Solution.

The induced metric

Non-negativity and \(d(x,y)=0\iff x=y\) come directly from the corresponding norm axioms applied to \(x-y\). Symmetry follows from homogeneity with \(\alpha =-1\): \[d(y,x)=||y-x||=||-(x-y)||=|-1|\,||x-y||=d(x,y).\] For the triangle inequality, write \(x-z=(x-y)+(y-z)\) and apply the norm triangle inequality: \[d(x,z)=||x-z||\leq ||x-y||+||y-z||=d(x,y)+d(y,z).\]

Not every metric comes from a norm

Take the discrete metric on any non-trivial vector space. A norm-induced metric is homogeneous, \(d(\alpha x,\alpha y)=|\alpha |\,d(x,y)\), but the discrete metric gives \(d(2x,0)=1=d(x,0)\) for \(x\neq 0\), so no norm can induce it.

Translation invariance, \(d(x+z,y+z)=d(x,y)\), is the other property a norm-induced metric always has; a metric failing either cannot come from a norm.

Solution.

Bounded \(\implies \) continuous

Suppose \(||L(x)||\leq M||x||\) for all \(x\). Then for any \(x,x'\), by linearity \[||L(x)-L(x')||=||L(x-x')||\leq M||x-x'|| ,\] so \(L\) is Lipschitz and hence continuous — indeed uniformly continuous, with \(\delta =\varepsilon /M\).

Continuous \(\implies \) bounded

Suppose \(L\) is continuous, in particular at \(0\). Taking \(\varepsilon =1\), there is \(\delta >0\) with \(||L(z)||\leq 1\) whenever \(||z||\leq \delta \). For any \(x\neq 0\) put \(z=\dfrac {\delta x}{||x||}\), so \(||z||=\delta \) and \[\left \|L\!\left (\frac {\delta x}{||x||}\right )\right \|\leq 1 \implies \frac {\delta }{||x||}\,||L(x)||\leq 1 \implies ||L(x)||\leq \frac {1}{\delta }||x|| .\] So \(L\) is bounded with \(M=\frac 1\delta \).

The equivalence is a genuine economy: continuity is a condition at every point, boundedness a single inequality, and for linear maps they say the same thing. It fails badly without linearity.

Solution.

The inequality

For any \(x\), applying the defining bound twice, \[||(LT)(x)||=||L\big (T(x)\big )||\leq ||L||\,||T(x)||\leq ||L||\,||T||\,||x|| .\] So \(||L||\,||T||\) is one of the constants \(M\) with \(||(LT)(x)||\leq M||x||\), and since the operator norm is the least such constant, \(||LT||\leq ||L||\,||T||\).

Strictness

On \(\mathbb {R}^2\) take the projections \[L(x_1,x_2)=(x_1,0),\qquad T(x_1,x_2)=(0,x_2).\] Each has operator norm \(1\). But \(LT=0\), since \(T\) lands in the second axis and \(L\) annihilates it, so \(||LT||=0<1=||L||\,||T||\).

The inequality can therefore be very far from equality; what it guarantees is only that composing bounded operators keeps them bounded, which is what makes \(\mathcal {L}(X)\) an algebra.

Solution.

Construction

Let \(M=\{\alpha x:\alpha \in \mathbb {F}\}\), the one-dimensional subspace spanned by \(x\), and define \(v:M\rightarrow \mathbb {F}\) by \[v(\alpha x)=\alpha ||x|| .\] This is well defined because \(x\neq 0\) makes the representation unique, and it is linear. Its norm is \(1\): \[|v(\alpha x)|=|\alpha |\,||x||=||\alpha x|| ,\] so the ratio \(|v(y)|/||y||\) equals \(1\) for every non-zero \(y\in M\).

By the Hahn–Banach theorem \(v\) extends to \(u\in X^*\) with \(||u||=||v||=1\), and \(u(x)=v(x)=||x||\).

Separation

If \(x\neq y\) then \(x-y\neq 0\), so the above gives \(u\in X^*\) with \[u(x)-u(y)=u(x-y)=||x-y||\neq 0 ,\] hence \(u(x)\neq u(y)\). So the bounded functionals distinguish any two distinct points.

This is the practical content of Hahn–Banach: without it there would be no guarantee that \(X^*\) contains anything beyond the zero functional, and duality arguments would be vacuous.

Solution.

Complete \(\implies \) absolutely convergent series converge

Suppose \(\sum _n ||x_n||<\infty \) and let \(S_N=\sum _{n=1}^{N}x_n\). For \(M<N\), \[||S_N-S_M||=\left \|\sum _{n=M+1}^{N}x_n\right \|\leq \sum _{n=M+1}^{N}||x_n|| ,\] which is a tail of a convergent series of reals and so is small for \(M\) large. Hence \((S_N)\) is Cauchy and converges by completeness.

Converse

Suppose every absolutely convergent series converges, and let \((y_n)\) be Cauchy. Choose a subsequence with \[||y_{n_{k+1}}-y_{n_k}||<2^{-k} ,\] possible by the Cauchy condition. The series \[y_{n_1}+\sum _{k=1}^{\infty }\big (y_{n_{k+1}}-y_{n_k}\big )\] is absolutely convergent, since \(\sum 2^{-k}<\infty \), so it converges; but its partial sums telescope to \(y_{n_{k+1}}\), so the subsequence converges. A Cauchy sequence with a convergent subsequence converges, so \((y_n)\) does.

The criterion is useful because absolute convergence is checked with real series, for which the standard tests are available.