1. Let $(X,{‖⋅‖})$ be a normed space.

  2. Let $X=C([-1,1])$ equipped with the sup-norm and define $f: X→ℝ$ by $$ f(ϕ)=∫_0^1 ϕ(t) d t-∫_{-1}^0 ϕ(t) d t $$ Prove that
  3. Let $X$ be the subspace of $ℓ^2$ that is defined by $$ X:=\{x∈ℓ^2:(j x_j)∈ℓ^1\}⊂ℓ^2 $$ and define $P(x_1, x_2, …)=\left(\sum_{j=1}^∞ j x_j, 0,0, …\right)$.
  4. Let $X=C[a, b]$ equipped with the sup norm. For $x$ in $X$ define $T x$ by $$ (Tx)(t)=∫_a^t x(s) \mathrm{~d} s  (t ∈[a, b]) $$
  5. Let $t_{j k}∈ℝ\;(j, k∈ℕ)$ and denote by $e^{(k)}, k∈ℕ$ the sequences $e^{(k)}=\left(δ_{k j}\right)_{j∈ℕ}$. Show that if $\sup_{k, j}\left|t_{j k}\right|<∞$ then there exists a bounded linear operator $T: ℓ^1→ℓ^∞$ so that

    \[\leqalignno{\left(T e^{(k)}\right)_j&=t_{j k} .&(\style{font-family:"Latin Modern Math"}{\hbox{⋆⋆}})}\]

    Conversely, show that if there is a bounded linear operator $T: ℓ^1→ℓ^∞$ so that (⋆⋆) holds, then we must have that $\sup_{k, j}\left|t_{j k}\right|<∞$ and indeed $\sup_{k, j}\left|t_{j k}\right|={‖T‖}$.

    Let $$X=ℓ^1,Z=ℓ^∞,Y=\operatorname{span}(e_n)⊂X,\bar Y=X$$ By Theorem 4.1.3 we can extend $T:Y→ℓ^∞$ to $\tilde T:X→ℓ^∞$ such that ${‖T‖}={‖\tilde T‖}$ $$y_1e_1+y_2e_2+⋯↦\left(\sum_{j=1}^∞y_jt_{j1},\sum_{j=1}^∞y_jt_{j2},…\right)$$ by triangle inequality \[{|(T x)_j|}≤{‖x‖}_1⋅\sup_{k, j}\left|t_{j k}\right|\] so $T x∈ℓ^∞$ and \[{‖T x‖}_∞≤{‖x‖}_1⋅\sup_{k, j}\left|t_{j k}\right|\] so $T$ is bounded. Conversely, if there is a bounded linear operator $T: ℓ^1→ℓ^∞$ so that (⋆⋆) holds, \[\sup_j{|t_{j k}|}={‖T e^{(k)}‖}_∞≤{‖T‖}⋅{‖e^{(k)}‖}_1={‖T‖}\] taking supremum over $k$,$$\sup_{k, j}\left|t_{j k}\right|≤{‖T‖}$$ On the other hand, by triangle inequality \[{\color{red}{|(T x)_j|}=\left|\sum_{k∈ℕ}x_k(Te^{(k)})_j\right|}=\left|\sum_{k∈ℕ}x_kt_{jk}\right|≤{‖x‖}_1⋅\sup_{k, j}\left|t_{j k}\right|\] taking supremum over $j$, \[{‖T x‖}_∞≤{‖x‖}_1⋅\sup_{k, j}\left|t_{j k}\right|\] Therefore $\sup_{k, j}\left|t_{j k}\right|={‖T‖}$.
    red equality is not obvious: linearity can only be applied to finite sums.
    we need approximation argument: take first $n$ terms
    Theorem 4.1.3. Let $\left(X,{‖⋅‖}_X\right)$ be a normed space, let $Y$ be a dense subspace of $X$ (which we equip with the norm of $X)$ and let $\left(Z,\|⋅\|_Z\right)$ be a Banach space. Then any $T ∈ ℬ(Y, Z)$ has a unique extension $\tilde{T} ∈ ℬ(X, Z)$, i.e. there exists a unique bounded linear operator $\tilde{T}: X → Z$ so that $\tilde{T} y=T y$ for every $y ∈ Y$ and we furthermore have that $$ \|\tilde{T}\|_{ℬ(X, Z)}=\|T\|_{ℬ(Y, Z)} $$
    Finite dimensional analog: $$\pmatrix{t_{11}&t_{12}&t_{13}\\ t_{21}&t_{22}&t_{23}\\ t_{31}&t_{32}&t_{33}}\pmatrix{x_1\\x_2\\x_3}=\pmatrix{\sum_{k=1}^3x_kt_{1k}\\\sum_{k=1}^3x_kt_{2k}\\\sum_{k=1}^3x_kt_{3k}}$$

  6. Let $X$ be the space $C[0,1]$ (over the reals) equipped with the supremum norm. Fix an integer $n ≥ 2$ and $(n+1)$ points $0 ≤ x_0<x_1 …<x_n ≤ 1$ which will be called nodes. The associated $i$ th Lagrange polynomial is defined as $$ y_i(x)=\prod_{\substack{j=0 \\ j ≠ i}}^n \frac{x-x_j}{x_i-x_j}   \text { for }   0 ≤ i ≤ n, x ∈[0,1] . $$ It is easy to see that each $y_i$ is a polynomial with degree $n, y_i\left(x_i\right)=1, y_i\left(x_j\right)=0$ for $j ≠ i$, and $\sum_{i=0}^n y_i(x)=1$, so that $\left\{y_i\right\}_{i=0}^n$ is a basis for the subspace $𝒫_n⊂X$ of polynomials of degree at most $n$. For a function $f∈X$, define its Lagrange interpolant $L f$ by $$ L f(x)=\sum_{i=0}^n f\left(x_i\right) y_i(x) $$ The function $Λ(x)=\sum_{i=0}^n\left|y_i(x)\right|$ is called the Lebesgue function of $L$.
  7. Equip $X=L^2((0,1))$ and $X × X$ with their standard inner products so that both are Hilbert spaces. Let $Y=C^1([0,1])$ and consider $Y$ as a subspace of $X$. Define an operator $T: Y → X$ by $$ T f=f' . $$ Let $Γ(T)=\{(f, T f): f ∈ Y\} ⊂ X × X$ be the graph of $T$, which is a subspace of $X × X$.