Inner product space $\langle v, x \rangle = \langle w, x \rangle \implies v = w$ on total set

Question

Let $M$ be a total subset of an inner product space $X$. Show that $\forall x \in M$: $$ \langle v, x \rangle = \langle w, x \rangle \implies v = w $$

My idea was to calculate $$ \langle v, x \rangle - \langle w, x \rangle = \langle v - w, x \rangle = 0 $$ I know that no vector $a \in X$ can be perpendicular to all $x \in M$. So therefore $\exists x \in M$ such that $v-w \not\perp x$. So in order for this inner product to be $0$ $\forall x \in M$, one of the operands must be $0$. And $x$ can't be it since it is part of a total set. $$ \implies v = w $$ Is this proof correct? It seems too easy to be true

Answer 1

Since $M$ is a total subset, the closure of $\text{span} (M)$ is the whole space $X$. Now let $(u_n)$ be a sequence of $\text{span}(M)$ converging to $v-w$. For all $n$, we have

$$ \langle v-w, u_n \rangle = 0 $$

by linearity, and the inner product is of course continuous so letting $n$ goes to infinity we get

$$ \|v - w\|^2 = \langle v - w, v - w\rangle = 0 $$

and $v = w$.

Answer 2

Observing that the equality $\langle v, x \rangle - \langle w, x \rangle = \langle v - w, x \rangle = 0$ holds for all $x\in M$, we deduce that $v-w$ belongs to the orthogonal complement $M^\perp$. Given that $M$ is a total subset, it follows that the closure of the span of $M$, denoted as $\overline{\text{span}(M)}$, equals $X$. This, in turn, implies that the orthogonal complement $M^\perp$ consists solely of the zero vector, i.e., $M^\perp=\{0\}$. Consequently, the fact that $v-w\in\{0\}$ leads us to the conclusion that $v$ equals $w$.