Let $(X,{‖⋅‖})$ be a normed space.

• Let $S, T∈ℬ(X)$ be invertible. Prove that also $S T$ is invertible.

• Let $S∈ℬ(X)$ be algebraically invertible, i.e. so that there exists a map $T: X→X$ so that $T S=S T=\text{Id}$ which we call the algebraic inverse and denote by $S^{-1}$. Prove that $S^{-1}∈ℬ(X)$ if and only if

  (⋆)	\[∃ δ>0\text{ so that } ∀ x∈X \text{ we have }{‖S(x)‖} ≥ δ{‖x‖} .\]

  Conversely show that (⋆) is violated if and only if there exist $x_n∈X$ with $\left‖x_n\right‖=1$ so that $S x_n→0$.

• $S^{-1},T^{-1}∈ℬ(X)$, so the composition $T^{-1}S^{-1}∈ℬ(X)$.
  $(T^{-1}S^{-1})(ST)=(ST)(T^{-1}S^{-1})=\text{Id}$, so $ST$ is invertible.

• If $S^{-1}∈ℬ(X)$, we have ${‖x‖}={‖S^{-1}(S(x))‖}≤{‖S^{-1}‖}{‖S(x)‖}$, let $δ^{-1}={‖S^{-1}‖}$ in (⋆).
  Conversely (⋆) holds, ${‖S^{-1}(x)‖}≤δ^{-1}{‖S(S^{-1}(x))‖}=δ^{-1}{‖x‖}$, so $S^{-1}∈ℬ(X)$.

  If (⋆) is violated, for $δ=n^{-1},∃x∈X:{‖S(x)‖}<δ{‖x‖}$, let $x_n={x\over‖x‖}$ so $\left‖x_n\right‖=1$.
  They satisfy $\left‖S x_n\right‖<δ$, so $S x_n→0$.
  Conversely if $∃x_n∈X,\left‖x_n\right‖=1,S x_n→0$, then $δ=δ{‖x_n‖}≤{‖S(x_n)‖}→0$, contradicts $δ>0$ in (⋆).

• $f∈X^*=ℬ(X, 𝔽)$.
• ${‖f‖}=2$.
• There exists no $ϕ∈X$ so that ${‖ϕ‖}_\sup=1$ and $f(ϕ)=2$.

• $∫_0^1(ϕ+αψ)-∫_{-1}^0(ϕ+αψ)=∫_0^1ϕ-∫_{-1}^0ϕ+α(∫_0^1ψ-∫_{-1}^0ψ)$, so $f$ is linear. ${|f(ϕ)|}≤∫_{-1}^1{|ϕ|}≤2{‖ϕ‖}_\sup$, so $f∈ℬ(X, 𝔽)$.
• We proved ${‖f‖}≤2$ in (a). To prove ${‖f‖}≥2$, let $ϕ_n∈X$ such that ${‖ϕ_n‖}_\sup=1$ \begin{alignat*}2 ϕ_n&=1&&\text{ on }\left[\frac1n,1\right]\\ ϕ_n&∈[0,1]&&\text{ on }\left[0,\frac1n\right]\\ ϕ_n&∈[-1,0]&&\text{ on }\left[-\frac1n,0\right]\\ ϕ_n&=-1&&\text{ on }\left[-1,-\frac1n\right] \end{alignat*} so ${|f(ϕ_n)|}≥2-\frac2n→2$ as $n→∞$. So ${‖f‖}=2$.
• $∫_0^1ϕ=∫_{-1}^0{-ϕ}={‖ϕ‖}_\sup$, then $\begin{aligned}[t]ϕ=1&\text{ on }[0,1]\\ϕ=-1&\text{ on }[-1,0]\end{aligned}$, so $\begin{aligned}[t]&ϕ(0)=1\\&ϕ(0)=-1\end{aligned}$ contradiction.

