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Can you check if my proof is correct?

Let $X$ be a normed linear space. Prove that if $f(x)=f(y)$ for all $f\in X^{*},$ then $x=y$

Let $f\in X^{*}$, then $f$ is a bounded linear functional. Assume that $x,\in X$ such that \begin{align}f(x)=f(y)&\iff f(x)-f(y)=0, \\& \iff f(x-y)=0, \;\text{since}\;f \;\text{is a linear functional}\;\\& \iff x-y\in \ker f =\{0\}\\& \iff x=y\end{align}

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    $\begingroup$ Does $X^*$ stand for the set of bounded linear functionals, or for the set of all linear functionals? And do you really believe $\ker f = \{0\}$ when $f\colon X\to\Bbb C$ (or $\Bbb R$)? $\endgroup$ – Ted Shifrin Jan 10 at 19:25
  • $\begingroup$ @Ted Shifrin: $X^{*}$ stands for the set of bounded linear functionals $\endgroup$ – Omojola Micheal Jan 10 at 19:27
  • $\begingroup$ @Ted Shifrin: However, I am not certain if $\ker f=\{0\}$ when $f:X\to \Bbb{R}.$ What do you think? $\endgroup$ – Omojola Micheal Jan 10 at 19:29
  • $\begingroup$ LOL, What if $X=\Bbb R^n$? What's the kernel then? $\endgroup$ – Ted Shifrin Jan 10 at 19:29
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    $\begingroup$ Do you know the nullity-rank theorem? ... The key thing you're not using here is that if holds for all $f\in X^*$. Try this in $\Bbb R^n$: Suppose $x\cdot v = 0$ for all $v\in\Bbb R^n$. Why must $x=0$? $\endgroup$ – Ted Shifrin Jan 10 at 19:31
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The statement is true, but the proof is not. In the proof, it is implicitly assumed that if $f\in X^*$, then $\ker f=\{0\}$, which is not true in general (think of the zero functional).

A correct proof can be constructed using the Hahn–Banach theorem. In particular, if $x-y\neq 0$, then there exists some $f\in X^*$ such that $f(x-y)=\|x-y\|\neq 0$, so that $f(x)\neq f(y)$. For details, see Theorem 5.8(b) in Folland (1999, p. 159).

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  • $\begingroup$ Thanks for your quick response. I'll read through (+1) $\endgroup$ – Omojola Micheal Jan 10 at 19:38
  • $\begingroup$ I can't access the book. Is there anyway of getting it? Or can you share with me on my email? $\endgroup$ – Omojola Micheal Jan 10 at 19:48
  • $\begingroup$ The relevant portions may be accessible as a preview through Google Books or Amazon. There is also a wide variety of electronic resources available free of charge. See, for example, Proposition 6.5 in this handout. $\endgroup$ – triple_sec Jan 10 at 20:05
  • $\begingroup$ Yes, that's true! $\endgroup$ – Omojola Micheal Jan 10 at 20:10
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Your proof is not correct, because, unles $\dim X\leqslant1$, $\ker f$ cannot possibly be $\{0\}$.

Let $z=x-y$ and let $Z$ be the vector space spanned by $z$. Consider the linear map$$\begin{array}{rccc}g\colon&Z&\longrightarrow&\mathbb R\\&\lambda z&\mapsto&\lambda.\end{array}$$Then $g$ is bounded and therefore, by the Hahn-Banach theorem, you can extend it to an element $f\in X^*$. But\begin{align}f(x)-f(y)&=f(x-y)\\&=f(z)\\&=g(z)\\&=1\\&\neq0.\end{align}

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  • $\begingroup$ Thanks for your quick response. (+1) $\endgroup$ – Omojola Micheal Jan 10 at 19:37

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