Is this a necessary and sufficient condition of a continuous linear functional to be injective? Is it true that a continuous linear functional $f$ is injective if and only if there exist a constant $M\gt 0$, and for any given $x \in (X,||\cdot||)$,  such that $|f(x)|\ge M||x||$?
I know it is obviously from the right hand side we can get the left hand side.
But I want to know is the opposite direction still true？
Many thanks for your help.
 A: No, it's not true. 
Take the space $\mathcal{l}^1(\mathbb{R})$, the space of the absolutely summable sequences, with the norm $\| u\|_1 = \sum_{n=0}^\infty |u_n|$
Now define the operator $T: l^1 \to l^1$ by 
$T(u)_k = \frac{u_k}{k+1}$
It's injective, but there is no $M$ such that $\forall u \in l^1, \|T(u)\|_1 \geq M\|u\|_1$. Indeed, let's define $u^{(n)}$ by $u^{(n)}_k = 1$ if $k=n$, $0$ if $k\neq n$.
Then we have $\| u^{(n)}\|_1 = 1$, but $\| T(u^{(n)}) \|_1 = \frac{1}{n+1}$
So if there was such an $M$, it would need to be lower or equal to $\frac{1}{n+1}$ for all $n$. ie. equal to $0$
A: My other answer show that it's not true in the general case. But what if $X$ is finite dimensional? 
In this case, it's true. A possible argument :
Suppose there is no such $M$. Then there exist a sequence $(x_n) \in X$ with $\|x_n\| = 1$ such that $\| f(x_n) \| \to 0$
But, as $X$ is finite dimensional, the unit sphere of $X$ is compact, so we can extract a subsequence $x_{\sigma(n)}$ that converge in $X$ toward $x$.
And then $f(x) = f( \lim x_{\sigma(n)} ) = \lim f(x_{\sigma(n)}) = 0$ (by continuity of $f$)
But $f$ is injective, so $x=0$
But a sequence of elements with norm 1 cannot converge to zero : absurd
