Recently I was reading a proof of the following proposition,

Any two norms on a finite dimensional normed linear space are equivalent.

To prove this result the author used the following result without proof,

Result. Let $X$ be a finite dimensional normed linear space of dimension $n$. Suppose that $\mathcal{B}:=\{u^{(1)},\ldots,u^{(n)}\}$ be a basis of $X$ and $x\in X$. Furthermore, suppose that $x=\alpha_1u^{(1)}+\ldots+\alpha_nu^{(n)}$. If $\lVert\cdot\rVert$ be any norm on $X$ then prove that, $$\lVert x\rVert\ge d\left(\alpha_ju^{(j)},X_j\right)$$ for all $j\in \{1,\ldots,n\}$ where, $$X_j=\operatorname{span}\{u_i:i\ne j\}$$and $$d\left(\alpha_ju^{(j)},X_j\right)=\inf\{\lVert \alpha_ju^{(j)}-z\rVert:z\in X_j\}$$

I tried to prove this result but couldn't. Can anyone give me a proof of the above theorem?


Observe that clearly $z=-(x - \alpha_ju^{(j)})\in X_j$, and $$\|\alpha_ju^{(j)} - z\| = \| x\|$$ which by the definition of the infimum and the distance means that $\|x\|\geq d(\alpha_ju^{(j)},X_j)$ :)

  • $\begingroup$ I don't understand why $z=-(x-\alpha_ju^{(j)})$. Can you elaborate it? $\endgroup$ – user 170039 Apr 14 '17 at 8:07
  • $\begingroup$ This is just the definition of $z$; I'm picking this particular $z$ to show the inequality you want $\endgroup$ – amakelov Apr 14 '17 at 8:08
  • $\begingroup$ I don't understand. Your equation implies that $z=-(a_1u^{(1)}+\ldots+a_{j-1}u^{(j-1)}+a_{j+1}u^{(j+1)}+\ldots+a_{n}u^{(n)})$. Why should $z$ be of this particular form? $\endgroup$ – user 170039 Apr 14 '17 at 8:15
  • $\begingroup$ It is because I chose it this way. If that helps you can just forget about $z$ and think about the vector $x-\alpha_ju^{(j)}$. This vector is an element of $X_j$, which implies that the distance between this vector and $\alpha_ju^{(j)}$ is lower bounded by the distance between $\alpha_ju^{(j)}$ and the whole subspace $X_j$ $\endgroup$ – amakelov Apr 14 '17 at 8:17
  • $\begingroup$ Got it, thanks. $\endgroup$ – user 170039 Apr 14 '17 at 8:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.