Linear functional $f$ is continuous at $x_0=0$ if and only if $f$ is continuous $\forall x\in X$?

by Daniel   Last Updated June 13, 2019 12:20 PM

Let $f$ be a linear functional on a normed space $(X, \|\cdot\|)$. Prove that $f$ is continuous at $x_0=0$ if and only if $f$ is continuous at every $x\in X$.

I understand that the $\Leftarrow$ is trivial but what about the other way?

Answers 4

Suppose $f$ is continuous at some $x_0\in X$ and let $x_n\to x^*$, for some $x^*\in X$. Then the sequence $x_n-x^*+x_0\to x_0$ and thus $\|f(x_n)-f(x^*)\|=\|f(x_n-x^*+x_0)-f(x_0)\|\to 0$. Hence $f$ is continuous at $x^*$ as well.

November 25, 2014 20:37 PM


$$\lVert T(x-x_0)\rVert=\lVert T(x)-T(x_0)\rVert$$

So if $\lVert x-x_0\rVert<\delta$ we have that $\lVert T(x)-T(x_0)\rVert=\lVert T(x-x_0)\rVert<\epsilon$ by continuity at $0$.

Adam Hughes
Adam Hughes
November 25, 2014 20:37 PM

if $f$ be linear function on a normed space $(X,\| \cdot\|)$ then these conditions are equivalent.

$(1)$ : $\|f\| < \infty $

$(2)$ : $f$ is uniformly continuous

$(3)$ : $f$ is continuous

$(4)$ : $f$ is continuous in $0$

$(1)$ $\to$ $(2)$ : $\forall x,y \in X$ and $\forall \epsilon> 0 $ $\|f(x)-f(y)\|$= $\|f(x-y)\|$ $\leq$ $\|f\|$ $\|x-y\|$, then if $\delta$ = $\frac{\| \epsilon\|}{\|f\| +1}$ we claim that $f$ is uniformly continuous

$(2)$ $\to$ $(3)$ : obvious

$(3)$ $\to$ $(4)$ : obvious

$(4)$ $\to$ $(1)$ : let $f$ be continuous in $0$. the there exist $\epsilon = 1$ and $\delta > 0$ such that if $\| u \|< \delta$ then $\|f(u)\|< 1$

$\|\frac{x}{\frac{2}{\delta}\|x\|}\|$ < $\delta$ $\longrightarrow$ $\frac{\|f(x)}{\|x\|}$= $\frac{\|f(u)}{\|u\|}$ < $\frac{1}{\frac{2}{\delta}}$= $\frac{\delta}{2}$ $\longrightarrow$ $\|f\|$ < $\infty$

erfan soheil
erfan soheil
November 27, 2014 20:40 PM


$\implies$ obviously


Assume L is continuous at $x_0 \in E$. We define sequence $x_1,x_2,\dotsm \in E$ and for any point $x\in E$ such that $||x_n-x ||\rightarrow 0$.

Since L is linear mapping, we have $$ ||L(x_n-x+x_0-x_0)||=||L(x_n-x+x_0)-L(x_0)|| $$ Since L is continuous at $x_0\in E$, for every sequence converging to $x_0$, the $L$ converges $L(x_0)$. We have $$ ||L(x_n-x+x_0)-L(x_0)||\rightarrow 0 $$ Q.E.D.

Zilong Cheng
Zilong Cheng
June 13, 2019 12:02 PM

Related Questions

Updated April 23, 2017 11:20 AM

Updated November 11, 2018 05:20 AM

Updated July 07, 2017 03:20 AM