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Let $A$ be real symmetric and $D$ shall contain the eigenvalues of $A$. I've learned that $\|A\|_{\text{max}}< \|D\|_{\text{max}}$, where $\|A\|_{\text{max}}$ means the Max norm.

I want to get a sharper bound $\|A\|_{\text{max}}$ by using the knwoledge about more than one eigenvalue, let say two. Ky-Fan norms seem appropriate, so I'm looking for something like $$ \|A\|_{\text{max}}\not <\frac12\|D\|_2, $$ where $\|D\|_2=\lambda_0+\lambda_1$ sums up the largest eigenvalues. Numerics showed that it doesn't hold, ven if I use absoulute values $|\lambda_0|+|\lambda_1|$.

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What you want is impossible, unless you put further restrictions on which $A$ are allowable. To see why, start with $A$ diagonal, so that $D=A$. In this example, your original inequality is sharp. Then any linear combination of the diagonal of $D$ with coefficients less than $1$ in absolute value will fail your desired inequality.

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Sounds a little like Luke Skywalker: 'You want the impossible.' Anyway, so are you saying that the knowledge about more than one eigenvalue doesn't help at all when I want to know something about $\|A\|_\max$? Help me Obi-Wan Kenobi, you are my only hope. –  draks ... Nov 10 '12 at 12:55
I gave you an example where $\|A\|_{\max}$ is exactly the biggest eigenvalue, and all the information about the remaining eigenvalues is irrelevant. You cannot expect to improve on that. –  Martin Argerami Nov 10 '12 at 14:23
Would it help to say that $A$ is positive definite? –  draks ... Nov 13 '12 at 6:36
Nope, the bound is still sharp for positive definite, take $$A=D=\begin{bmatrix}2&0\\0&1\end{bmatrix}.$$ –  Martin Argerami Nov 13 '12 at 10:59
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