- Define the term orthonormal sequence. State the Pythagorean theorem for a finite orthonormal sequence.
- Let $(x_n)_{n=1}^{∞}$ be an orthonormal sequence in $H$. Prove for every $x ∈ H$ that the sum $\sum_{n=1}^{∞}⟨x, x_n⟩ x_n$ converges in the sense of norm to some $P(x) ∈ H$ satisfying $⟨x-P(x), x_n⟩=0$ for all $n$. Show further that $P: H → H$ is bounded linear.
- Let $(x_n)_{n=1}^{∞}$ and $P$ be as in (a)(ii). Let $I$ be the identity operator on $H$. Show that $P^2=P$. Deduce that $\operatorname{Im} P$ and $\operatorname{Im}(I-P)$ are closed and orthogonal complementary spaces for one another.

- An orthonormal sequence is a sequence $(x_n)_{n=1}^{∞}$ in $H$ such that $⟨x_n, x_m⟩=δ_{n m}$ for all $n, m$. The Pythagorean theorem states that if $(x_n)_{n=1}^{N}$ is a finite orthonormal sequence, then for every $x ∈ H$, we have $\|x\|^2=\sum_{n=1}^{N}|⟨x, x_n⟩|^2+\|x-P x\|^2$, where $P(x)=\sum_{n=1}^{N}⟨x, x_n⟩ x_n$.
- Bessel's inequality $\sum_{n=1}^∞|⟨x, x_n⟩|^2≤\|x\|^2$ and $H$ is complete (absolute convergence implies convergence), so $\sum_{n=1}^{N}⟨x, x_n⟩ x_n$ converges to some $P(x) ∈ H$. We have $⟨x-P(x), x_n⟩=⟨x, x_n⟩-⟨P(x), x_n⟩=0$ for all $n$. The linearity of $P$ follows from the linearity of the inner product. The boundedness of $P$ follows from the Pythagorean theorem: $\|P(x)\|^2=\sum_{n=1}^{∞}|⟨x, x_n⟩|^2≤\|x\|^2$.
- By construction, $⟨P x, x_n⟩=⟨x, x_n⟩$. Hence $P(P x)=P x$, i.e. $P^2=P$.

From the above, we have $(I-P) P=0$, which implies that $\operatorname{Im} P ⊂ \operatorname{Ker}(I-P)$. Also, if $x ∈ \operatorname{Ker}(I-P)$, then $x=P x$, and so $x ∈ \operatorname{Im} P$. Thus $\operatorname{Im} P=\operatorname{Ker}(I-P)$, which is closed. Likewise, using $P(I-P)=0$, $\operatorname{Im}(I-P)=\operatorname{Ker} P$ is closed.

It is clear that $H=\operatorname{Im} P+\operatorname{Im}(I-P)$. Suppose that $x ∈ \operatorname{Im} P ∩ \operatorname{Im}(I-P)$. From the above, $x ∈ \operatorname{Ker}(I-P) ∩ \operatorname{Ker} P$ and so $x-P x=0$ and $P x=0$, which implies that $x=0$. Hence $H=\operatorname{Im} P ⊕ \operatorname{Im}(I-P)$. Lastly, for every $x ∈ H,⟨x-P x, x_n⟩=0$ and so, for every $y ∈ H,⟨x-P x, \sum⟨y, x_n⟩ x_n⟩=0$, i.e. $x-P x ⟂ P y$. This shows that $\operatorname{Im} P ⟂ \operatorname{Im}(I-P)$. We conclude that $\operatorname{Im} P$ and $\operatorname{Im}(I-P)$ are orthogonal complementary spaces.

- Let $(X,\|⋅\|_\sup)$ denote the Banach space of complex-valued, continuous, $2 π$-periodic functions on $ℝ$, equipped with the supremum norm. Let $(c_0,\|⋅\|_{∞})$ denote the Banach space of sequences $(c_n)_{n ∈ ℤ}$ such that $c_n → 0$ as $|n| → ∞$.

Show that $T$ is an injective bounded linear operator from $X$ into $c_0$ and compute its operator norm. - Let $X$ and $c_0$ be as in (b)(i). By considering how $T$ acts on the Dirichlet kernels
\[
k_N(x)=\frac{1}{2 π} \frac{\sin ((N+\frac{1}{2}) x)}{\sin (\frac{1}{2} x)}=\frac{1}{2 π} \sum_{|n| ⩽ N} e^{\mathrm{i} n x}, N=1,2, …,
\]
or otherwise, show that $T: X → c_0$ is not surjective.

[You may use without proof any result about bounded linear operators between Banach spaces provided you state it correctly and verify that all conditions are met.] - Does there exist a $2 π$-periodic and continuously differentiable function $f: ℝ → ℂ$ such that\[ \limsup _{|n| → ∞}|n||T f(n)|>0 ? \]

- The linearity of $T$ is clear. The boundedness of $T$ follows from an easy estimate: \[ |T f(n)| ⩽ \frac{1}{2 π} ∫_{-π}^π|f(t)| d t ⩽\|f\|_\sup \] The injectivity of $T$ follows from (completeness of the trigonometric series) the hint $\{\frac{1}{\sqrt{2 π}} e^{\mathrm{i} n x}\}_{n ∈ ℤ}$ is an orthonormal basis: Indeed, if $f$ is such that $T f=0$, then the Fourier series of $f$ is identically zero, and hence $f=0$ a.e., by completeness. Since $f$ is continuous, $f ≡ 0$. Finally, from the above $\|T\| ⩽ 1$, but $T(1)=(δ_{n 0})_{n ∈ ℤ}$ has norm $1$, so $\|T\|=1$.
Inverse mapping theorem: Let $X, Y$ be Banach spaces and let $T ∈ B(X, Y)$ be bijective. Then $T$ is invertible.

Suppose $T$ is surjective, but (i) showed $T$ is injective, by the inverse mapping theorem, $T$ is invertible, so $T$ is bounded below \[ \inf_{f∈X}\frac{\|T f\|_∞}{\|f\|_\sup} >0 \] Now consider the Dirichlet kernel \begin{align*} k_N(t)&=\frac{1}{2 π} \sum_{|n| ⩽ N} e^{\mathrm{i} n x}\\ Tk_N(n)&=\begin{cases}\frac{1}{\sqrt{2 π}}&|n|≤N\\0&|n|>N\end{cases} \end{align*} We have $\|T k_N\|_{∞}=\frac{1}{\sqrt{2 π}}$. On the other hand, \[ \|k_N\|_\sup⩾ k_N(0)=\frac{1}{2 π}(2 N+1) → ∞, \] as $N → ∞$. This is a contradiction.

Similar question: Fourier transform is not surjective from $ℒ^1(ℝ)$ to $C_0(ℝ)$- The Fourier coefficients $d_n$ of $f'$ are given by \[ d_n=\frac{1}{2 π} ∫_{-π}^π f' e^{-i n x} d x=\frac{i n}{2 π} ∫_{-π}^π f e^{i n x} d x . \] Since $f' ∈C[-π,π]⊂L^2[-π, π]$, its Fourier coefficients are square summable, and so $d_n=n c_n$ converges to zero as $|n| → ∞$. The answer to the question is negative.