I'm trying to prove that if a linear operator is continuous, then it is bounded.

Let $T:V\to W$. Let us assume it is continuous. Then for any $\epsilon>0$, $\|T(x-x_0)\|<\epsilon$ if $\|x-x_0\|<\delta$ for some $\delta\in \Bbb{R}$.

If $T$ is bounded, then $\sup \frac{\|T(x-x_0)\|}{\|x-x_0\|}$ exists, where $x$ and $x_0$ may be any vectors in $V$.

The proof in my book "Functional Analaysis" by Kreyszig (pg.97) proceeds by stating

Let $x=x_0+\delta\frac{y}{\|y\|}$, where $y$ is any vector in $V$

Aren't we artificially restricting the possible values of $x$ in comparison with $x_0$?

Thanks in advance!

  • 2
    $\begingroup$ Any vector $x$ can be written in that form with $y = x-x_0$ and $\delta = \Vert y \Vert$. $\endgroup$ Sep 24, 2013 at 10:25
  • $\begingroup$ True. But at the end of the proof, it is stated $\frac{\|Ty\|}{\|y\|}=\frac{\epsilon}{\delta}$. If $\delta=\|y\|$, then $\frac{\epsilon}{\delta}$ is dependant on $y$, and hence not $\sup_{y\in V}\frac{\|Ty\|}{\|y\|}$ $\endgroup$
    – user67803
    Sep 24, 2013 at 10:53

3 Answers 3


Here's how I'd prove this. Since $T$ is continuous, in particular there is a $\delta>0$ such that $\Vert T x \Vert < 1$ whenever $\Vert x \Vert < \delta$. For any $y \in V$ let $x(y) = \frac \delta {2 \Vert y \Vert} y$ - then $\Vert x(y) \Vert = \delta/2 < \delta$ so $\Vert T (x(y)) \Vert < 1$. Furthermore since $T$ is linear, $$\Vert T(x(y))\Vert = \frac \delta {2 \Vert y \Vert} \Vert Ty\Vert < 1,$$ and thus $$\sup_{y\in V} \frac {\Vert T y \Vert}{\Vert y \Vert}\le \frac 2 \delta<\infty.$$ I can't help you with interpreting Kreyszig's proof without having the whole thing in front of me.

  • $\begingroup$ Nice proof. But could you tell me why "Since $T$ is continuous, in particular there is a $\delta > 0$ s.t. $||Tx|| < 1$" is true? Would really help me. $\endgroup$
    – psyph
    Sep 10, 2019 at 22:05
  • 1
    $\begingroup$ @psyph this is exactly the ε-δ definition of continuity (at the point $T(0)=0$) with ε=1. $\endgroup$ Sep 10, 2019 at 22:09
  • $\begingroup$ I actually didn't see that, thank you. $\endgroup$
    – psyph
    Sep 10, 2019 at 22:13

Here's how I prove this:

Take $x_0 = 0 $ in the definition of continuity. Then for any $\epsilon > 0$ we have $\delta' > 0$ such that

$\Vert x \Vert < \delta' \Longrightarrow \Vert T(x) \Vert < \epsilon; \tag 1$

furthermore we may choose $\delta$, $0 < \delta < \delta'$, and then

$\Vert x \Vert = \delta \Longrightarrow \Vert T(x) \Vert < \epsilon = \delta^{-1}\epsilon \delta = \delta^{-1} \epsilon \Vert x \Vert; \tag{1.1}$

for any $0 \ne z \in V$ we may choose a real $r > 0$ such that

$\Vert rz \Vert = r\Vert z \Vert =\delta, \tag 2$

simply take

$r = \delta \Vert z \Vert^{-1}; \tag 3$

then by (1.1) and (2),

$r \Vert T(z) \Vert = \Vert rT(z) \Vert = \Vert T(rz) \Vert < (\delta^{-1}\epsilon) \Vert rz \Vert = r(\delta^{-1}\epsilon) \Vert z \Vert; \tag 4$

dividing by $r$ yields

$\Vert T(z) \Vert < (\delta^{-1}\epsilon) \Vert z \Vert, \tag 5$

holding for all $z \ne 0$; to handle the case $z = 0$ we simply replace "$<$" by "$\le$" in (5), so

$\Vert T(z) \Vert \le (\delta^{-1}\epsilon) \Vert z \Vert, \; \forall z \in V, \tag 6$

which shows that $T$ is bounded with bound no greater than $\delta^{-1}\epsilon$.

Kreyszig's assumption that $x = x_0 + \delta \dfrac{y}{\Vert y \Vert}$ does not impose an artificial restriction on $x$, since the linearity of $T$ allows us to re-scale $x - x_0$ (as is done above with $z$), which addresses any restriction imposed.

  • 1
    $\begingroup$ There is a mistake in the proof. Inequality $(\delta^{-1} \epsilon) \delta <(\delta^{-1} \epsilon) ||x||$ in (1) is equivalent to $\delta<||x||$ which isn't true. $\endgroup$ Oct 27, 2020 at 7:21

In the proof, at last it was shown $\|T(y)\|$ is bounded for any $y$ in $V$. So it depends on y, in this case $x$ is just a parameter, so it does not come into the picture.


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