Exercise 12, Section 3.5 of Hoffman’s Linear Algebra 
Let $V$ be a finite-dimensional vector space over the field $F$ and let $W$ be a subspace of $V$. If $f$ is a linear
functional on $W$ , prove that there is a linear functional $g$ on $V$ such that $g(\alpha)= f(\alpha)$ for each $\alpha$ in the subspace W.

We can prove a stronger result:

Let $V$ be a finite-dimensional vector space over field $F$. If $W\leq V$ and $f\in L(W,Z)$, then $\exists g\in L(V,Z)$ such that $f(\alpha)=g(\alpha)$, $\forall \alpha \in W$.

My attempt: Let $\{\alpha_1,…,\alpha_m\}$ be basis of $W$. Let $\{\alpha_1,…,\alpha_m,\alpha_{m+1},…,\alpha_n\}$ be basis of $V$. By theorem 1 section 3.1, $\exists !g\in L(V, Z)$ such that $g(\alpha_i)=f(\alpha_i)$, $\forall i\in J_m$ and $g(\alpha_i)=0_Z$, $\forall i\in J_n\setminus J_m$. Let $\alpha \in W$. Since $\mathrm{span}(\{\alpha_1,…,\alpha_m\})=W$, we have $\alpha=\sum_{i\in J_m}a_i\cdot_V \alpha_i$. Since $f$, $g$ are linear map, we have $f(\alpha)$ $=f(\sum_{i\in J_m}a_i\cdot_V \alpha_i)$ $=\sum_{i\in J_m}a_i\cdot_Z f(\alpha_i)$ $= \sum_{i\in J_m}a_i\cdot_Z g(\alpha_i)$ $= g(\sum_{i\in J_m}a_i\cdot_V \alpha_i)$ $=g(\alpha)$. Hence $f(\alpha)=g(\alpha)$, $\forall \alpha \in W$. Is my proof correct?
 A: 
My attempt:


Let $\{\alpha_1,…,\alpha_m\}$ be basis of $W$. Let $\{\alpha_1,…,\alpha_m,\alpha_{m+1},…,\alpha_n\}$ be basis of $V$. By theorem 1 section 3.1, $\exists !g\in L(V, Z)$

What is $Z$?

such that $g(\alpha_i)=f(\alpha_i)$, $\forall i\in J_m$ and $g(\alpha_i)=0_F$, $\forall i\in J_n\setminus J_m$.

I'm not familiar with this use of $J_m$. Although one can guess that you mean $J_m=\{1,...,m\}$ and $J_n=\{1,...,n\}$, it might be considerate to take the trouble to define $J_n$ and $J_m$ somewhere.
Theorem 1 of section 3.1 states


*

*Theorem 1 Let $V$ be a finite-dimensonal vector space over the field $F$ and let $\{\alpha_1,..., \alpha_n\}$ be an ordered basis for $V$. Let $W$ be a vector space over the same field $F$ and let $\beta_, ..., \beta_n$ be any vectors in $W$. Then there is precisely one linear transformation $T$ from $V$ into $W$ such that $$T\alpha_j=\beta_j, j=1,...,n.
$$

so you seem to have decided to choose the $\beta_i$ in the above to be $f(\alpha_1),...,f(\alpha_m),0_F,...0_F$ in the vector space consisting of the scalar field, right?

Let $\alpha \in W$. Since $\mathrm{span}(\{\alpha_1,…,\alpha_m\})=W$, we have $\alpha=\sum_{i\in J_m}a_i\cdot_V \alpha_i$. Since $f$, $g$ are linear map, we have $f(\alpha)$ $=f(\sum_{i\in J_m}a_i\cdot_V \alpha_i)$ $=\sum_{i\in J_m}a_i\cdot_W f(\alpha_i)$ $= \sum_{i\in J_m}a_i\cdot_W g(\alpha_i)$ $= g(\sum_{i\in J_m}a_i\cdot_V \alpha_i)$ $=g(\alpha)$. Hence $f(\alpha)=g(\alpha)$, $\forall \alpha \in W$.

This is a reasonable argument. But since $\alpha \in W$, but $g$ operates over $V$, you might want to use the mappings to $O_F$ which we saw earlier somewhere in the above.