• Check that $P$ is a linear map from $X$ to $X$ such that $P^2=P$.
• Is $X$ (equipped with the $ℓ^2$-norm) a Banach space?
• Is $P$ bounded?

• For all $x,y∈X$, λ∈ℝ, P(x+λy)$=\left(∑j (x_j+λy_j),0,…\right)=$Px+λPy, so $P$ is linear.
  For all $x∈X$, ${‖Px‖}_2={‖( jx_j)‖}_1<∞$, so $Px∈X$.
  $P\left(∑ j x_j, 0,0, …\right)=\left(1∑ j x_j, 0,0, …\right)$, so $P^2x=Px$, so $P^2=P$.
• $\sum_{j>n}\frac1{j^2}→0$, so $(1,\frac12,…,\frac1n,0,…)→(1,\frac12,…)$ in $ℓ^2$.
  $P(1,\frac12,…,\frac1n,0,…)=(n,0,…)$, so $(1,\frac12,…,\frac1n,0,…)∈X$ but the limit $(1,\frac12,…)∉X$, so $X$ is not closed, so not a Banach space.

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  More general method: let $e_n=(δ_{jn})_j∈ X$ $$\left.\begin{array}r\operatorname{span}(e_n)⊂ X⊂ℓ^2\\\operatorname{span}(e_n)=ℓ^2\end{array}\right\}⇒\bar X=ℓ^2$$ $X=(1,\frac12,\frac13,…)∈ℓ^2∖ X$ so $X≠\bar X$, $X$ is not Banach.
• For $e^{(k)}=\left(δ_{k j}\right)_{j∈ℕ}$, ${‖e^{(k)}‖}_2=1$, ${‖Pe^{(k)}‖}_2=k$, so $P$ is unbounded.

• Prove that this defines a bounded linear operator $T∈ℬ(X)$ and ${‖T‖}=b-a$.
• Find a function $k:\{(s, t) ∣ a ≤ s ≤ t ≤ b\}→ℝ$ so that $$ \left(T^2 x\right)(t)=∫_a^t k(s, t) x(s) \mathrm{~d} s $$ and determine $\left‖T^2\right‖$.
• Determine $k_n:\{(s, t) ∣ a ≤ s ≤ t ≤ b\}→ℝ$ so that $$ \left(T^n x\right)(t)=∫_a^t k_n(s, t) x(s) \mathrm{~d} s $$

• $∀x∈X$, by Fundamental Theorem of Calculus, $Tx$ is continuous, so $Tx∈X$.
  $∀x,y∈X,k∈ℝ:$T(x+ky)(t)$=∫_a^t x(s)+ky(s) \mathrm{~d} s=$(Tx+kTy)(t), so $T$ is linear.
  ${|(Tx)(t)|}≤∫_a^t{|x(s)|}\mathrm{~d}s≤(b-a){‖x‖}_∞$, so ${‖Tx‖}_∞≤(b-a){‖x‖}_∞$, ${‖T‖}≤b-a$.
  $(T1)(t)=∫_a^t1=t-a$, so ${‖T1‖}_∞=b-a$, so ${‖T‖}=b-a$.
• Since $$ ∫_{[a,t]×[a,u]} {|x(s)|}\mathrm{~d}(s×u) ⩽ {‖x‖}_∞(b-a)^2<∞, $$ by Fubini's theorem, for $k(s,t)=t-s$ we have \begin{align*} T^2x(t) &= ∫_a^t∫_a^u x(s)\mathrm{~d}s\mathrm{~d}u\\ &= ∫_a^t x(s) ∫_s^t \mathrm{~d}u \mathrm{~d}s\\ &= ∫_a^t k(s,t)x(s)\mathrm{~d}s. \end{align*} For any $x∈X$, \begin{align*} {‖T^2x‖}_∞&≤{‖x‖}_∞∫_a^t{|k(s,t)|}\mathrm{~d}s\\ &={‖x‖}_∞∫_a^t(t-s)\mathrm{~d}s\\ &={‖x‖}_∞\frac{(t-a)^2}2 \end{align*} equality when $x$ is constant. So ${‖T^2‖}=\frac{(b-a)^2}2$
• Assume $T^nx(t) = ∫_a^t \frac{(t-s)^{n-1}}{(n-1)!} x(s)\mathrm{~d}s$ for some $n⩾1$. By Fubini's theorem \begin{align*} T^{n+1}x(t) &= ∫_a^t ∫_a^u \frac{(u-s)^{n-1}}{(n-1)!} x(s)\mathrm{~d}s\mathrm{~d}u\\ &= ∫_a^t x(s)∫_s^t \frac{(u-s)^{n-1}}{(n-1)!}\mathrm{~d}u \mathrm{~d}s\\ &= ∫_a^t \frac{(t-s)^n}{n!}x(s)\mathrm{~d}s. \end{align*} It follows by induction that $$k_n(s,t) = \frac{(t-s)^{n-1}}{(n-1)!}.$$ For any $x∈X$, \begin{align*} {‖T^nx‖}_∞&≤{‖x‖}_∞∫_a^t{|k_n(s,t)|}\mathrm{~d}s\\ &={‖x‖}_∞∫_a^t\frac{(t-s)^{n-1}}{(n-1)!}\mathrm{~d}s\\ &={‖x‖}_∞\frac{(t-a)^n}{n!} \end{align*} equality when $x$ is constant. So ${‖T^n‖}=\frac{(b-a)^n}{n!}$.

