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RBega2
RBega2
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Let $(V, \langle, \rangle)$ be an $n$ dimensional innerproduct space and let $S^k(V)$ denote the space of $k$-fold symmetric tensors. The inner product naturally extends to $S^k(V)$. Denote the associated norms by $\Vert \cdot \Vert_2$.

Let $S_1^k(V)=\{v\otimes\cdots \otimes v: v\in V, \}\subset $ be the set of rank one symmetric $k$-tensors. One can show that $S^k(V)=\mathrm{span}(S_1^k(V))$.

For $S\in S^k(V)$, set $$ \Vert S \Vert_*^2= \inf\{ \Vert S_1\Vert_2^2+\ldots+\Vert S_r \Vert_2^2: S=\sum_{i=1}^r S_i, S_i\in S_1^k(V)\}. $$$$ \Vert S \Vert_*= \inf\{ \Vert S_1\Vert_2+\ldots+\Vert S_r \Vert_2: S=\sum_{i=1}^r S_i, S_i\in S_1^k(V)\}. $$ Clearly, $\Vert \cdot \Vert_*$ is a norm on $S^k(V)$.

It's immediate that, when $k=1$, $\Vert S\Vert_2=\Vert S\Vert_*$. When $k=2$, you can use the spectral theorem to see that $\Vert S\Vert_*\leq \Vert S\Vert_2$ (and I suspect one has equality but didn't bother to check)$\Vert S\Vert_*\leq \sqrt{rank(S)}\Vert S\Vert_2\leq \sqrt{n} \Vert S\Vert_2$.

My question is whether the bound $\Vert S\Vert_*\leq \Vert S\Vert_2$$\Vert S\Vert_*\leq C(n) \Vert S\Vert_2$ continues to hold for $k\geq 3$. Or if there is at least a bound $$ \Vert S\Vert_* \leq C(n,k) \Vert S\Vert_2 $$ for some reasonably explicit $C(n,k)\geq 1$.

Let $(V, \langle, \rangle)$ be an $n$ dimensional innerproduct space and let $S^k(V)$ denote the space of $k$-fold symmetric tensors. The inner product naturally extends to $S^k(V)$. Denote the associated norms by $\Vert \cdot \Vert_2$.

Let $S_1^k(V)=\{v\otimes\cdots \otimes v: v\in V, \}\subset $ be the set of rank one symmetric $k$-tensors. One can show that $S^k(V)=\mathrm{span}(S_1^k(V))$.

For $S\in S^k(V)$, set $$ \Vert S \Vert_*^2= \inf\{ \Vert S_1\Vert_2^2+\ldots+\Vert S_r \Vert_2^2: S=\sum_{i=1}^r S_i, S_i\in S_1^k(V)\}. $$ Clearly, $\Vert \cdot \Vert_*$ is a norm on $S^k(V)$.

It's immediate that, when $k=1$, $\Vert S\Vert_2=\Vert S\Vert_*$. When $k=2$, you can use the spectral theorem to see that $\Vert S\Vert_*\leq \Vert S\Vert_2$ (and I suspect one has equality but didn't bother to check).

My question is whether the bound $\Vert S\Vert_*\leq \Vert S\Vert_2$ continues to hold for $k\geq 3$. Or if there is at least a bound $$ \Vert S\Vert_* \leq C(n,k) \Vert S\Vert_2 $$ for some reasonably explicit $C(n,k)\geq 1$.

Let $(V, \langle, \rangle)$ be an $n$ dimensional innerproduct space and let $S^k(V)$ denote the space of $k$-fold symmetric tensors. The inner product naturally extends to $S^k(V)$. Denote the associated norms by $\Vert \cdot \Vert_2$.

Let $S_1^k(V)=\{v\otimes\cdots \otimes v: v\in V, \}\subset $ be the set of rank one symmetric $k$-tensors. One can show that $S^k(V)=\mathrm{span}(S_1^k(V))$.

For $S\in S^k(V)$, set $$ \Vert S \Vert_*= \inf\{ \Vert S_1\Vert_2+\ldots+\Vert S_r \Vert_2: S=\sum_{i=1}^r S_i, S_i\in S_1^k(V)\}. $$ Clearly, $\Vert \cdot \Vert_*$ is a norm on $S^k(V)$.

It's immediate that, when $k=1$, $\Vert S\Vert_2=\Vert S\Vert_*$. When $k=2$, you can use the spectral theorem to see that $\Vert S\Vert_*\leq \sqrt{rank(S)}\Vert S\Vert_2\leq \sqrt{n} \Vert S\Vert_2$.

My question is whether the bound $\Vert S\Vert_*\leq C(n) \Vert S\Vert_2$ continues to hold for $k\geq 3$. Or if there is at least a bound $$ \Vert S\Vert_* \leq C(n,k) \Vert S\Vert_2 $$ for some reasonably explicit $C(n,k)\geq 1$.

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RBega2
RBega2
  • 2.5k
  • 1
  • 9
  • 18

An two-norm estimate for symmetric $k$-tensors

Let $(V, \langle, \rangle)$ be an $n$ dimensional innerproduct space and let $S^k(V)$ denote the space of $k$-fold symmetric tensors. The inner product naturally extends to $S^k(V)$. Denote the associated norms by $\Vert \cdot \Vert_2$.

Let $S_1^k(V)=\{v\otimes\cdots \otimes v: v\in V, \}\subset $ be the set of rank one symmetric $k$-tensors. One can show that $S^k(V)=\mathrm{span}(S_1^k(V))$.

For $S\in S^k(V)$, set $$ \Vert S \Vert_*^2= \inf\{ \Vert S_1\Vert_2^2+\ldots+\Vert S_r \Vert_2^2: S=\sum_{i=1}^r S_i, S_i\in S_1^k(V)\}. $$ Clearly, $\Vert \cdot \Vert_*$ is a norm on $S^k(V)$.

It's immediate that, when $k=1$, $\Vert S\Vert_2=\Vert S\Vert_*$. When $k=2$, you can use the spectral theorem to see that $\Vert S\Vert_*\leq \Vert S\Vert_2$ (and I suspect one has equality but didn't bother to check).

My question is whether the bound $\Vert S\Vert_*\leq \Vert S\Vert_2$ continues to hold for $k\geq 3$. Or if there is at least a bound $$ \Vert S\Vert_* \leq C(n,k) \Vert S\Vert_2 $$ for some reasonably explicit $C(n,k)\geq 1$.