Computing $||T||$ for linear operators

Question

Letting $(X,||\cdot ||)$ be a normed linear space, I want to compute ||T|| for the linear operators $T:X \to X$ when

a) $(X, ||\cdot ||)=(C[0,1],||\cdot ||_{\infty})$ and $$ (Tx)(t) = t^2 \cdot x(0).$$

b) $(X, ||\cdot ||)=(\ell_1,||\cdot ||)$, the space of absolutely summable sequences with norm $$||x||_1=\sum_{n=1}^{\infty}|{x_n}|.$$ and $$T(\{x_1,x_2,x_3,\ldots\})=\{0,x_1,x_2,\ldots\}.$$

Essentially, I think the notation is what confusing me. I am aware to compute the operator norm,$ ||T||$, I need to show that $||Tx|| \leq c||x||, \forall x \in X$.

Then find $x_0 \in X$ such that $||Tx_0||=c||x_0||$, then I conclude that $||T||=c$.

However, when I begin with a), I am a little confused how to address it. Recalling that $||x||_{\infty}=\max\{|x|\}$, does this mean I need to compute

$||(Tx)(t)||_{\infty}=\max\{|t^2\cdot x(0)|\}$? And if so what could I even bound that by? Would it be the case that it is actually $$ ||(Tx)(t)||_{\infty}=t^2\max\{|x(0)|\} $$ and imply that $c = ||T||=t^2?$

These are honestly just guesses and any explaination would be hugely appreciated.

Answer 1

For the first part you seem to be confused about what $\|Tx\|_\infty$ is. Remember that $Tx$ is a function on $[0,1]$ and, using your notation, $t$ is its argument. That means that you would like to compute $\max_{t \in [0,1]} |Tx(t)|$. In particular, it does not make sense to pull $t$ out of the maximum.

Instead we have that $\| Tx \|_\infty = \max_{t \in [0,1]} |t^2 x(0)| = |x(0)|$ since $x(0)$ is a constant. In particular, $\| Tx \|_\infty = |x(0)| \leq \|x\|_\infty$ and if $|x|$ is maximised at $0$ then $\|Tx \|_\infty = \|x\|_\infty$. This means that $\|T\| = 1$.

For the second part you should try directly computing $\|Tx\|_1$ for an arbitrary vector $x \in \ell^1$ and see what this gives you.

Answer 2

a) $\|Tx\|_{\infty}=\max_{t\in[0,1]}|t^{2}x(0)|\leq\max_{t\in[0,1]}|x(0)|=|x(0)|\leq\max_{u\in[0,1]}|x(u)|=\|x\|_{\infty}$, so $\|T\|\leq 1$.

Now by letting the $x$ to be the constant function $x=1$, then $\|x\|_{\infty}=1$, so $\|Tx\|_{\infty}=\max_{t\in[0,1]}|t^{2}x(0)|=\max_{t\in[0,1]}t^{2}=1$, so $\|T\|=1$.

b) $\|Tx\|_{1}=\displaystyle\sum_{i=1}^{\infty}|(Tx)_{i}|=0+\sum_{i=2}^{\infty}|(Tx_{i})|=\displaystyle\sum_{i=1}^{\infty}|x_{i}|=\|x\|_{1}$, so $\|T\|=1$.

Answer 3

Let $T:X\to Y$ where $X,Y$ are both normed linear spaces.

$$||T||:=\sup\{||x||_X\leqslant 1:||Tx||_Y\}$$ In the first case: $X=Y=C[0,1]$ and $||\cdot||_X=||\cdot||_{\infty}$. $$\sup\{||x||_{\infty}\leqslant 1:||Tx||_{\infty}\}=\sup\{||x||_{\infty}\leqslant 1: \sup_{t\in[0,1]}|(Tx)(t)|\}\\= \sup\{||x||_{\infty}\leqslant 1: \sup_{t\in[0,1]}|t^2x(0)|\}\\=\sup\{||x||_{\infty}\leqslant 1: |x(0)|\}=\sup\{\sup_{t\in[0,1]}|x(t)|\leqslant 1: |x(0)|\}\leqslant 1$$ Hence $||T||\leqslant 1$. Now let $\hat{x}(t)\in C[0,1]$ be such that $\hat{x}(0)=1$ and $\hat{x}(t)\leqslant 1$ for all $t\in[0,1]$. Then $$||(T\hat{x})(t)||_{\infty}=\sup_{t\in[0,1]}|(T\hat{x})(t)|=\sup_{t\in[0,1]}|t^2\hat{x}(0)|=|\hat{x}(0)|=1=\sup_{t\in[0,1]}|\hat{x}(t)|=||\hat{x}||_{\infty}$$ Hence $||T||=1$. Analogue arguments for the other example.