Let $(X,\lVert \cdot\lVert)$ be a normed space. Let $T$ be a function given by \begin{equation} \begin{array}{cccc} T:&X&\longrightarrow&X\\ &x&\longmapsto&T(x)=x\lVert x\lVert. \end{array} \end{equation} Is $T$ uniformly continous?
My answer is yes. Let $(x_n),(y_n)$ sequences in $X$ such that $\lVert x_n-y_n\lVert\xrightarrow[n\to+\infty]{}0$. I need to prove that $\lVert T(x_n)-T(y_n)\lVert\xrightarrow[n\to+\infty]{}0$. Indeed, \begin{equation} \lVert T(x_n)-T(y_n)\lVert=\lVert \lVert x_n\lVert x_n-\lVert y_n\lVert y_n\lVert\leq (\lVert x_n\lVert+\lVert y_n\lVert)\lVert x_n-y_n\lVert\xrightarrow[n\to+\infty]{}(2\lVert x_n\lVert)\cdot0=0. \end{equation} However, I could find a sequence $(x_n)$ in $X$ such that $\lVert x_n\lVert\xrightarrow[n\to+\infty]{}+\infty$ but I'm not sure if that sequence exists.
Thank you so much!
Suppose $\|x_0\|=1$. Let $x_n =nx_0, y_n=(n+\frac 1 n)x_0$. Show that $\|x_n-y_n\|=\frac 1 n$ and $\|Tx_n-Ty_n\|=\|2x_0+\frac 1{n^{2}}x_0\| \to 2$. What does this say about uniform continuity?