12 Normed spaces[0ZT]

Let in the following $X$ be a vector space based on the real field ${\mathbb {R}}$.

Definition 323 Norm

[0ZV] A norm is an operation that maps a vector $v∈ X$ in a real number $\| v\| $, which satisfies

$\| v\| ≥ 0$ and $\| v\| =0$ if and only if $v=0$;
for every $v∈ X$ and $t∈ {\mathbb {R}}$ we have $|t|\, \| v\| =\| tv\| $ (we will say that the norm is absolutely homogeneous);
(Triangle inequality) for every $v,w∈ X$ we have

\[ \| v+w\| ≤ \| v\| +\| w\| \quad ; \]

this says that one side of a triangle is less than the sum of the other two.

Remark 324

[0ZW]Many of the results in subsequent exercises generalize to the case of ”asymmetic norms”, where the second request will be replaced by this: for every real $t≥ 0$ you have $t\| v\| =\| tv\| $. (In this case we will say that the norm is positively homogeneous).

E324

[0ZX] Let $X$ be a vector space and $f:V→ {\mathbb {R}}$ a function that is positively homogeneous, that is: for every $v∈ X$ and $t≥ 0$ you have $t f(v)=f (tv)$.

Show that $f$ is convex if and only if the triangle inequality holds: for every $v,w∈ X$ you have

\[ f(v+w) ≤ f(v)+f(w)\quad . \]

In particular, a norm is always a convex function.

E324

[0ZY] Note that if $v,w∈ X$ are linearly dependent and have the same direction (i.e. you can write $v=𝜆 w$ or $w=𝜆 v$, for $𝜆≥ 0$), then you have

\[ \| v+w\| = \| v\| +\| w\| \quad . \]

In particular, a norm is not a strictly convex function, because

\[ \| \frac{v} 2+\frac{v} 2\| = \frac 1 2 \| v\| +\frac 1 2\| v\| \quad . \]

E324

[0ZZ] Prerequisites:4, 372, 1.Difficulty:*.We will say that the normed space $(X,\| ⋅\| )$ is strictly convex ¹ if the following equivalent properties apply.

The disc $D=\{ x∈ X:\| x\| ≤ 1\} $ is strictly convex. ²
The sphere $\{ x∈ X,\| x\| =1\} $ does not contain non-trivial segments (that is, segments of positive length).
For $v,w∈ D$ with $\| v\| =\| w\| =1$ and $v≠ w$, for every $t$ such that $0{\lt}t{\lt}1$, we have that $\| t v+(1-t)w\| {\lt}1$.
For every $v,w∈ X$ that are linearly independent we have $ \| v+w\| {\lt} \| v\| +\| w\| \quad .$

Show that the previous four clauses are equivalent.

[UNACCESSIBLE UUID ’101’]

Hidden solution: [UNACCESSIBLE UUID ’102’] $\includegraphics[width=\textwidth ]{UUID/1/0/4/blob_und.png}$ [105]Let $X$ be a normed vector space with norm $\| ⋅\| $. Show that the sum operation $'+':X× X→ X$ is continuous. [106] Prerequisites:286.

Let again $X$ be a normed vector space with norm $\| ⋅\| $. Let $B(x,r){\stackrel{.}{=}}\{ y∈ X : \| x-y\| {\lt} r\} $ be the ball. Let $D(x,r){\stackrel{.}{=}}\{ y∈ X : \| x-y\| ≤ r\} $ be the disk. Let $S(x,r){\stackrel{.}{=}}\{ y∈ X : \| x-y\| = r\} $ be the sphere. Show that $\overline{B(x,r)}= D(x,r)$, that $B(x,r)= {{D(x,r)}^\circ }$, and that $∂{B(x,r)}=∂{D(x,r)}= S(x,r)$. Also show that $B(x,r)$ is not closed and $D(x,r)$ is not open. [107] Prerequisites:15.Let $X$ be a vector space, let $𝜙,𝜓$ be two norms on it. Show that the topologies generated by $𝜙$ and $𝜓$ coincide, if and only if there exist $0{\lt}a{\lt}b$ such that

\begin{equation} ∀ x , ~ ~ a𝜓(x)≤𝜙(x)≤ b 𝜓(x)~ ~ .\label{eq:norme_ equiv} \end{equation}

325

(When the relation ?? holds, we will say that the norms are ”equivalent”).

Hidden solution: [UNACCESSIBLE UUID ’108’] [109] We want to show that ”the norms in $ℝ^ n$ are all equivalent.”

Let $\| x\| =\sqrt{∑_{i=1}^ n x_ i^ 2}$ be the Euclidean norm. Let $𝜙:ℝ^ n→[0,∞)$ be a norm: it can be shown that $𝜙$ is a convex function, see 1; and therefore $𝜙$ is a continuous function, see 4. Use this fact to prove that there exist $0{\lt}a{\lt}b$ such that

\begin{equation} ∀ x , ~ ~ a\| x\| ≤𝜙(x)≤ b\| x\| ~ ~ .\label{eq:Rn_ norma_ equiv} \end{equation}

326

Hidden solution: [UNACCESSIBLE UUID ’10B’]