prove that if a linear map is continuous at a point then it is bounded

Question

I want to prove that, if a linear map is continuous at a point, then it must be bounded. Here's what I've tried so far. I think there's a hold in my logic when I divide $\epsilon$ by $\delta$...

A function is continuous if there exists $\epsilon>0$ and $\delta>0$ such that $||f(x_1) - f(x_2)||_Y < \epsilon$ whenever $||x_1 - x_2||_X < \delta$, where $x_1, x_2 \in X$. If $T$ is continuous at a point $x_0\in X$, then by definition, there exists $\epsilon$ and $\delta$ such that $||Tx - Tx_0|| < \epsilon$ whenever $||x - x_0|| < \delta$.

Notice that

$||T(x_n - x)|| = ||T||\ ||x_n - x|| < \epsilon$.

Since $||x_n - x|| < \delta$, we can say that $||T||<\frac{\epsilon}{\delta}$ and the linear map must be bounded.

Answer 1

A simple but central result in the theory of linear operators.

In the following, let $V$ be a normed space, and let

$T: V \to V \tag 1$

be a linear map on $V$ which is continuous at the point $x \in V$; we will prove that

1.) $T$ is continuous at $0 \in V$;

2.) $T$ is bounded;

3.) $T$ is continuous everywhere on $V$.

For suppose $T$ is continuous at $x \in V$; then for any $\epsilon > 0$ there exists $\delta > 0$ such that

$\Vert y - x \Vert < \delta \Longrightarrow \Vert Ty - Tx \Vert < \epsilon; \tag{2}$

the linearity of $T$ yields

$T(y - x) = Ty - Tx; \tag{3}$

thus (2) becomes

$\Vert y - x \Vert < \delta \Longrightarrow \Vert T(y - x) \Vert < \epsilon; \tag{4}$

now for

$\Vert z \Vert < \delta \tag{5}$

set

$y = x + z; \tag{6}$

then

$z = y - x, \tag{7}$

$Tz = T(y - x); \tag{8}$

when (7) and (8) are substituted into (4) we see that

$\Vert z \Vert < \delta \Longrightarrow \Vert Tz \Vert < \epsilon, \tag{9}$

which is simply a statement that $T$ is continuous at $0$. (1) is thereby proved.

As for (2), given $\epsilon$ and $\delta$ as in (9), we may choose $0 < \delta^\ast < \delta$ and then affirm that

$\Vert z \Vert \le \delta^\ast \Longrightarrow \Vert Tz \Vert < \epsilon, \tag{10}$

and given $0 \ne y \in V$ we have

$\left \Vert \dfrac{\delta^\ast y}{\Vert y \Vert} \right \Vert = \delta^\ast \dfrac{\Vert y \Vert }{\Vert y \Vert } = \delta^\ast; \tag{11}$

it then follows from (9) that

$\left \Vert T \left ( \dfrac{\delta^\ast y}{\Vert y \Vert} \right ) \right \Vert < \epsilon, \tag{12}$

which may be re-written as

$\Vert Ty \Vert < \dfrac{\epsilon}{\delta^\ast} \Vert y \Vert, \tag{13}$

and we see that $\epsilon / \delta^\ast$ is a bound for $T$.

Finally, (3) is immediate from (13) by taking

$y = z - x; \tag{14}$

then if

$\Vert z - x \Vert < \delta^\ast, \tag{15}$

it follows that

$\Vert Tz - Tx \Vert < \epsilon, \tag{16}$

and we see that $T$ is continuous at every $x \in V$.

Answer 2

Here is a better proof: suppose $T$ is not bounded. Then there exists vectors $x_n$ of norm $1$ such that $\|Tx_n\| >n$ for each $n$. Let $y_n=\frac 1 n x_n+x_0$. Then $y_n \to x_0$ so ( by linearity) $\frac 1 n T(x_n)+T(x_0) \to T(x_0)$. Hence $\frac 1 n \|Tx_n\| \to 0$ which is a contradiction to the choice i=of $x_n$'s.