Subset of Banach space is bounded

Question

I am currently having problems understanding the following problem:

Let X be a Banach space and M be a subset of X such that $$\forall l \in X' \exists c \gt 0: \sup_{m\in M}l(m) \le c$$ Show that M is bounded. (X' is the dual space of X)

My approach would have simply been to assume M is not bounded. Then, we can find a sequence $(x_n)\subseteq M$ with $\Vert x_n\Vert \ge n$. Now, let $l:X\to \mathbb{R}, l(x)=\Vert x\Vert$. Then, $l(x_n)\ge n$. Thus, $\sup_{m\in M}l(m) = \infty$.

But this solution seems to be too easy. Also, I did not even use the fact that X is a Banach space. Where did I make a mistake? How should I approach this problem correctly?

Answer 1

Hint: for every $m\in M$, you can consider the linear functional $\Lambda_m \colon X' \to \mathbb{R}$ defined by $\Lambda_m ( l ) := l(m)$. Clearly $\Lambda_m \in X''$ and $\|\Lambda_m\| = \|m\|$ for every $m$.

Now you can try to use the Uniform Boundedness Principle on the family $(\Lambda_m)_{m\in M}$.