0

Here is the question link:

Show that there is $m>0$ such that $\|T(x)\|\geq m\|x\|$ for all $x\in X$ iff $T$ is injective and $T(X)$ is closed.

And here is the part in the OP question I am speaking about:

Let $y\in \overline{T(X)}$. Then choose $(y_n)=(T(x_n))$ in $T(X)$ such that $y_n\to y$. For each $m,n\in\mathbb{Z^+},\|y_m-y_n\|=\|T(x_m)-T(x_n)\|=\|T(x_m-x_n)\|\geq m\|x_m-x_n\|$, and therefore since $(y_n)$ is cauchy $(x_n)$ is Cauchy and hence converges to some $x\in X$ as $X$ is complete. Now by continuity of $T$ we have $y=T(x)\in T(X)$. Hence $T(X)$ is closed.

My question is:

I do not understand this statement " Now by continuity of $T$ we have $y=T(x)\in T(X)$ " is there a proof for it?

2 Answers2

2

Recall that a function $T$ between normed linear spaces (or indeed, metric spaces) is continuous at $x$ if and only if, for every $x_n \to x$, we have $T(x_n) \to T(x)$. Also in such spaces, limits are unique.

So, we have $x_n$ is convergent to some $x$. Since $T$ is continuous, it is continuous at this $x$, so $T(x_n) \to T(x)$. But, at the same time, we have $T(x_n) = y_n \to y$. Thus, by uniqueness of limits, $T(x) = y$.

Now, $T(X)$, the range of $T$, is defined to be vectors of the form $T(v)$ where $v \in X$. This obviously includes $T(x) = y$.

user759562
  • 822
1

We have $x_n \to x$. Since $T$ is continuous, we get $T(x_n) \to T(x).$ Since $y_n =T(x_n)$, it follows that $y_n \to T(x).$ From $y_n \to y$ we derive $y=T(x).$ Hence $y \in T(X).$

Fred
  • 77,394