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If $N(\bullet)$ is a matrix norm, it is unitarily invariant if and only if for any unitary matrices $U$, $V$, and for any matrix $A$ it satisfies: $$N(A)=N(UAV).$$

To say that they form a convex set, is to say that for any unitarily invariant matrix norm $N$, $M$, and for any $0\le \lambda\le 1$,$$P(A)=\lambda N(A)+(1-\lambda)M(A)$$ is a unitarily invariant norm.

It's easy to show that $P(A)\ge0$, $P(A)=0$ if and only if $A=0$, $P(cA)=|c|P(A)$ and $P(A+B)\le P(A)+P(B)$. The only property yet to prove is sub-multiplicative: $$P(AB)\le P(A)P(B).$$

To prove this, I notice that $$\begin{align}P(AB)&=\lambda N(AB)+(1-\lambda)M(AB)\\&\le\lambda N(A)N(B)+(1-\lambda)M(A)M(B)\end{align}$$and that $$\begin{align}P(A)P(B)&=\bigg(\lambda N(A)+(1-\lambda)M(A)\bigg)\bigg(\lambda N(B)+(1-\lambda)M(B)\bigg)\end{align}$$

but is it possible to prove that $$\lambda N(A)N(B)+(1-\lambda)M(A)M(B)\le \bigg(\lambda N(A)+(1-\lambda)M(A)\bigg)\bigg(\lambda N(B)+(1-\lambda)M(B)\bigg)$$?

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