Bounded linear operator in a normed space, $Tf=f(0)$ on $C[0,1]$ Let $E = C[0,1]$ with the norm $\|\cdot\|_\infty$. Define $T:E\rightarrow\mathbb{R}$ as $Tf=f(0)$. Prove that $T$ is a bounded linear operator and calculate $\|T\|$.
I already tried to prove that T is linear. I just need to show that $T$ is bounded and to calculate $||T||$, but I do not know how to do that. I appreciate your support.
This is my proof that T is linear. Is it correct?
If $f, g\in E$ and $\alpha \in \mathbb{R}$, then $$T(\alpha f(t)+g(t))=\alpha f(0)+g(0)=\alpha Tf+Tg.$$
 A: Your proof that $T$ is linear is correct except that $T$ is applied on function so you should remove the $(t)$ in $f(t)$. And I would've added a step to say that $(\alpha f+g)(0)=\alpha f(0) + g(0)$ (which is simply the definition of addition on functions).
To prove it is bounded, you just need to notice that $\left|Tf\right|=\left|f(0)\right|\le 1 \cdot \sup\limits_{[0,1]}\left|f\right| = 1 \cdot \left\|Tf\right\|_\infty$ so that for $f\not= \Bbb 0, \cfrac{\left|Tf\right|}{\left\|Tf\right\|_\infty} \le 1$, which means that $T$ is bounded.
It also tells you that $\|T\| \le 1$
Since for $f=\Bbb 1$, $\cfrac{\left|Tf\right|}{\left\|Tf\right\|_\infty}=1$, you get that $\|T\| \ge 1$
So $\|T\|=1$
A: To show the function is bounded we need to show that there exists $M>0$
$ \left\| T(f)\right\|\leq M\left\| f\right\|_\infty$.
So we need to find a way to relate $\sup\{f(x)\, :\,x\in [0,1]\}$ to $f(0)$. However clearly $f(0)\leq\sup\{f(x)\, :\,x\in [0,1]\}=\left\| f\right\|_\infty$ so pick $M=1$ and we have
$ \left\| T(f)\right\|=f(0)\leq\sup\{f(x)\, :\,x\in [0,1]\}=\left\| f\right\|_\infty=M\left\| f\right\|_\infty$
$\left\|T\right\|$ is defined to be the infimum $M$ which works in the above, so clearly $\left\|T\right\|\leq 1$, can we find a function where it equals 1? Well that would mean that $f(0)=\sup\{f(x)\, :\,x\in [0,1]\}$ and we can arrange this easily by taking a constant function.
A: $|Tf| = |f(0)| \le \|f\|_\infty$. It follows that $\|T\| = \sup_{\|f\|_\infty \le 1} |Tf| \le 1$. Choosing $f(t) = 1$ gives $\|Tf\| = 1$, Hence the bound is attained which gives $\|T\| = 1$.
Your linearity proof is correct, although the notation is a little sloppy. I would prefer $T(\lambda f) = \lambda f(0) = \lambda Tf$ and $T(f+g) = f(0)+g(0) = Tf + Tf$, but this is purely a personal preference for showing linearity.
