# An Upper Bound of $\left<H,X\right>_F$ with Constrainted Rank

We have $$X=[\mathbf{x}_1,\mathbf{x}_2...,\mathbf{x}_n]\in\mathbb{R}^{d\times n}$$, $$H=[\mathbf{h}_1,\mathbf{h}_2...,\mathbf{h}_n] \in\mathbb{R}^{d\times n}$$, and $$d. $$H$$ has rank $$r\leq d$$ and $$X$$ has rank $$d$$.

Assume we have $$\|H\|_F\leq C$$ and $$\|\mathbf{x}_i\|_2\leq R$$, where $$\|\cdot\|_F$$ and $$\|\cdot\|_2$$ are the Frobenius norm and vectro 2-norm respectively.

What do we know about the upper bound of Frobenius inner product $$\left_F$$ of $$H$$ and $$X$$?

Hint: I know by Cauchy-Schwarz inequality $$\left_F < \sqrt nCR$$. The equality will not hold because we can not guarantee $$\mathbf{x}_i = k\mathbf{h}_i, \forall i$$. This is because $$H$$ is rank deficient. I think there is a tighter upper bound for $$_F$$ in terms of singular values of $$X$$ and rank $$r$$ of $$H$$.

I have also posted the question here.

You can try the von Neuman trace inequality $$\langle H,X\rangle\equiv\mathrm{trace}(H^{T}X)\le\sum_{i=1}^{d}\sigma_{i}(H)\sigma_{i}(X)\equiv\langle\sigma(H),\sigma(X)\rangle$$ where $$\sigma_{i}(H)$$ is the $$i$$-th largest singular value of $$H$$ and $$\sigma(H)$$ is the length-$$d$$ vector of singular values in decreasing order. From here, you can apply the Hölder inequality $$\langle\sigma(H),\sigma(X)\rangle\le\|\sigma(H)\|_{1}\|\sigma(X)\|_{\infty}\equiv\sigma_{\max}(X)\;\sum_{i=1}^{d}\sigma_{i}(H).$$ If $$H$$ is rank-deficient, then we expect the Nuclear norm $$\sum_{i=1}^{d}\sigma_{i}(H)$$ to be small, and the bound to be relatively tight.