Isometric Isomorphism From $\mathbb{C}^n \oplus M([a,b])$ Onto $(C^n[a,b])^*$

Question

I have the following norm in $C^n[a,b]$: $$\| f \| = \sum_{k=0}^{n-1}|f^{(k)}(a)| + \sup_{[a,b]} |f^{(n)}|$$ I want to show that: $$(\lambda_0,...,\lambda_{n-1},\mu)\mapsto \left( f\mapsto\sum_{k=0}^{n-1}\lambda_kf^{(k)}(a) + \int_a^b f^{(n)}(t)d\mu(t)\right)$$ Is an isometric isomorphism from $\mathbb{C}^n\oplus M([a,b])$ onto $(C^n[a,b])^*$, with the norm: $$\|(\lambda_0,...,\lambda_{n-1},\mu \| = \max \{|\lambda_0|,...,\lambda_{n-1},\|\mu \|\}$$ Where $M([a,b])$ is the space of complex Lebesgue-Stieltjes measures on $[a,b]$, and $\| \mu \|$ is the total variation of $\mu$.
I think that it's clear that the mapping is surjective, but I don't know how to show that it's an isometry. At first, I thought to show that the functional linear described is bounded, But I don't know how this can help me.
I do however know that the norm: $$\| f\| = \max_{k=0,...,n} \sup_{[a,b]}|f^{(k)}|$$ is an equivalent norm on $C^n[a,b]$, but I don't know if I need to use that to show the isometry.
I'm struggling with this question too much, I could use a hint to figure out what I'm supposed to do. Thank you for the help!

Answer 1

First, let's go through the same process as you would use to show that the map of interest is bounded. I'll call it $T$, and I'll write $\lambda := (\lambda_0, \cdots, \lambda_{n-1})$.

We have: $$\begin{split} \left\|T(\lambda,\mu)\right\| &= \sup_{\|f\| \leq 1} \left|T(\lambda, \mu)(f)\right|\\ &= \sup_{\|f\| \leq 1} \overbrace{\left|\sum_{k=0}^{n-1}\lambda_kf^{(k)}(a) + \int_a^b f^{(n)}(t)\mathrm{d}\mu(t)\right|}^{=:\, G_4(f)}\\ &\leq \sup_{\|f\| \leq 1} \overbrace{\left(\sum_{k=0}^{n-1}\left|\lambda_kf^{(k)}(a)\right| + \int_a^b \left|f^{(n)}(t)\right|\mathrm{d}|\mu|(t)\right)}^{=:\,G_3(f)}\\ &\leq \sup_{\|f\| \leq 1} \overbrace{\left(\|\lambda\|_\infty\sum_{k=0}^{n-1}\left|f^{(k)}(a)\right| + \sup_{x \in [a,b]} \left|f^{(n)}(x)\right| \underbrace{\int_a^b \mathrm{d}|\mu|(t)}_{= \|\mu\|}\right)}^{=:\,G_2(f)}\\ &\leq \sup_{\|f\| \leq 1} \overbrace{\max(\|\lambda\|_\infty, \|\mu\|)\|f\|}^{=:\, G_1(f)} = \|(\lambda, \mu)\|\end{split}$$

Now, since we want to prove that $T$ is an isometry, we'll a fortiori have that all the inequalities above are actually equalities.

Conversely, and most interesting to us, if they are equalities then we'll obtain that $T$ is an isometry, thus this gives us a good opportunity to reach our goal. Not only that, but here it suffices to show that a single well-chosen function $f$ maximises all the $\sup$s.
Indeed, assume a function $f \in \mathcal{C}^n(a,b)$ of norm $1$ is built such that: $$\|(\lambda, \mu)\| = G_1(f) = G_2(f) = G_3(f) = G_4(f) =: K$$ From our previous endeavour, we know that: $$\|T(\lambda, \mu)\| = \sup_{\|h\| \leq 1} G_4(h) \leq \sup_{\|h\| \leq 1} G_3(h) \leq \sup_{\|h\| \leq 1} G_2(h) \leq \sup_{\|h\| \leq 1} G_1(h) \leq K$$ Thus, by a finite induction argument using that $G(f) = K \geq \sup_h G(h)$ implies that $K = \max_h G(h)$: $$\|T(\lambda, \mu)\| = \|(\lambda, \mu)\|$$ and $T$ will be an isometry.

You are able to define a good function $f$ for all the cases below:

• If $\|\lambda\| \geq \|\mu\|$, then the best course of action looking at our previous inequalities would be to create an $f$ which satisfies the following conditions, for $k_0$ such that $|\lambda_{k_0}| = \|\lambda\|_\infty$: $$\begin{cases} \forall k \leq n-1, k\neq k_0,\,\, f^{(k)}(a) = 0\\ f^{(k_0)}(a) = 1\\ f^{(n)} \equiv 0 \text{ on } [a,b]\end{cases}$$ which is obtainable through Hermite interpolation.
• If $\|\mu\| \geq \|\lambda\|_\infty$, then you'd rather look for the following: $$\begin{cases} \forall k \leq n-1,\,\, f^{(k)}(a) = 0\\ f^{(n)} \equiv 1 \text{ on }[a,b]\end{cases}$$ that is, the function $f : x \mapsto \displaystyle\frac{(x - a)^n}{n!}$.

This lets you demonstrate that $T$ is indeed an isometry, and thus an isometric isomorphism by surjectivity (I don't know if surjectivity is that obvious but if it is for you then you'll be done).