This follows from the fact that \(h(Kx) \in \mathcal{P}_q\) and \((T_j)_{0 \leq j \leq q}\) is a basis of \(\mathcal{P}_q\), since for all \(0 \leq j \leq q\), \(\deg T_j = j\).

By definition of \(f\) and \((a_j)_{0 \leq j \leq q}\), for all \(\theta \in \mathbb {T}\),

By definition of \((T_j)_{j \geq 0}\), \(T_j(\cos \theta ) = \cos (j \theta )\). The evenness comes from the fact that \(\cos \) is even.

This is probably already in Mathlib.

By Lemma 5.1.2 and 5.1.3, we have

Hence by the triangle inequality

I think this might be in Mathlib? If not this is a classic result and should be added. This is proven by interpolation between Parseval and the \(L^\infty \)-\(L^1\) trivial estimate, using Riesz-Thorin.

Assume \(1 {\lt} \beta \leq 2\). By the Hölder inequality,

We shall bound these two sums successively.

• Since \(\beta {\gt}1\) the first sum converges. By parity it is equal to twice the sum for \(j \geq 1\). Since \(t \mapsto 1/t^\beta \) is decreasing on \([1,\infty )\), we can compare the sum and integral which yields

  \begin{equation*} \sum _{j =1}^\infty \frac{1}{j^\beta } \leq 1 + \int _1^\infty \frac{\d t}{t^\beta } = 1 - \left[ \frac{1}{(\beta -1)t^{\beta -1}} \right]_1^\infty = 1 + \frac{1}{\beta - 1} = \frac{\beta }{\beta - 1} = \frac{1}{1 - 1/\beta } = \beta _*. \end{equation*}

  As a conclusion,

  \begin{equation*} \left(\sum _{j \neq 0} \frac{1}{|j|^\beta } \right)^{1/\beta } \leq 2 \beta _*^{1/\beta } \leq 2 \beta _* \end{equation*}

  since \(0 \leq 1/\beta \leq 1\) and \(\beta _* \geq 1\).

• For the second sum, we observe that the Fourier coefficients of \(f'\) are \((i j c_j)_{j \in \mathbb {Z}}\). Since \(1 {\lt} \beta \leq 2\), we can apply Theorem 5.1.5 to \(f'\), which yields

  \begin{equation*} \left(\sum _{j \neq 0} |j c_j|^{\beta _*} \right)^{1/\beta _*} \leq \left(\frac{1}{2 \pi } \int _0^{2 \pi } |f’(\theta )|^\beta \d\theta \right)^{1/\beta } \leq \frac{1}{(2 \pi )^{1/\beta }} \| f’\| _{L^\beta (\mathbb {T})} \leq \frac{1}{\sqrt{2 \pi }} \| f’\| _{L^\beta (\mathbb {T})} \end{equation*}

  since \(\beta \leq 2\).

Putting everything together we obtain

which implies the claim in the case \(1 {\lt} \beta \leq 2\) since \(\sqrt{2/\pi } \leq 4\).

Now assume \(\beta \geq 2\). By the Hölder inequality, since \(\mathbb {T}\) has finite measure \(2 \pi \), \(L^\beta (\mathbb {T}) \subseteq L^2(\mathbb {T})\) and

In particular, we can apply the result for \(\beta =2\) (in which case \(\beta _*=2\)), which yields

since \(\beta _* \geq 1\).

First, we apply Lemma 5.1.3 to express the Fourier coefficients \((c_j)_{j \in \mathbb {Z}}\) in terms of \((a_j)_{0 \leq j \leq q}\). Let \(m \geq 0\). By differentiation, the Fourier coefficients of \(f^{(m)}\) satisfy, for all \(j\), \(c_j^{(m)} = i^m j^m c_j\). In particular,

The conclusion follows by applying Lemma 5.1.6 to the function \(f^{(m)}\), of derivative \(f^{(m+1)}\) in \(L^\beta (\mathbb {T})\) as it is a trigonometric polynomial and in particular smooth.

We apply Lemma 5.1.7 with \(\beta = \infty \), in which case \(\beta _*=1\) and hence

We then use the chain rule to find a constant \(C_{m,K}\) such that \(\| f^{(m+1)}\| _{L^\infty (\mathbb {T})} \leq C_{m,K} \| h\| _{C^{m+1}([-K,K])}\) and conclude with \(c_{m,K}=4C_{m,K}\).

We first prove that the restriction is an element of \(\mathcal{D}'(\Omega ,F)\). By definition, for all \(h \in \mathcal{E}(\Omega ,F)\), we have \(|\nu (h)| \leq C \| h\| _{C^m(K)}\), with \(K\), \(C\), \(m\) as in the statement. Now take \(K'\) a compact set of \(\Omega \). Let \(h \in \mathcal{D}(\Omega ,F)\) such that \(\mathrm{supp} \, h \subset K'\). In particular, \(h \in \mathcal{E}(\Omega ,F)\), so we have \(|\nu (h)| \leq C \| h\| _{C^m(K)}\). But since \(h\) is identically equal to \(0\) outside of \(K'\), \(\| h\| _{C^m(K)} \leq \| h\| _{C^m(K')}\), which is what we needed to prove.

