1. Let $X$ be a real or complex vector space and assume that $x↦{‖x‖}_0$ is a semi-norm, i.e. a function from $X$ to $[0, ∞)$ which satisfies

    (N2) ${‖λx‖}_0={|λ|}{‖x‖}_0$ for all $λ∈𝔽$ and all $x∈X$;

    (N3) ${‖x+y‖}_0≤{‖x‖}_0+{‖y‖}_0$ for all $x,y∈X$.

    Let $X_0=\left\{x∈X\middle|{‖x‖}_0=0\right\}$. Show that $X_0$ is a subspace of $X$.
    Let $X/X_0$ be the associated quotient space (as defined in Part A Linear Algebra). “Define” $\left‖x+X_0\right‖={‖x‖}_0$ for $x∈X$.

    By (N2) $X_0$ is closed under scaling, by (N3) closed under addition, so $X_0$ is a subspace of $X$.

  2. We let $$ Z:=\{f:[-1,1]→ℝ \text{ Lipschitz continuous}\} $$ and define for $f∈Z$ $$ \operatorname{Lip}(f):=\inf\{L∈ℝ:{|f(s)-f(t)|}≤L{|s-t|} \text{ for all } s, t∈[-1,1]\} . $$ We furthermore consider the subspace $X:=\{f∈Z:f(0)=0\}$ of $Z$.
    The infimum is not always achieved for $s≠t$, e.g. $\operatorname{Lip}(x^2)=1$ achieved at $s=t=±1$, but we still have ${|f(s)-f(t)|}≤\operatorname{Lip}(f){|s-t|}$ because taking limit preserves ≤ inequality
  3. Let $(X,{‖⋅‖})$ be a real norm vector space satisfying the parallelogram law: $$ {‖x+y‖}^2+{‖x-y‖}^2=2{‖x‖}^2+2{‖y‖}^2 \text { for all } x, y ∈ X . $$ Define $$ f(x, y)=\frac14\left({‖x+y‖}^2-{‖x-y‖}^2\right) \text { for } x, y ∈ X . $$ Show that Conclude that $f(x, y)$ defines an inner product on $X$.
    [Hint: The tricky part is (c). Prove it, successively, for $α$ being an integer, a rational number, and finally a real number.]

  4. To prove $Z=Y^⟂$, we need to check two conditions:
    1. $∀y∈Y,z∈Z:⟨y,z⟩=0$
    2. $∀x∈ X:∃ y∈ Y,z∈ Z:x=y+z$
    Note that (i) equivalent to $Z⊂Y^⟂$, (ii) implies $Y^⟂⊂Z$.
  5. Let $X=c_0$ be the space of sequences $\left(α_j\right)$ that converge to zero, equipped with the sup norm. Consider the subsets $$ Y=\left\{\left(α_j\right)\middle|α_{2 j-1}=0, j=1,2, …\right\} \text{ and } Z=\left\{\left(α_j\right)\middle|α_{2 j}=j^2 α_{2 j-1}, j=1,2, …\right\} $$ Show that $Y$ and $Z$ are closed subspaces of $X$ and that the element $x=\left(1,0, \frac14, 0, \frac19, 0, …\right)$ lies in the closure of $Y+Z$ but not in $Y+Z$.
    To prove $x∉Y+Z$, suppose $x=y+z,y∈Y,z∈Z$, then $z_{2j-1}=x_{2j-1}-y_{2j-1}=\frac1{j^2},z_{2j}=j^2z_{2j-1}=1$ contradicting $z_j→0$.
    To prove $x∈\overline{Y+Z}$, let $x_n=\sum_{j=1}^n\frac1{j^2}e_{2j-1}$ then $x_n→x$ since \[{‖x-x_n‖}=\left‖\sum_{j=n+1}^∞\frac1{j^2}e_{2j-1}\right‖=\frac1{(n+1)^2}→0\] For all $j$, $x_{2j}=(\frac1{j^2}e_{2j-1}+e_{2j})-e_{2j}$, so $x_n∈Y+Z$.
  6. (Question 1b from 2020 exam.) Let $I=[0,2]$ and consider $$ X=\left\{f∈L^1(I):∫_I \frac1x{|f(x)|} d x<∞\right\} \text{ equipped with }{‖f‖}_X:=∫_I \frac1x{|f(x)|} d x $$ as well as $Y=C(I) ∩ X$. You may use that $\left(X,{‖⋅‖}_X\right)$ is a normed space, and that $\left(L^1(I),{‖⋅‖}_{L^1}\right)$ and $\left(C(I),{‖⋅‖}_\sup\right)$ are Banach spaces.
  7. Let $X$ be a Banach space, $\overline{𝔹}$ be its closed unit ball and $𝕊$ its unit sphere. A retraction from $\overline{𝔹}$ to $𝕊$ is a continuous map $r:\overline{𝔹}→𝕊$ such that $\left.r\right|_𝕊$ is the identity map on $𝕊$. When $X$ has finite dimension, no such retraction exists. This statement is wrong in any infinite dimensional space $X$. A well-known example given by Kakutani in 1943 for $X=ℓ^2$ is a retraction of the form $r(x)=x+λ(x) \frac{T x-x}{‖T x-x‖}$ where $T(x)=\left(ε(1-{‖x‖}), x_1, x_2, …\right)$ and $λ$ is such that ${‖r(x)‖}=1$. Here we give a different example.

    Let $X$ be the space of continuous functions on $[0,1]$, equipped with its supremum norm.