# Linear transformation $T$ such that for every extension $\overline{T}$, $\|\overline{T}\|>\|T\|$.

Let $E$ and $F$ be normed spaces such that $\dim F < \infty$, $G$ a subspace of $E$ and $T:G\rightarrow F$ a continuous linear map. I know that there exists a continuous linear extension $\overline{T}:E\rightarrow F$. Also, if $E$ is a Hilbert space, then $\overline{T}$ can be chosen in the way that $\|\overline{T}\|=\|T\|$.

Problem: Find an example of $E$, $F$, $G$ and $T$ (like above) such that every continuous linear extension $\overline{T}$ has a greater norm, i.e. $\|\overline{T}\|>\|T\|$.

Now, $F$ must be at least a 2-dimensional space, otherwise I could use Hahn-Banach to find an extension with equal norm. My professor told me it could be done with $E$ of finite dimension. Of course, I tried to come up with an example of $E$ with a norm that doesn't satisfy the parallelogram law. For example, $E=\left(\mathbb{R}^3,\|\cdot\|_{\infty}\right)$ and $F=\left(\mathbb{R}^2,\|\cdot\|_1\right)$. But I couldn't prove that it works with any example I tried using those spaces.

Can somebody help me to find an example and assure that it really has that property?

EDIT: Apparently, it can't be done with $E$ of finite dimension nor with $F$ equipped with the $\sup$ norm, as @Hamza proved below.

• "otherwise I could use Hahn-Banach to find an extension with equal norm" ---> Are you assuming that $T$ is not $0$? Jun 24 '16 at 22:07
• @ajotatxe I'm afraid i'm not. I don't know how it helps, but I'm sure not looking for trivial examples, if that's the case (i'm still a newbie in functional analysis). =) Jun 24 '16 at 22:15
• @JoãoVictorBateliRomão You probably should start by looking for trivial examples :)
– Neal
Jun 24 '16 at 22:22
• @ajotatxe But using $T=0$, then $\overline{T} = 0$ is a counterexample of an extension, right? Jun 24 '16 at 22:43
• @Neal I'm really having difficulty with this problem. I couldn't find any trivial examples at all. I can't even prove that this example exists. I going to start a bounty. Jun 27 '16 at 3:12

Let $$E=\mathbb R^3$$, equipped with the sup norm, namely $$\|(x,y,z)\|_\infty = \max\{|x|, |y|, |z| \},$$ and let $$F=G=\{(x, y, z)\in E : x+y+z=0\},$$ equipped with the induced norm from $$E$$. Also let $$T:G\to F$$ be the identity function. Then clearly $$\|T\|=1$$, but any extension $$\bar T:E\to F$$ has norm strictly bigger than one. The reason is that any such $$\bar T$$ is necessarily a projection from $$E$$ to $$G$$ and there is no such projection with norm $$1$$. The best way to convince oneself of this fact is to make a cardboard model of this cube

cut it along the red line, place one of the two halves on top of the table with the red hexagon down, and attempt to shine a flashlight so that the shadow is restricted to within the hexagon. It is impossible!

This is based on another answer I recently gave to this question.

• You made those diagrams yourself! Frankly , magnificent stuff from you in both answers, +1 Feb 17 '21 at 9:46
• Thank you @Teresa!
– Ruy
Feb 18 '21 at 21:40

If $G$ have a topological complimentary in $E$ (in particular if $E$ is finite dimensional space) i don't think that such construction can be done, because prolonging by $0$ on the complementary space, will have the same norme : $$\|\hat{T}\|=\sup_{\begin{array}{c} x\in E \\ x\neq 0\end{array}} \frac{\|\hat{T}x\|}{\|x\|}=\sup_{\begin{array}{c} x=z+y \\ y,z\neq 0\end{array}} \frac{\|\hat{T}z+\hat{T}y\|}{\|z+y\|}=\sup_{\begin{array}{c} z\in G\\y\in E-G \\ y,z\neq 0\end{array}} \frac{\|Tz\|}{\|z+y\|}=\sup_{\begin{array}{c} z\neq 0\end{array}} \frac{\|Tz\|}{\|z\|}=\|T\|$$ If now $E$ is infinite dimensional and $G$ a closed subspace but with no complaiment in $E$, let for example : $$\begin{array}{} I&:& c_0& \to &l_\infty\\ & & x &\mapsto & x \end{array}$$ so by a famous Phillips's lemma $I$ can't be extended to a continuous linear maps in $l_\infty$.

but if $F$ is finite dimensional space i don't have an explicit example but as remarks you can't take the infinite norm as norm to $F$ in fact, let $T :G \to \mathbb{R}^n$, and let $(T_1,T_2,\dots,T_n)$ it composite then it's clear that $\|Tx\|=\sup_{i\leq n}|T_ix|$, so let $\hat{T}=(\hat{T_1},\dots,\hat{T}_n)$ be the preserving norm extension given by Hahn-Banach theorem : $$\|\hat{T}\|=\sup_{ x\in S_E } \|\hat{T}x\|=\sup_{ x\in S_E } (\sup_{i\leq n}\|\hat{T}_ix\|)=\sup_{i\leq n}(\sup_{ x\in S_E } \|\hat{T}_ix\|)=\sup_{i\leq n} \|T_i\|)=\|T\|$$ we call this propriety the injectivity of Banach space, and it prove that a finite dimensional space is injective if and only if he is isomorphe isometrically to $l^\infty_n$

• Interesting! Specially the fact that $I$ can't be extended to a continuous linear map. I have some notes of Functional Analysis with an exercise that I solved asking to prove that there is a version of Hahn-Banach Theorem for linear maps into $l_{\infty}$. I must have suppose that the subspace have a complement somewhere in the passages of my proof. Jul 9 '16 at 20:30
• for the case of $l_\infty$ it's not necessary to suppose that you have a complement but you can use the second remark about decomposition, but your decomposition will be infinite and the same reasoning will work in this case Jul 9 '16 at 20:43
• @João, I think Hamza's mistake is in the penultimate step of the first displayed calculation.
– Ruy
Feb 15 '21 at 19:54
• I think @Ruy is correct. Think to $E=\mathbb R^2$. If $G$ is the $X$-axis, and $L$ is any complent different from $Y$-axis, say the line with slope $(10000000,1)$, then the projection of $(0,1)$ on $G$ along $L$ has huge norm. So extending $T$ by setting it zero on $L$ does not work. Feb 17 '21 at 12:04