Now to prove the support claim, let us prove that \(\nu \) vanishes on \(K^c\), ie for all \(h \in \mathcal{D}(\Omega ,F)\) such that \(\mathrm{supp} \, h \subseteq K^c\), we have \(\nu (h)=0\). Indeed, for such a \(h\), we have \(|\nu (h)| \leq C \| h\| _{C^m(K)}\). But \(\| h\| _{C^m(K)} = 0\) since \(h\) is identically equal to \(0\) on \(K\), and hence \(\nu (h)=0\). Hence, the support of \(\nu \) is included in \(K\). Since the support is closed and \(K\) is compact, \(\mathrm{supp} \, \nu \) is compact.

This follows from Leibniz’s formula.

The support of \(\nu \) is compact, we denote it as \(K_0\). We take a function \(\phi \in \mathcal{D}(\Omega ,F)\) such that \(\phi \) is identically equal to \(1\) on \(K_0\). We then define \(\mu : \mathcal{E}(\Omega ,F) \rightarrow F\) by letting \(\mu (h) := \nu (h \phi )\) for \(h \in \mathcal{E}(\Omega ,F)\).

• This is well-defined by Lemma 5.2.5. The linearity is immediate.

• We prove that \(\mu \) is an extension of \(\nu \). Let \(h \in \mathcal{D}(\Omega ,F)\). We observe that \((1-\phi ) h\) belongs in \(\mathcal{D}(\Omega ,F)\) and is identically equal to \(0\) on \(K_0 = \mathrm{supp} \, \nu \). This implies (by a simple lemma on supports) that \(\nu ((1-\phi )h) = 0\), hence \(\nu (h) = \nu (\phi h) = \mu (h)\).

• Let us now prove the norm estimate for \(\mu \).

  • The function \(\phi \) has compact support, which we denote by \(K\). By definition of \(\mathcal{D}'(\Omega ,F)\) applied to \(K\), we obtain a constant \(C_1\) and an integer \(m\) such that, for all \(h \in \mathcal{D}(\Omega ,F)\), if \(\mathrm{supp} \, h\) is included in \(K\), then we have \(|\nu (h)| \leq C_1 \| h\| _{C^m(K)}\).

  • Lemma 5.2.5 grants a constant \(C_2\) (depending on \(\phi \)) such that, for all \(h \in \mathcal{E}(\Omega ,F)\), \(\| \phi h\| _{C^m(K)} \leq C_2 \| h\| _{C^m(K)}\).

  Now take \(h \in \mathcal{E}(\Omega ,F)\). By Lemma 5.2.5, \(\phi h \in \mathcal{D}(\Omega ,F)\) and its support is included in \(K\). Hence,

  \begin{equation*} |\mu (h)| = |\nu (\phi h)| \leq C_1 \| \phi h\| _{C^m(K)}. \end{equation*}

  We then note that \(\| \phi h\| _{C^m(K)} \leq C_2 \| h\| _{C^m(K)}\). This proves the claim with \(C := C_1 C_2\).

• The uniqueness is then obtained by the density of \(\mathcal{D}(\Omega ,F)\) in \(\mathcal{E}(\Omega ,F)\).

We will prove \(1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 1\).

Let us assume \(1\). Let \(j \geq 1\). Let \(h (x) := T_j(x/K)\). We note that \(h\) is an element of \(\mathcal{P}_j\), so can apply the hypothesis to it, which yields \(|\nu (h)| \leq c j^m \| h\| _{C^0([-K,K])}\). We conclude by observing that \(\| h\| _{C^0([-K,K])} = \| T_j\| _{C^0([-1,1])}=1\) using the definition of the Chebyshev polynomials.

Let us now assume \(2\), which gives us some parameters \(c, K, m\). Let \(q \geq 1\) and \(h \in \mathcal{P}_q\). We write \(h = \sum _{j=0}^q a_j T_j(\cdot / K)\) as before. Then by linearity, the triangle inequality and the assumption

By Lemma 5.1.4, \(|a_0| \leq \| h\| _{C^0([-K,K])}\). By Corollary 5.1.8, there exists a constant \(c_{m,K}\) depending only on \(m\) and \(K\) such that \(\sum _{j=1}^q j^m|a_j| \leq c_{m,K} \| h\| _{C^{m+1}([-K,K])}\). Then, taking \(C := |\nu (1)| + c c_{m,K}\), we have that for all \(h \in \mathcal{P}\), \(|\nu (h)| \leq C \| h\| _{C^{m+1}([-K,K])}\). By density of \(\mathcal{P}\) in \(\mathcal{E}(\mathbb {R}, \mathbb {C})\) for \(\| \cdot \| _{C^{m+1}([-K,K])}\), \(\nu \) extends uniquely to a compactly supported distribution.

To conclude, we prove \(3 \Rightarrow 1\). [todo]

Let \(i \geq 0\). By Lemma 4.3.3, for every \(q \geq 1\) and \(h \in \mathcal{P}_q\),

where \(c = (4 q_0(1+\log r))^{4i}\). By Proposition 5.2.7 (\(3 \Rightarrow 1\)) with \(K\), \(c\) and \(m=4i\), \(\nu _i\) can be uniquely extended to a compactly supported distribution.

Let us now take \(q \geq 1\) and \(h \in \mathcal{P}_q\). [todo]