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$H$ is a separable Hilbert space . $A:H\rightarrow H$ is a linear symmetric compact operator. Assume $$ \sigma(A) \le C $$ Whether we have $$ ||A|| \le C ~~~~? $$

What I think: first $H$ has a countable orthonormal basis consisting of eigenvectors of $A$, assume they are $\{u_i\}$ . Then $$ ||Au_i||=\frac{||Au_i||}{||u_i||}\le C $$ Assume $\forall v\in H,~v=v^iu_i $, sum over $i$. Then $$ \sup_{||v||=1}||Av||=\sup_{||v||=1}||v^iA(u_i)||\le \sum_{i=1}^\infty Cv^i $$ But I don't know how to show $$ \sum_{i=1}^\infty v^i =1 $$ And I think it is not right .

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    I don't see the implication in general, but if $A$ is a self-adjoint linear operator, then there exists an element of $\sigma(A)$ that achieves the norm. – 2017-02-24

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If $A$ is selfadjoint, then

$\max\{| \lambda|: \lambda \in \sigma(A)\}=||A||$.

Your "implication" $\sigma(A) \le C$, then $||A|| \le C$ is not true in general !

Example: $A=diag(-1,-1)$, then $\sigma(A) \le -1$, but $||A||=1$

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    I think your self-adjoint is same with symmetric. Where I can find the prove ? Thanks. – 2017-02-26