# Hermitian Projections on $C[0,1]$

If $X$ is a normed linear space and $S(X)$ its unit sphere, $X′$ its dual space and $Π=\{(x,f)∈S(X)×S(X′) \ | \ f(x)=1\}$, then for an operator $T$ on $X$, the numerical range $V(T)$ is defined by $V(T)=\{f(Tx) \ | \ (x,f)∈Π\}$. A projection on a complex Banach space $X$ is said to be hermitian if its numerical range is real.

I tried to find an example of an hermitian operator on the Banach space $C_{\mathbb C}[0,1]$ but could not find it. I have some intiutions that the decomposition $E\oplus F$ of $C_{\mathbb C}[0,1]$ where $E=\{f=\alpha \ | \ \mbox{for some } \alpha \in \mathbb C\}$ and $F=\{g=g_1+i g_2 \in C_{\mathbb C}[0,1] \ | \ \int g=\int g_1 + i \int g_2 =0\}$ is an hermitian decomposition, i.e., the projection $P: C_{\mathbb C}[0,1] \to E$ is hermitian, but could not prove it.

Does anyone know an example of an hermitian projection on $C_{\mathbb C}[0,1]$?

This question also asked in stackexchange, https://math.stackexchange.com/questions/1610433/hermitian-projections-on-c0-1

• To mselcuk: It looks like Ostrovskii has answered your question. You did not accept the answer, does this mean that you find it incomplete? Jan 24, 2016 at 19:24

As I understand, the fact that for $C(0,1)$ we have too many supporting pairs $(x,f)$ implies that there are no nontrivial Hermitian projections in the following way. Let $t\in[0,1]$, observe that $f=\delta_t$ and any $x$ with $||x||=1$ and $x(t)=1$ form a supporting pair. Let $T$ be a Hermitian projection. Then $f(Tx)=T^*f(x)$ should be real for any such pair. We claim that this implies that $T^*\delta_t$ ($T^*$ Banach-space conjugate) is a multiple of $\delta_t$. Suppose not, then $T^*\delta_t$ is a measure which is supported on a set which contains something outside singleton $\{t\}$. But then one can find a vector $x$ as above such that $T^*\delta_t(x)$ is not real. This can be done as follows: Our assumption implies that the support of $T^*\delta_t$ contains a nontrivial part in some open interval $(a,b)$ for which $[a,b]$ does not contain $t$. Now pick $\hat x$ satisfying $\hat x(t)=1$, $||\hat x||=1$, $\hat x(a)=\hat x(b)=0$ and $\mu(\hat x)\ne 0$, where $\mu$ is the restriction of $T^*\delta_t$ to $(a,b)$. To get the desired $x$ multiply $\hat x$ on $(a,b)$ by a suitable complex number with absolute value $1$.
Since $T^*\delta_t$ is a multiple of $\delta_t$, one can conclude that $T$ is an operator of a multiplication by a continuous function. Since it is also a projection, it is easy to conclude that it is either $0$ or the identity.