# Bounded linear operator on Banach spaces determines a bounded linear functional on its range.

I am trying to prove the following proposition, but so far yielding only unsatisfactory results. Anyone got any ideas?

Let $$X$$ and $$Y$$ be Banach spaces and define a bounded linear operator $$T:X\to Y$$whose range $$T(X)$$ is closed $$Y$$. Then, assume $$\varphi:X\to\mathbb{C}$$ is a bounded linear functional on $$X$$, such that $$Tx=0\implies \varphi(x)=0 \text{ for }x\in X.$$

Then the assignment $$\psi(Tx)=\varphi(x),\:x\in X$$ is a bounded linear functional on $$T(X)$$.

The set $$T(X)$$ is closed in $$Y$$ that is a Banach Space so $$T(X)$$ is a Banach Space.

Your map is well defined because for every $$y\in T(X)$$ such that $$y=T(x_1)$$ and $$y=T(x_2)$$ you have that

$$T(x_1-x_2)=T(x_1)-T(x_2)=y-y=0$$

and so

$$\phi(x_1-x_2)=0$$ then $$\phi(x_1)=\phi(x_2)$$

Oviously your map is linear on $$T(X)$$.

If you want prove that is also bounded you can see that it is continuos on $$T(X)$$.

I think that your condition must be that if $$T(x_n)\to 0$$ then $$\phi(x_n)\to 0$$

Let $$(y_n=T(x_n))_n$$ a sequence convergent to $$y=T(x)$$ in $$T(X)$$. Then

$$T(x-x_n)\to 0$$

and so

$$\phi(x-x_n)\to 0$$ and $$\phi(x_n)\to \phi(x)$$

Due to the open mapping theorem, $$T(X)$$ is isomorphic to the quotient $$X/L$$ where $$L$$ is the kernel of $$T$$. The universal property of quotients (in the category of Banach spaces) says that a continuous linear $$S:X\to Z$$ factorizes continuously over the quotient if and only if it vanishes on $$L$$.

If you don't like this argument you can proceed more directly to see the continuity of the induced map (using that the qutient map $$X\to X/L$$ maps the unit ball of $$X$$ onto the unit ball of the quotient).