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Let $f:X \rightarrow \mathbb R$ be a continuous linear functional. Prove $f$ is bounded.

Since $f$ is continuous, $\forall \varepsilon >0$, there exists $\delta >0$ such that $|f(x)-f(y)|=|f(x-y)|=|f(z)|< \varepsilon$ whenever $|x-y|<\delta$. We let $z=x-y$.

Can we just now let $\varepsilon = C \|x\|_X$ for some $C>0$ and then it is bounded?

Maximilian Janisch
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snowman
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  • Your question does not make sense for me... $\varepsilon$ is an arbitrary positive number: you can't say "let's take $\varepsilon = \mbox{something}$". – Crostul Apr 15 '16 at 21:59
  • Im guessing you mean my answer doesn't make sense? Or is it the question? I copied it out correctly. I was pretty unsure about that... hence why i made the thread. – snowman Apr 15 '16 at 22:08
  • let $T : X \to Y$ two normed vector spaces. if $\sup_{|v|=1} |T v| = \infty$ there is a sequence $(v_n) \in X$ such that $|v_n| \to 0$ but $|T v_n| \not\to 0$ (proof : for a fixed $|u|$, $|T u|$ can be arbitrary large) – reuns Apr 15 '16 at 22:26

2 Answers2

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Since $f$ is continuous (at $0$), there is a neighbourhood $U$ of $0$ such that $f(U)\subset(-1,1)$. Choose $\delta>0$ such that $\{x\in X|\|x\|\leq\delta\}\subseteq U$. Then, if $x\in X$ is such that $\|x\|\leq \delta$, we have $x\in U$, and hence, $|f(x)|\leq 1$. Since $\|\frac{\delta x}{\|x\|}\|=\delta$, it follows that for all $x\in X$ we have $$ 1\geq\left|f\left(\frac{\delta x}{\|x\|}\right)\right|=\frac{\delta}{\|x\|}|f(x)|\implies|f(x)|\leq\frac{1}{\delta}\|x\|. $$ Therefore, $f$ is bounded.

ervx
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  • so are you basically saying that you made epsilon equal to 1? Since the continuity of f a zero means for all epsilon, there exists delta positive such that $|f(x)|<\varepsilon $ whenever $|x|<\delta $. So you just let epsilon equal 1? – snowman Apr 20 '16 at 13:47
  • Yes. The continuity of $f$ at $0$ tells us that $\forall\epsilon>0,\exists\delta>0$ such that $|x|<\delta\implies |f(x)|<\epsilon$. So, we are allowed to use $\epsilon=1$. – ervx Apr 20 '16 at 14:01
  • OK how does $|\frac{\delta x}{|x|}| = \delta$? – snowman Apr 20 '16 at 14:02
  • $|\frac{\delta x}{|x|}|=|\frac{\delta}{|x|}||x|=|\delta|\frac{|x|}{|x|}=\delta$. – ervx Apr 20 '16 at 14:05
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Your estimate holds only in a neighborhood of $x$. Hence, you need can not control whether you may choose such a $C$ independent of $x$. You need to rely on the linearity of your operator in the following way:

By continuity of $f$ in $0$, there exists $\delta>0$ such that for all $x\in X$ with $\Vert x \Vert_X <\delta$ holds

$$\vert f(x) - f(0) \vert \leq 1.$$

Using the linearity of $f$ we obtain

$$\vert f(x) \vert = \vert f(x) - 0 \vert =\vert f(x) - f(0) \vert \leq 1.$$

For $y\in X$, $y\neq0$ we get again by linearity of $f$

$$ \vert f(y) \vert = \frac{\Vert y \Vert_X}{\delta} \cdot \underbrace{\Vert f\left(\frac{\delta y}{\Vert y \Vert_X} \right) \Vert_X}_{\leq 1} \leq \frac{1}{\delta} \Vert y \Vert_X.$$

This is presicely the definition of bounded operator.

Severin Schraven
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