I have to determine the operator norms, the kernels and the images of the following 2 maps:

1) $F_1 :\{x\in C^0([0,10],\mathbb R)|x(0)=0\}\rightarrow C^0([0,10],\mathbb R)$

$(F_1x)(t)=\left\{\begin{array}{ll} x(t-1), & t\in [1,10] \\ 0, & t\in [0,1)\end{array}\right.$

2) $F_2:C_b^0(\mathbb R,\mathbb R)\rightarrow C_b^0(\mathbb R,\mathbb R)$

$(F_2x)(t)=x(t-1), t\in \mathbb R$

I also have to determine the operator norm of the following:

3) $F:C^0([0,2]),\mathbb R)\rightarrow\mathbb R$


Where the norm on the spaces of continuous functions is the usual $||\circ ||_{\infty}$

Since I am very new to this topic (operator norm) I might need some help.

1) I have to find the smallest $M\geq0$ such that $||F_1x||<M||x||$ for all x.

For $t \in [0,1)$ its 0, for $t\in [1,10]$ its 9,because t-1 is monotone increasing on that interval, is this correct?

The kernel should be just the zero function. But whats the image here?

2) Since the term (t-1) is unbounded for all real numbers, M should be equal to $\infty$?

The kernel should be again zero. And again I don't know how to determine the image..

3) No ideas here.

Thanks in advance

  • $\begingroup$ You seem very confused. For example, $F_2$ is a bounded linear operator satisfying $||F_2(x)||_{\infty} = ||F_2(x)||_{\infty}$ for all $x \in C^0_b(\mathbb{R}, \mathbb{R})$. Infact, $\sup_t |x(t)| = \sup_{1-t} |x(1-t)|$ since $t \mapsto (1-t)$ is a bijection. So $M=1$. $\endgroup$ – Crostul Nov 4 '14 at 23:05

You seem a bit confused in your reasoning for the first part. I'll give the solution for part 1, see if that helps.

The operator norm of $F_1$ is one. The norm of $F_1(x)$ is $\sup x([0,9])$. The norm of $x$ is $\sup x([0,10])$. Because $x([0,9])$ is contained in $x([0,10])$, we have that the norm of $F_1(x)$ is no greater than the norm of $x$, so the operator norm of $F_1$ is no greater than one. Because there exist functions (any constant function, for example, will do) such that $\sup x([0,9])$ = $\sup x([0,10])$, the operator norm can be no less than one. Thus the operator norm of $F_1$ is exactly one.

The kernel is the set of all functions $k$ such that $F_1(k)$ is the zero function. $F_1(k)(t)$ is by definition zero for $t$ between $0$ and $1$, so we only need to consider $t \in[1,10]$. On this interval, $F_1(k)(t) = k(t-1)$. So we need that $k(t-1)$ equals zero on $[1,10]$, meaning $k(t)$ equals zero on $[0,9]$. The kernel is thus the set of continuous functions on $[0,10]$ that are zero on $[0,9]$.

Lastly, the image is the set of all continuous functions on $[0,10]$ that are zero on $[0,1]$.

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Bounds for 2. and 3.:

$2)$ Yes, $(t-1)$ is unbounded in $\mathbb{R}$, but your operator shift by $-1$ the values at where the function is applied $F(f)(t) = f(t-1)$. (And these functions are bounded, remember $ f \in C_b^0$), so it is well defined. Now to see it has a norm, $|F(f)(t)| = |f(t-1)| \leq \sup_t |f(t-1)| = sup_t|f(t)| = ||f||_\infty$. Hence $||F|| \leq 1$ and using a constant function you can see $||F|| = 1$.

$3)$ They're continous functions again, so the integral is well defiend an is a value in $\mathbb{R}$ so $F$ is well defined. Next note that:

$$|F(f)| = \bigg| \int_0^1 f - \int_0^2 f\bigg| \leq \bigg|\int_0^1 f\bigg| + \bigg|\int_1^2 f\bigg|$$ $$\leq \int_0^1 |f| + \int_1^2 |f| \leq \int_0^1 ||f||_{\infty} + \int_1^2 ||f||_{\infty} = 3||f||_{\infty} $$ To see this is the minimum norm, fix an $\varepsilon>0$ and consider a function with constant value 1 up to $1 - \varepsilon$, then from there extend it linearly to $-1$ at $1 + \varepsilon$ and then youll see that if you apply $F$ you will get $|F(f)| = (3 - \varepsilon)||f||_{\infty}$ And because $\varepsilon$ is abritrary you can conclude that $||F|| = 3$.

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