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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 $T(X)$.

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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 $\phi(x_n)\to \phi(x)$

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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).

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