If a norm on a vector space is given by an inner product then we have $\lVert x + y\rVert = \lVert x \rVert + \lVert y \rVert$ only when $x=\lambda y$ for some constant $\lambda$ (this follows from the Cauchy-Schwarz inequality). This does not hold for general (unitary invariant) norms. E.g. when considering generalised Ky Fan norms: $$ \lVert x\rVert_s = \sum_i s_i \lvert x\rvert_i^\downarrow$$ where $s_i\geq 0$ and $\lvert x\rvert^\downarrow$ is $\lvert x\rvert$ rearranged with the components from biggest to smallest. So for example the operator (sup) norm has $s_1 = 1$ and $s_i=0$ for $i>1$, while the trace norm has $s_i=1$ for all $i$. For these norms it is in general possible to find $x$ and $y$ so that the norm is additive for these vectors, while $x$ and $y$ are not multiples of each other.

Now my question is the following. Given a unitarily invariant norm and vectors $x$ and $y$ such that $\lVert x + y\rVert = \lVert x\rVert + \lVert y \rVert$, can we then conclude that $\lVert tx + (1-t)y\rVert = t\lVert x\rVert + (1-t)\lVert y \rVert$ for $0\leq t \leq 1$?

This is true for these generalised Ky Fan norms and for all norms that come from inner products. It is also true for all $p$-norms as this additivity still only holds for $x$ and $y$ that are multiples of each other (even though $p$-norms don't come from an inner product).

My intuition behind this is that this additivity property (when $x$ and $y$ aren't multiples of each other) forces the norm to be "linear" in a certain way, such that the norm has to be very similar to a generalised Ky Fan norm. So as a bonus question: If there are $x$ and $y$ that aren't multiples of each other, such that the norm is additive for these $x$ and $y$, show that the norm is a generalised Ky Fan norm. Note that this can only be true when we assume the norm to be unitary invariant since otherwise we can construct a counter-example of this (take $(X_1,\lVert\cdot\rVert_1)\oplus (X_2,\lVert\cdot\rvert_2)$ with the norm $\sup\{\lVert\cdot\rVert_1,\lVert\cdot\rVert_2\}$ where $\lVert\cdot\rVert_1$ is a Ky Fan norm and $\lVert\cdot\rVert_2$ is some other norm).


1 Answer 1


The conclusion holds for any vector norm. If $\lVert x + y\rVert = \lVert x\rVert + \lVert y \rVert$ then for $0 \le t \le 1$: $$ \lVert x\rVert + \lVert y \rVert = \lVert x + y\rVert \\ = \lVert (1-t)x + tx + (1-t)y + ty\rVert \\ \le (1-t)\lVert x \rVert + \lVert tx + (1-t)y \rVert + t \lVert y\Vert \\ \stackrel{(*)}{\le} (1-t)\lVert x \rVert + t \lVert x \rVert + (1-t)\lVert y \rVert + t\lVert y\rVert \\ = \lVert x\rVert + \lVert y \rVert $$ so that equality holds everywhere, in particular at $(*)$: $$ \lVert tx + (1-t)y\rVert = t\lVert x\rVert + (1-t)\lVert y \rVert $$

It follows even that for all $s, t > 0$ $$ \lVert sx + ty\rVert = s\lVert x\rVert + t\lVert y \rVert \, , $$ as can be seen by replacing $t$ by $\frac{s}{s+t}$ in the previous identity.

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    $\begingroup$ Interestingly, by homogeneity, this is also true for the whole positive cone spanned by $x$ and $y$. $\endgroup$
    – thomas
    Commented Jun 28, 2018 at 16:39
  • $\begingroup$ Thank you, Martin! $\endgroup$
    – Botond
    Commented Oct 29, 2019 at 13:07

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