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I want to prove the following estimate (I'm not really sure if it really holds, but I'm interested in proving it):

$$(1,...,1)UDU^T\left(\begin{array}{c} 1 \\ \vdots\\1 \end{array}\right) < n$$

where $U\in\mathbb{R}^{n\times n}$ is an orthogonal matrix and $D\in\mathbb{R}^{n\times n}$ is a diagonal matrix whose diagonal elements $d_i$ are all in the interval $\lbrack 0, 1) $. I began the following way: Let $w:=U^T\left(\begin{array}{c} 1 \\ \vdots\\1 \end{array}\right)$ then we have $$(1,...,1)UDU^T\left(\begin{array}{c} 1 \\ \vdots\\1 \end{array}\right) = \langle w, Dw\rangle = \sum_{i=1}^n d_iw_i^2.$$ Further we have $w_i = \sum_{k=1}^nu_{k,i}$ where $U = (u_{i,j})_{i,j=1,...,n}$. I now want to exploit that the the columns of $U$ are orthonormal. Unfortunately I didn't find the right way to do so.

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As you have written, $$(1,...,1)UDU^T\left(\begin{array}{c} 1 \\ \vdots\\1 \end{array}\right) = \langle w, Dw\rangle = \sum_{i=1}^n d_iw_i^2.$$ Since $0\leq d_i<1$ for all $i$, we have $$\sum_{i=1}^n d_iw_i^2<\sum_{i=1}^n w_i^2=\|Ux\|^2$$ where $x=(1,\ldots,1)^T$. Since $U$ is orthogonal, it is an isometry, so $\|Ux\|=\|x\|=\sqrt{n}$. Combining all of these facts, we obtain $$(1,...,1)UDU^T\left(\begin{array}{c} 1 \\ \vdots\\1 \end{array}\right)= \sum_{i=1}^n d_iw_i^2<\|Ux\|^2=\|x\|^2=n.$$

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Since $d_i<1$ the matrix $I-D$ is positive definite, i.e. $D<I$, hence $$ \begin{pmatrix}1 & ... &1\end{pmatrix}UDU^T\begin{pmatrix}1 \\ \vdots\\1 \end{pmatrix}<\begin{pmatrix}1 & ... &1\end{pmatrix}UU^T\begin{pmatrix}1 \\ \vdots\\1 \end{pmatrix}=\begin{pmatrix}1 & ... &1\end{pmatrix}\begin{pmatrix}1 \\ \vdots\\1 \end{pmatrix}=n. $$

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