminorm or norm is necessarily a sublinear functional, which means that any proposition that holds for an arbitrary sublinear functional also holds for an arbitrary seminorm or norm.

Lemma 1 We extend a linear functional , where is a subspace of , defined such that for all , for an arbitrary sublinear functional to the subspace such that for all .

Proof: must of course also be nonzero. Let be our extended functional, with and . We require that for any , for any ,

The case of is trivial.

In the case of , the inequality in is equivalent to

In the case of , the inequality in is equivalent to

Here, we notice that the product of any by any scalar is also in , by the closure property of subspace. Thus, if we show that for arbitrary ,

we have shown the existence of the desired . follows from

in which we used linearity of , the fact that on , and the triangle inequality on . This completes our proof.

The proof of the Hahn-Banach theorem, of which Lemma 1, is the most difficulty part, can come across as coming out of the blue. I certainly developed a better idea of how to derive it by “working backwards”, as done above, first assuming the existence of the desired property, then finding a condition that implies it, and finally proving that that condition is indeed satisfied.

Theorem 1 (Hahn Banach theorem) Let be a real vector space on which is defined a sublinear functional . Let be any subspace of it and be some linear functional such that for all . Then, there exists a linear functional such that for all , and for all , .

Proof: Lemma 1 tells us that if , we can always extend onto some subspace , which is a proper extension of subspace such the extension of is bounded above by on and agrees with on . Let be the collection of two-tuples such that is a linear functional defined on and bounded above by on . Let . We say that iff and on . One easy verifies that this is a partial order on . For any chain in , we take the union of all sets in the chain and define a function with for any , for some in the chain that is defined on a domain that contains . It is apparent that for any in the chain, . Thus, we can apply Zorn’s lemma to derive the existence of a maximal element in with respect to this partial order. The set associated with any maximal element must be itself in order for Lemma 1 to not be violated.

Now, we will go about generalized the Hahn-Banach theorem to complex vector spaces.

Lemma 2 Let be a complex vector space and let be a linear functional on . If is a complex linear functional on and , then is a real linear functional, and for all . Conversely, if is a real linear functional on and is defined by , then is complex linear. In this case, if is normed, we have .

Proof: Let Then, for any , we write , where . We have and . Thus, . For any , . That for any , is also easily verified.

For the converse, one easily verifies that , and for ,

For any , we have that . This shows that . With for some , , since is linear, we have that . This shows that for any , there exists a of the same norm such that , which shows that . This completes our proof.

In the proof of the above lemma, we omitted the case of . [1] introduced the notation

Using this we can define the polar decomposition of any as

Applying to gives us . We note that in the proof of Lemma 2, we multiplied by .

Lemma 3 For any complex vector space , there exists a real vector space and a function that is bijective and linear with respect to real but not complex coefficients.

Proof: For any , we must have and also for all . We also stipulate that for any , .Take any basis of . Then we have as a set of basis elements defining , . To verify that no non-trivial linear combination of basis elements of can equal , one can simply use linearity to derive violation of the definition of basis of in the case of linear dependence of a subset of basis elements of .

We have defined to be linear with respect to real coefficients. It is not at all linear with respect to complex coefficients as and are basis elements of a real vector space; since the underlying field is multiplying a vector of it by an imaginary number is simply not defined here.

Theorem 2 (Complex Hahn-Banach Theorem) Let be a complex vector space, a seminorm on , a subspace of , and a complex linear functional on such that for . Then there exists a complex linear functional on such that for all and

Proof: Let . By Lemma 3, there exists a real vector space with linear with respect to real coefficients and bijective. Let be defined by . One easily verifies that is real linear functional on . Moreover, on , which is a subspace of , , with a seminorm on easily verified as well. By the Hahn-Banach theorem for real vector spaces (Theorem 1), there is a real linear functional defined on that agrees with on such that for all , . From this we also derive an analogous extension of to , which is . Now, let on . As in the proof of Lemma 2, if , we have . Since on , we also have on . This completes our proof.

Theorem 3 Let be a normed vector space (over ). Then,

1. If is a subspace of and , there exists such that and . In fact, if , can be defined such that , in which case .
2. If , there exists such that and .
3. The bounded linear functionals on separate points.

Proof: We wish to define on , which is, by linearity, done simply by prescribing the value of . The function on is such that if and only if . We need to prescribe an and we try the largest value which satisfies the requirement as given in the Hahn-Banach theorem, where the take to be the upper bounding seminorm. We let . Take arbitrary . We now wish to show that

The inequality in this is equivalent to

which is true from the definition of .

We have by the definition of operator norm that

Set and pick a sequence such that from above which exists by definition of . We have thus shown that . Obviously, from this we also have . Applying Hahn-Banach theorem to with as the seminorm to the extension of shows that we can extend the extension of all of , which completes the proof of (1).

(2) is a special case of (1) with . As for (3), if , ether is a complex multiple of or not. If yes, would suffice. If not, we can define on to be the constant zero function, and then, by (1), we can extend to a function on the entire space such that .

References

• [1] Gerald B. Folland. Real Analysis – Modern Techniques and their Applications. John Wiley & Sons, Inc., 1999.