# 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?

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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.

JohnD
November 25, 2014 20:37 PM

Consider

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

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 $0$.

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
November 27, 2014 20:40 PM

Proof.

$$\implies$$ obviously

$$\Longleftarrow$$

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
June 13, 2019 12:02 PM

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