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  $T^n→0$ agree with Banach fixed point theorem (the function 0 is a fixed point).

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

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 $α=(α_j)$ be a fixed bounded sequence. Define $M_α$ by $$ M_α:(x_1, x_2, x_3, …) ↦(α_1 x_1, α_2 x_2, α_3 x_3, …) . $$ Prove that $M_α∈ℬ(ℓ^∞)$. For which values of $α$ is $M_α$ injective? Prove that $M_α$ has a bounded inverse if and only if $\inf _j\left|α_j\right|>0$.
• Let $X$ be the space of real polynomials on $[0,1]$ regarded as a subspace of the Banach space $C[0,1]$ of continuous functions equipped with the sup norm. For $k=0,1,2, …$, let $m_k$ be the $k$-th monomial, i.e. $$ m_k(t)=t^k  (t ∈[0,1]) . $$ Define $T: X→X$ by letting $T m_k=\frac1{k+1} m_k$ and extending $T$ to $X$ by linearity.
  • Give an integral expression for $T x$ for a general $x∈X$ and use this expression to prove that $T$ can be extended to a bounded linear operator $\tilde T∈ℬ(C[0,1])$ with ${‖\tilde T‖}=1$.
  • Prove that $T: X→X$ is a bijection but that its inverse is unbounded.
  • Is $\tilde T: C[0,1]→C[0,1]$ still injective? Is it still surjective?

