If the restriction of two continuous linear functions are equal, then are these two functions equal? Let $X$ be a normed space and let $M$ be a linear subspace of $X$. Let $f,g\in X^*$ be such that $f|_M=g|_M$. Then can we get $f=g$? 
Or do we need $M$ being dense to get $f=g$? If so, how to prove it? 
Thank you in advance!
 A: Take $f,g\colon \Bbb R^2\to \Bbb R$ where $f(x,y)=xy$ and $g(x,y)=0$. Then $f|_{\Bbb R\times\{0\}}=g|_{\Bbb R\times\{0\}}$ since both are identically $0$, but $f\neq g$ since $f(1,1)\neq g(1,1)$.
A: The first statement is not true, as we can take any $f\in X^*$ nonzero, let $g=\lambda f$ for some nonunit scalar $\lambda$, and let $M=\ker(f)$.  
The second statement is true, and follows by continuity.  Suppose $M$ is dense, and $f=g$ on $M$.  Then $f-g=0$ on $M$, and by continuity $f-g=0$ on $X$.  To be more precise, fix $x\in X$ and $\varepsilon>0$.  Since $f-g$ is continuous, there is some $\delta>0$ such that $|(f-g)(x-y)|<\varepsilon$ whenever $\|x-y\|<\delta$.  Since $M$ is dense, there is some $y\in M$ with $\|x-y\|<\delta$, and thus $|(f-g)(x)|=|(f-g)(x-y)|<\varepsilon$.  Since $\varepsilon>0$ was arbitrary, $(f-g)(x)=0$, and since $x\in X$ was arbitrary, $f-g=0$.
A: Consider the normed space $(C[-1,1], \|\cdot\|_\infty)$ and two functionals $F,G \in C[-1,1]^*$ given by
$$F(f) = \int_{-1}^1 f(x)\,dx, \qquad \quad G(f) = 2\int_{0}^1 f(x)\,dx$$
for all $f \in C[-1,1]$.
Then $F \ne G$ but $F$ and $G$ coincide on the subspace of even continuous functions on $[-1,1]$.
