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.

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    $\begingroup$ Try to show that the projection $\pi \colon X \to X/\ker T$ maps the open unit ball of $X$ onto the open unit ball of $X/\ker T$. $\endgroup$ – Daniel Fischer Sep 12 '13 at 9:14
  • $\begingroup$ That is equivalent to prove $||\hat T||=||T||$ I don't know how $\endgroup$ – Shanks Sep 12 '13 at 9:19
  • $\begingroup$ Right, it's equivalent. But methinks easier to handle, since you know exactly what the quotient map does. How to prove it: recall first what the definition of the norm on $X/N$ is. $\endgroup$ – Daniel Fischer Sep 12 '13 at 9:22
  • $\begingroup$ That's the difficult part, prove that if $||x||\le 1$ then $||\pi (x)||=||\bar x|| \le 1$ $\endgroup$ – Shanks Sep 12 '13 at 9:27
  • $\begingroup$ That one is the easy part, that part is even true for the closed unit ball. What was the definition of the norm on the quotient again? $\endgroup$ – Daniel Fischer Sep 12 '13 at 9:34

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 $\xi \in \pi(U)$.

  • $\begingroup$ Why can't we just argue that $\pi: X \to X/K$ is continuous and linear and therefore so is $T \circ \pi$? $\endgroup$ – user167889 Feb 7 '15 at 0:55
  • $\begingroup$ We don't have $T\circ \pi$ here, we have $T = \hat{T}\circ \pi$. We know that $\pi$ and $\hat{T} \circ \pi$ are continuous, but that alone does not imply that $\hat{T}$ is continuous, we need some special properties of $\pi$ to conclude that. $\endgroup$ – Daniel Fischer Feb 7 '15 at 1:04
  • $\begingroup$ Thank you, I understand. But how does $0 \in N$ imply that $\|\pi(x)\|\le \|x\|$? $\endgroup$ – user167889 Feb 7 '15 at 2:19
  • $\begingroup$ Oh I see. Never mind my previous comment. It's the definition of the norm on the quotient. $\endgroup$ – user167889 Feb 7 '15 at 2:24
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    $\begingroup$ @student Pick $\varepsilon = \frac{1-\nu}{2}$, then $\nu + \varepsilon = \frac{1+\nu}{2}$. It's the argument of your last comment, but with a different $\delta$. $\endgroup$ – Daniel Fischer Feb 7 '15 at 12:54

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