• ${‖M_αx‖}_\sup≤\sup_j{|α_j|}{‖x‖}_\sup⇒M_α$ is bounded $$M_α\text{ is injective}⇔\ker M_α=\{0\}⇔α_j≠0∀j≥0$$ $β=1/α,M_α∘M_β=M_β∘M_α=\text{id},\sup{|β_j|}=\frac1{\inf{|α|}}<∞$
• • $$Tx(t)=\cases{t^{-1}\int_0^tx(s)\,\text ds&for $t>0$\\x(0)&for $t=0$}$$ ${|t^{-1}\int_0^tx(s)\,\text ds|}≤{‖x‖}_\sup⇒{‖T‖}≤1$, also $T(m_0)=m_0$, so ${‖T‖}=1$.
    $\bar X=C[0,1]$, so $T$ extends to $\tilde T:C[0,1]\to C[0,1]$ with $‖\tilde T‖=‖T‖=1$.
    For every continuous function $f$ we have $\lim_{t → 0} t^{-1} ∫_0^t f(s) \mathrm{d} s=f(0)$ by mean value theorem which shows $\tilde{T}f$ is continuous at $0$ from the right.
  • Define $T^{-1}$ by $T^{-1}(m_k)=(k+1)m_k$ and extend it to $X$ by linearity. so $T$ has an inverse, so $T$ is a bijection. $$\|T^{-1}(m_k)\|/\|m_k\|=k+1\to\infty$$so $T^{-1}$ is not bounded.
  • still injective. not surjective. MSE
    $\tilde T$ is injective, because we can recover $f$ from $\tilde Tf$:$$\frac{d}{dt}\bigl(t\cdot\tilde Tf(t)\bigr)=f(t)$$ $\tilde T$ is not surjective: notice that $t⋅\tilde Tf(t)$ is differentiable on $[0,1]$, so $\tilde Tf$ is differentiable on $(0,1]$, but not all functions in $C[0,1]$ are differentiable on $(0,1]$.

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    A more educated (though admittedly extremely overkill) argument: Since $C[0,1]$ is a Banach space and $\tilde T$ doesn't have a bounded inverse, by open mapping theorem, $\tilde T$ is not surjective.

• Show that $L∈ℬ(X), L(X)=𝒫_n$ and $L f=f$ if $f∈𝒫_n$.
• Show that $$ {‖L‖}={‖Λ‖} . $$ Hint: for each $x$, choose some $f$ with ${‖f‖}=1$ such that $L f(x)=Λ(x)$.
• Show that $Λ(x) ≥ 1$ in $[0,1]$, and equality holds only at nodes.
• Show that for each $1 ≤ i ≤ n, Λ$ is a polynomial in $\left[x_{i-1}, x_i\right]$ and has a unique local maximum in $\left(x_{i-1}, x_i\right)$. Hint: Consider the polynomial extension of $\left.Λ\right|_{\left[x_{i-1}, x_i\right]}$, its values at the nodes and the location of its critical points.
• Suppose now that the nodes are equidistant: $x_i=\frac in$. Show that $$ y_i(x)=(-1)^{n+i}\binom{nx}i\binom{n x-i-1}{n-i} $$ where $\binom ab=\frac{a(a-1) ⋯(a-b+1)}{b(b-1) ⋯ 1}$ for real $a$ and positive integer $b$. Deduce that $$ {‖Λ‖}→∞ \text { as } n→∞ $$ Hint: Consider for example the value of $y_i\left(\frac1{2 n}\right)$. You may need to use Sterling's formula $$ n !=\sqrt{2 π n}(n / e)^n(1+O(1 / n)) \text { as } n→∞ . $$

• Let $α ∈(1 / 2,1)$ and $g(x)=x^α$. Show that $\left(g, g'\right)$ does not belong to $Γ(T)$ but there exists a sequence $\left(\left(f_n, T f_n\right)\right)_{n ≥ 1} ⊂ Γ(T)$ which converges to $\left(g, g'\right)$ in $X × X$. Deduce that $Γ(T)$ is not closed in $X × X$.
• Show that if $\left(f_n\right) ⊂ Y$ is such that $f_n → 0$ and $T f_n → h$ in $X$, then $h=0$.
• Show that the set $\overline{Γ(T)}$ has the following property: For every $f ∈ X$, the set $\{(f, h) ∈ \overline{Γ(T)}\}$ contains at most one element.
• Let $π: X × X → X$ be the projection onto the first factor and let $W=π(\overline{Γ(T)})$. Show that $Y ⊊ W$ and use (iii) to deduce that there exists a linear map $\tilde{T}: W → X$ which extends $T$.
  One can in fact show that $\tilde{T}(W)=X$, that is, $W$ is the space of functions in $X$ whose 'derivatives' are also in $X$. The space $W$ is the Sobolev space $W^{1,2}((0,1))$