# Prove that if $f(x)=f(y)$ for all $f\in X^{*},$ then $x=y$

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}

• 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 $\Bbb R$)? – Ted Shifrin Jan 10 at 19:25
• @Ted Shifrin: $X^{*}$ stands for the set of bounded linear functionals – Omojola Micheal Jan 10 at 19:27
• @Ted Shifrin: However, I am not certain if $\ker f=\{0\}$ when $f:X\to \Bbb{R}.$ What do you think? – Omojola Micheal Jan 10 at 19:29
• LOL, What if $X=\Bbb R^n$? What's the kernel then? – Ted Shifrin Jan 10 at 19:29
• 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 $x=0$? – Ted Shifrin Jan 10 at 19:31

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).

• Thanks for your quick response. I'll read through (+1) – Omojola Micheal Jan 10 at 19:38
• I can't access the book. Is there anyway of getting it? Or can you share with me on my email? – Omojola Micheal Jan 10 at 19:48
• 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. – triple_sec Jan 10 at 20:05
• Yes, that's true! – Omojola Micheal Jan 10 at 20:10

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}

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