# Extension of equivalent norm in subspace to the whole space

I'm trying to prove the following result:

Let $$(X, \|\,\cdot\,\|_X)$$ be a normed space and $$Y$$ a subspace of $$X$$. If $$\|\,\cdot\,\|_{YY}$$ is an equivalent norm to $$\|\,\cdot\,\|_Y$$ (the inherited norm from $$X$$) in $$Y$$, then there exists a norm $$\|\,\cdot\,\|_{XX}$$ in $$X$$ which is equivalent to $$\|\,\cdot\,\|_X$$ and induces the norm $$\|\,\cdot\,\|_{YY}$$ into $$Y$$.

I've checked the proof in several books and post in this forum and it is esentially the following argument:

• As $$\|\,\cdot\,\|_Y$$ and $$\|\,\cdot\,\|_{YY}$$ are equivalent in $$Y$$, there exists a $$C>0$$ such that $$B_X \cap Y \subset C B_{YY}$$, where $$B_X,B_X \cap Y$$ and $$B_{YY}$$ are respectively the closed unit ball in $$X$$, the closed unit ball in $$Y$$ with the inherited norm, and the closed unit ball in $$Y$$ for the norm $$\|\,\cdot\,\|_{YY}$$.
• Defining $$B_{XX}=\text{conv}\left\{\dfrac{1}{C} B_X \cup B_{YY}\right\}$$ it is easy to see that it is convex, balanced, bounded and it has zero as an interior point. So, Minkowski's functional for this set is an equivalent norm in $$X$$, which we will call $$\|\,\cdot\,\|_{XX}$$, such that is closed unit ball is the closure of $$B_{XX}$$.
• Lastly, it is seen that $$B_{XX} \cap Y = B_{YY}$$ and from that, every proof concludes that the norm $$\|\,\cdot\,\|_{XX}$$ induces in $$Y$$ the norm $$\|\,\cdot\,\|_{YY}$$.

I understand every setp in the proof but there is something in the last step that I don't see and I don't know if it is a mistake. To prove that the norm defined by Minkowski's functional induces in $$Y$$ the desiered norm we should check that $$\text{cl}(B_{XX}) \cap Y = B_{YY}$$ Because $$\text{cl}(B_{XX})$$ is the closed unit ball in $$(X, \|\,\cdot\,\|_{XX})$$, not $$B_{XX}$$.

If $$B_{XX}$$ was a close set everything would be fine but I cannot prove it and, in fact, I don't see why it should be true in the case of infinite-dimensional spaces. Another option would be that $$B_{XX} \cap Y = \text{cl}(B_{XX}) \cap Y$$, but I cannot proof this (without assuming that $$B_{XX}$$ is closed of course).

In no proof of this fact is there any mention to this problem in the proof. Am I skipping something?

• Is $B$ the convex hull of a compact set? Sep 25 at 7:31
• Not in general, because if $X$ is infinite-dimensional then $B_X$ is not compact. Sep 25 at 7:36

[update: as I suspected this answer is more complicated than what you need (see Ryszard Szwarc‘s answer below for a simpler way to think about this), nonetheless here it is.]

We must show that $$\text{cl}(B_{XX}) \cap Y = B_{YY}$$. To this end, we suppose $$z_i\in B_{XX}$$ converges to $$y\in Y$$, and we will be done if we can show $$y\in B_{YY}$$.

Since $$\frac{1}{C}B_{X}$$ and $$B_{YY}$$ are convex, we can write each $$z_i$$ in the form $$z_i = t_i\frac{x_i}{C}+(1-t_i)y_i$$ for $$t_i\in [0,1]$$, $$x_i\in B_X$$, and $$y_i\in B_{YY}$$.

Passing to a subsequence if necessary (using compactness only of $$[0,1]$$), we may assume $$t_i\to t$$. It is easy to then check, by the boundedness of $$B_X$$ and $$B_{YY}$$, that this implies that $$t\frac{x_i}{C}+(1-t)y_i\to y\text{.}\tag{1}$$

Now we are done if $$t=0$$, so suppose otherwise. Then we must have $$\text{dist}_{X}(x_i,Y)\to 0$$ in order for the preceding sequence to converge. Thus we can write $$x_i=y_i'+e_i$$, with $$\|e_i\|_X\to 0$$ and $$y_i'\in Y$$. We may then upgrade (1) to $$z_i':= t\frac{y_i'}{C}+(1-t)y_i\to y\text{.}$$

But then since each $$x_i\in B_X$$, and $$e_i\to 0$$, we have $$\overline{\lim}_{i\to\infty} \|y_i'\|_X\leq 1$$, so that for each $$\epsilon>0$$, eventually $$\frac{y_i'}{C}\in \dfrac{(1+\epsilon)}{C}B_X\cap Y\subseteq (1+\epsilon)B_{YY}$$ for large enough $$i$$. Since $$y_i\in B_{YY}\subseteq (1+\epsilon)B_{YY}$$, and the ball is convex, we have $$z_i'\in(1+\epsilon) B_{YY}$$, and since the ball is closed and $$z_i'\to y$$, this means $$y\in (1+\epsilon)B_{YY}$$. Since $$\epsilon$$ is arbitrary, this completes the proof.

#### Remark

My knowledge of functional analysis is somewhat limited and very rusty, so I strongly suspect I'm either reinventing the wheel or doing a more convoluted argument than necessary, but in case no one else chimes in the above argument should be serviceable.

In my opinion the equality $$B_{XX}\cap Y=B_{YY}$$ suffices and there is no need to study the closure of $$B_{XX}.$$ Indeed, for $$y\in Y$$ we have $$\|y\|_{YY}=\inf\{r>0\,:\, y\in rB_{YY}\}=\inf\{r>0\,:\, y\in r[B_{XX}\cap Y]\}\\ =\inf\{r>0\,:\, y\in rB_{XX}\}=\|y\|_{XX}$$

It turns out that a stronger equality holds suggested in OP, namely $${\rm cl}\,(B_{XX})\cap Y=B_{YY}$$ A nice proof is presented in the answer by @MW .

• Much better, I had a nagging feeling I was being obtuse with my (now deleted) answer.
– M W
Sep 25 at 19:15
• @MW I suggest you undelete your answer, as you proved an interesting equality mentioned in OP. Sep 25 at 19:18