Does the sequence of operators $(A_nx)(t) = x(t^{1+\frac{1}{n}}) $converge by norm?

Question

I have a simple question:

The question is as follows:

Consider the sequence of operators $A_n:C[0,1] \rightarrow C[0,1]$ as follows:

$$ (A_nx)(t) = x(t^{1+\frac{1}{n}})$$ Prove that for each $n\in \mathbb{N}$, $A_n$ is a linear bounded operator. Show that $A_n \rightarrow I$ strongly. Does it converge by norm? Justify.

$\textbf{Notation and information}$

I refer you to the answer by "user438666", for to get information about the notations in the question.

I really cannot make connection with the defined operators!

Can someone help me to understand it?

Thanks!

Answer 1

1. $\lim_{n\to \infty} \|A_n(f) - f\|_\infty = 0$ for each $f\in C = C([0,1]).$

Proof: Verify that $t^{1 + 1/n} \to t$ uniformly on $[0,1].$ (You can use calculus to see the maximum value of $t-t^{1 + 1/n}$ occurs at $1/(1+1/n)^n.$)

Let $f \in C, \epsilon>0.$ Because $f$ is uniformly continuous on $[0,1],$ there exists $\delta > 0$ such that $x,y \in [0,1], |x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon.$ Because $t^{1 + 1/n} \to t$ uniformly, we can choose $N$ such that $|t^{1 + 1/n} -t | < \delta $ for $t\in [0,1]$ and $n\ge N.$ For such $n$ we have

$$\|A_n(f) - f\|_\infty = \sup_{t\in [0,1]} |f(t^{1 + 1/n})-f(t)| \le \sup_{x,y\in [0,1],|x-y| < \delta}|f(x)-f(y)| \le \epsilon.$$

This proves 1.

2. $A_n$ does not converge to $I$ in the operator norm. To prove this it's enough to show there exists a sequence $f_n$ in the closed unit ball of $C$ such that

$$\tag 1 \|A_n(f_n) - f_n\|_\infty \ge 1,\, n = 1,2,\dots$$

To do this, define $f_n$ to have the piecewise linear graph that joins the points $(0,0), (1/2^{n+1}, 0),$ $(1/2^n,1), (1,1).$ Then $\|f_n\|_\infty =1$ for all $n.$ We have the left side of $(1)$ at least as large as

$$|f_n((1/2^n)^{1+1/n}) - f_n(1/2^n)| = |f_n(1/2^{n+1}) - f_n(1/2^n)| = |0-1| = 1.$$

This proves $(1)$ and we are done.

Answer 2

This isn't a solution to the problem, but just a clear up of some definitions that wouldn't fit nicely in a comment. You're confusing notions of convergence. The one you call uniform convergence is really strong convergence. This is because uniform convergence is usually the convergence in the norm of the space of the sequence, which in this case is a sequence of operators, not functions. You're dealing with a sequence of operator on a Hilbert space. One defines a norm on the space of operator $L(H)$ on a Hilbert space $H$ as

$$ ||A||=\sup\left( ||Ax||_H:x\in H, ||x||_H\leq1 \right)$$

where $||\cdot||_H$ is the norm of the Hilbert space. If $||A||<\infty$ we say that $A$ is bounded.

A sequence $A_n$ is said to be strongly convergent to $A$ iff $$ \forall \,x\in H,\,\forall \,\varepsilon >0 \,\exists \, N\in\mathbb{N} : ||A_nx-Ax||_H<\varepsilon \,\forall \,n>N$$

notice how the convergence is defined using the norm of the Hilbert space, not of the operator space. Conversely, we say that the sequence converges uniformly to $A$ iff

$$ \forall \,\varepsilon >0 \,\exists \, N\in\mathbb{N} : ||A_n-A||<\varepsilon \,\forall \,n>N$$

Which uses directly the operator space norm.