I am having trouble proving the following result, which I want to do using the most elementary machinery (i.e., no Banach-Hahn theorem or Baire's or other functional analysis stuff).
Suppose $\mathrm{E}$ is a (real) normed space. We say that $\varphi$ is a linear form on $\mathrm{E}$ if it is scalar-valued and linear.
Suppose $x$ is a vector in $\mathrm{E}$ obeying the following relation: for every continuous lineal functional $\varphi,$ the image of $x$ is zero, that is $\varphi(x) = 0.$ Does it follow necessarily $x = 0$? I do not know any structure in $\mathrm{E}$ except it is normed (proving it for inner-product spaces is easy, we take $\varphi(z) = (x|z)$ "dot product against $x$"). Obviously, one would like to prove that the "projection onto $x$" is a continuous linear form; I explain this further. We assume $x \neq 0$ and extend it to a basis of $\mathrm{E},$ say $\mathrm{B}.$ Then, $$\mathrm{E} = \Bbb R x + \sum_{\substack{y \in \mathrm{B} \\ y \neq x}} \Bbb R y$$ the sum being direct. Every element $z$ in $\mathrm{E}$ can then be written uniquely as $\alpha_z x + \sum\limits_{y \in \mathrm{F}} \beta_y y,$ where $\alpha_z \in \Bbb R,$ $\mathrm{F}$ is finite and the $\beta_y$ are non-zero. One defines the projection to be $z \mapsto \alpha_z.$ The problem reduces to show that if it is true in general that this projection is continuous.