Let $X,Y$ be a normed vector spaces over $\mathbb K $, $T:X\to Y$ a $\mathbb K$-linear continuous map ($\mathbb K$ could be $\mathbb R$ or $\mathbb C$).
Let's consider $ \hat T: X/\ker T \to Y$ the induced linear map ( $\hat T (\bar x)=T(x)$. This map it's well defined and it's clearly injective. We consider the usual quotient norm.
I want to prove that $\hat T$ it's also continuous and also $\|\hat T\|=\|T\|$.
I only proved that $\|T\|\le \|\hat T\|$ if I prove the other inequality, then I'm done with the continuity but I don't know how.
Best Answer
Everything follows easily from the
Lemma: Let $X$ a normed space, $N \subset X$ a closed subspace, and $\pi \colon X \to X/N$ the canonical projection. Let $U = \{ x \in X : \lVert x\rVert < 1\}$ and $U_N = \{ y \in X/N : \lVert y\rVert < 1\}$ the open unit balls in $X$ resp. $X/N$. Then $\pi(U) = U_N$.
Assuming the lemma, we have
$$\begin{align} \lVert \hat{T}\rVert &= \sup \{ \lVert \hat{T}(\xi)\rVert : \xi \in U_N\}\\ &= \sup \{\lVert \hat{T}(\xi)\rVert : \xi \in \pi(U)\}\\ &= \sup \{\lVert \hat{T}(\pi(x))\rVert : x \in U\}\\ &= \sup \{ \lVert T(x)\rVert : x \in U\}\\ &= \lVert T\rVert. \end{align}$$
To prove the lemma, recall that the norm on the quotient space is given by
$$\lVert \pi(x)\rVert := \inf \{ \lVert x+n\rVert : n \in N\}.$$
Thus, since $0 \in N$, we trivially have $\lVert \pi(x)\rVert \leqslant \lVert x\rVert$, whence $\pi(U) \subset U_N$.
Conversely, if $\xi = \pi(x) \in U_N$, let $\nu := \lVert \xi\rVert$. By the definition of the quotient norm, there is an $n_x \in N$ with $\lVert x+n_x\rVert < \frac{1+\nu}{2} < 1$. But $\xi = \pi(x) = \pi(x+n_x)$, and $x+n_x \in U$, so $\xi \in \pi(U)$.