Existence of Directional Derivative in Normed Linear Space

[Gear300]

Given a finite-dimensional normed linear space $(L,\lVert \cdot \rVert)$, is there anything that suggests that at every point $x_0 \in L$, there exists a direction $\delta \in L$ such that that $\lVert x_0 + t\delta \rVert \geqslant \lVert x_0 \rVert$ for all $t \in \mathbb{R}$?

 

[fresh_42]

Given a finite-dimensional normed linear space $(L,\lVert \cdot \rVert)$, is there anything that suggests that at every point $x_0 \in L$, there exists a direction $\delta \in L$ such that that $\lVert x_0 + t\delta \rVert \geqslant \lVert x_0 \rVert$ for all $t \in \mathbb{R}$?

No. $1=|1|=|2+ (-1)\cdot 1|=\lVert x_0 + t\delta \rVert < \lVert x_0 \rVert = |2|=2$

 

[Gear300]

No. $1=|1|=|2+ (-1)\cdot 1|=\lVert x_0 + t\delta \rVert < \lVert x_0 \rVert = |2|=2$

Point taken. But what if the dimension of the space is $n \geqslant 2$? I figured there should be $n - 1$ such directions. The reason I ask is because of the attached question. Finding a minimum matrix $B$ has been simple enough for p-like norms, but I haven't found one for arbitrary norms in general. Proving the posted statement felt it should suffice to prove the inquiry. If the norm was differentiable, then I could take directions orthogonal to the gradient and inspect the Hessian, but I suspect differentiability is not granted for all $x_0 \in L$, such as with the supremum norm $p = \infty$.

 

[fresh_42]

Point taken. But what if the dimension of the space is $n \geqslant 2$? I figured there should be $n - 1$ such directions.

Same example: $||(2,2) - (1,1)||= \sqrt{2} < 2\sqrt{2} = ||(2,2)||\,.$

The reason I ask is because of the attached question. Finding a minimum matrix $B$ has been simple enough for p-like norms, but I haven't found one for arbitrary norms in general. Proving the posted statement felt it should suffice to prove the inquiry. If the norm was differentiable, then I could take directions orthogonal to the gradient and inspect the Hessian, but I suspect differentiability is not granted for all $x_0 \in L$, such as with the supremum norm $p = \infty$.

The point is, that regular matrices form a dense subset, so the distance to another regular one is arbitrary small, but to find the next singular one is not as trivial. The singular matrices form a closed subset and the condition number says how they are distributed. As we measure a distance, this depends a lot on the norm.

 

[Gear300]

The point is, that regular matrices form a dense subset, so the distance to another regular one is arbitrary small, but to find the next singular one is not as trivial. The singular matrices form a closed subset and the condition number says how they are distributed. As we measure a distance, this depends a lot on the norm.

Might this be provable by inspecting the continuity of the determinant as a polynomial function? The zeros would then correspond to a closed subset of singular matrices. I'm just wondering how this is normally proved.

 

[fresh_42]

Might this be provable by inspecting the continuity of the determinant as a polynomial function?

Yes. The singular matrices are the preimage of the closed set $\{\,0\,\}$ under the determinant.

The zeros would then correspond to a closed subset of singular matrices. I'm just wondering how this is normally proved.

 

[Gear300]

Yes. The singular matrices are the preimage of the closed set $\{\,0\,\}$ under the determinant.

Alright then. You don't mind recommending a book on this ? The book I'm using here is Rainer Kress's Numerical Analysis, and I've gotten up to exactly that question with what I know about Banach spaces. So I wouldn't mind additional references.

 

[fresh_42]

So that's how it is normally proved? (Apologies for the persistence, but just being sure.)

Well, we need the determinant to be continuous, and that $\{\,0\,\} \subseteq \mathbb{R}$ is closed. Continuity is equivalent to "preimages of closed sets are closed". So $\{\,M\in \mathbb{M}_n(\mathbb{R})\,|\,\det M = 0\,\} \subseteq \mathbb{M}_n(\mathbb{R})$ is closed. This means at the same time, that $GL_n(\mathbb{R}) = \{\,M\in \mathbb{M}_n(\mathbb{R})\,|\,\det M \neq 0\,\}$ is open. Density might be dependent on the norm which is used, but usually if we change a few matrix entries by arbitrary small amounts, we will always get a regular matrix. Maybe it won't work with the discrete norm.

 

[Gear300]

Well, we need the determinant to be continuous, and that $\{\,0\,\} \subseteq \mathbb{R}$ is closed. Continuity is equivalent to "preimages of closed sets are closed". So $\{\,M\in \mathbb{M}_n(\mathbb{R})\,|\,\det M = 0\,\} \subseteq \mathbb{M}_n(\mathbb{R})$ is closed. This means at the same time, that $GL_n(\mathbb{R}) = \{\,M\in \mathbb{M}_n(\mathbb{R})\,|\,\det M \neq 0\,\}$ is open. Density might be dependent on the norm which is used, but usually if we change a few matrix entries by arbitrary small amounts, we will always get a regular matrix. Maybe it won't work with the discrete norm.

An $n \times n$ matrix lives in a linear space of dimension $n^2$. Since all finite-dimensional norms are equivalent, we can say that all matrix norms are equivalent to the Frobenius norm.

That aside, like you intimated, I think my assertion is true. Funny it took me so long, but for a norm function $p(x)$, we can consider the surface of constant norm $p(x_0)$. Since it is convex, we should always be able to come up with a tangent hyperplane of dimension $n-1$ that satisfies the desired properties, correct?

 

[fresh_42]

I still doubt that it works for all $t\in \mathbb{R}$, but I'm not sure. You claim, that each point is a norm minimum for at least one direction, but with arbitrary $t$, these are two opposite directions. I cannot really imagine such a situation except for $x_0=0$.

There is a separation theorem for hypersurfaces in Hilbert spaces IIRC, but we don't have a norm induced by an inner product. I might have thought too much of a metric, which is too weak.

 

[Gear300]

I still doubt that it works for all $t\in \mathbb{R}$, but I'm not sure. You claim, that each point is a norm minimum for at least one direction, but with arbitrary $t$, these are two opposite directions. I cannot really imagine such a situation except for $x_0=0$.

There is a separation theorem for hypersurfaces in Hilbert spaces IIRC, but we don't have a norm induced by an inner product. I might have thought too much of a metric, which is too weak.

So I did a recourse through Kolmogorov's Real Analysis text and found Problem 5.c on pg 141. The interior of the unit sphere in any normed space is an open convex set, so for all points on the surface, a tangent hyperplane should exist in the sense that it never touches the interior.

 

[WWGD]

No. $1=|1|=|2+ (-1)\cdot 1|=\lVert x_0 + t\delta \rVert < \lVert x_0 \rVert = |2|=2$

I don't understand, doesn't he op ask for _a_ direction and not for _every_ direction? Then you canonsider || 2+1(1)||=||3||>2?

 

[WWGD]

An $n \times n$ matrix lives in a linear space of dimension $n^2$. Since all finite-dimensional norms are equivalent, we can say that all matrix norms are equivalent to the Frobenius norm.

That aside, like you intimated, I think my assertion is true. Funny it took me so long, but for a norm function $p(x)$, we can consider the surface of constant norm $p(x_0)$. Since it is convex, we should always be able to come up with a tangent hyperplane of dimension $n-1$ that satisfies the desired properties, correct?

I think this is one of the versions of Hahn-Banach.