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?

Thanks for the answers.

  • $\begingroup$ Is $B$ the convex hull of a compact set? $\endgroup$
    – AlvinL
    Sep 25 at 7:31
  • 1
    $\begingroup$ Not in general, because if $X$ is infinite-dimensional then $B_X$ is not compact. $\endgroup$
    – Eparoh
    Sep 25 at 7:36

2 Answers 2


[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.


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 .

  • $\begingroup$ Much better, I had a nagging feeling I was being obtuse with my (now deleted) answer. $\endgroup$
    – M W
    Sep 25 at 19:15
  • 2
    $\begingroup$ @MW I suggest you undelete your answer, as you proved an interesting equality mentioned in OP. $\endgroup$ Sep 25 at 19:18

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