bounded operator $T$ is not compact then there exists an orthonormal sequence $e_n$ and $d>0$ such that $\|T(e_n)\|>d$ for all $n\in\Bbb{N}$? I want to prove that if a bounded operator $T$ is not compact then there exists an orthonormal sequence $e_n$ and $d>0$ such that $\|T(e_n)\|>d$ for all $n\in\Bbb{N}$. Could someone helps me?

I think that the fact that T is not compact implies that there exists an arbitrary sequence in H that doesn't contain convergent subsequence, not for a orthonormal sequence. Moreover, we know that if T is a compact operator and en an orthonormal sequence then ||T(en)|| converges strongly to 0 but not the inverse. We want to show this. 
 A: By contraposition (and by taking a subsequence), it is sufficient
to prove that if $T:H\to H$ is a bounded linear operator with the
property that for each orthonormal sequence $\left(e_{n}\right)_{n\in\mathbb{N}}$
in $H$, the convergence $\left\Vert Te_{n}\right\Vert \to0$ holds,
then $T$ is compact.
Let us first prove this in the case that $T=T^{\ast}$ is self-adjoint.
In this case, we have the power of the spectral theorem at our disposal.
Let $\sigma\left(T\right)\subset\left[a,b\right]$ with $a<b$. For
each $n\in\mathbb{N}$, consider the operator
$$
T_{n}:=\left({\rm id}\cdot\chi_{\left[a,b\right]\setminus\left[-\frac{1}{n},\frac{1}{n}\right]}\right)\left(T\right),
$$
where $\chi_{E}$ is the indicator function (characteristic function)
of the set $E\subset\mathbb{R}$. Here, I used the bounded Borel-functional
calculus (see http://en.wikipedia.org/wiki/Borel_functional_calculus#The_bounded_functional_calculus)
for bounded self-adjoint operators.
Using $T={\rm id}\left(T\right)$, we conclude
$$
\left\Vert T-T_{n}\right\Vert =\left\Vert \left({\rm id}\cdot\chi_{\left[a,b\right]\cap\left[-\frac{1}{n},\frac{1}{n}\right]}\right)\left(T\right)\right\Vert \leq\left\Vert {\rm id}\cdot\chi_{\left[a,b\right]\cap\left[-\frac{1}{n},\frac{1}{n}\right]}\right\Vert _{\sup}\leq\frac{1}{n}\to0,
$$
so that it suffices to show that each $T_{n}$ is a compact operator,
because the set of compact operators is a closed ideal in $\mathcal{B}\left(H\right)$,
see e.g. http://en.wikipedia.org/wiki/Compact_operator_on_Hilbert_space#Some_general_properties.
To this end, it suffices to show that the range ${\rm Range}\left(T_{n}\right)$
is finite-dimensional (why?). Note that the functional calculus (a
product of functions translates to a composition of operators) yields
$$
T_{n}=\chi_{\left[a,b\right]\setminus\left[-\frac{1}{n},\frac{1}{n}\right]}\left(T\right)\circ T
$$
and hence ${\rm Range}\left(T_{n}\right)\subset{\rm Range}\left(\chi_{\left[a,b\right]\setminus\left[\frac{1}{n},\frac{1}{n}\right]}\left(T\right)\right)=:V_{n}$.
If $V_{n}$ was infinite-dimensional, there would be an orthonormal
sequence $\left(e_{k}^{\left(n\right)}\right)_{k\in\mathbb{N}}$ in
$V_{n}$. By assumption, $\left\Vert Te_{k}^{\left(n\right)}\right\Vert \xrightarrow[k\to\infty]{}0$.
But we can (for example, there is probably an easier way to see this)
use the unitary equivalence of $T$ to a multiplication operator on
some $L^{2}$-space, i.e. $T=UM_{f}U^{\ast}$ for some unitary map
$U:L^{2}\to H$ and $M_{f}:L^{2}\to L^{2},g\mapsto fg$ for some suitable
measure space $\left(X,\mathcal{M},\mu\right)$ and some function
$f\in L^{\infty}$, to conclude 
$$
\chi_{\left[a,b\right]\setminus\left[-\frac{1}{n},\frac{1}{n}\right]}\left(T\right)=UM_{\chi_{\left|f\right|\geq\frac{1}{n}}}U^{\ast}.
$$
This yields 
$$
U^{\ast}e_{k}^{\left(n\right)}=U^{\ast}\chi_{\left[a,b\right]\setminus\left[-\frac{1}{n},\frac{1}{n}\right]}\left(T\right)e_{k}^{\left(n\right)}=M_{\chi_{\left|f\right|\geq\frac{1}{n}}}U^{\ast}e_{k}^{\left(n\right)},
$$
so that the $L^{2}$-function $g_{k}:=U^{\ast}e_{k}^{\left(n\right)}$
vanishes almost everywhere on $\left|f\right|^{-1}\left(\left[-\frac{1}{n},\frac{1}{n}\right]\right)$.
Hence,
$$
\left\Vert Te_{k}^{\left(n\right)}\right\Vert _{H}=\left\Vert U^{\ast}TUU^{\ast}e_{k}^{\left(n\right)}\right\Vert _{L^{2}}=\left\Vert M_{f}g_{k}\right\Vert _{L^{2}}\geq\frac{1}{n}\cdot\left\Vert g_{k}\right\Vert _{L^{2}}=\frac{1}{n}\cdot\left\Vert e_{k}^{\left(n\right)}\right\Vert _{H}=\frac{1}{n},
$$
which contradicts the fact that $\left\Vert Te_{k}^{\left(n\right)}\right\Vert _{H}\xrightarrow[k\to\infty]{}0$.
This contradiction shows that ${\rm Range}\left(T_{n}\right)\subset V_{n}$
is finite-dimensional, so that each operator $T_{n}$, and hence also
$T$, is compact.
Now, let $T$ be as in the assumption (without assuming that $T$
is selfadjoint). Define $S:=T^{\ast}T$ and observe that this is a
self-adjoint operator. Even more, $S$ it positive semi-definite,
because of
$$
\left\langle Sx,x\right\rangle =\left\langle T^{\ast}Tx,x\right\rangle =\left\langle Tx,Tx\right\rangle =\left\Vert Tx\right\Vert ^{2}\geq0.
$$
Using (e.g.) the functional calculus again, we conclude that there
is a (unique) positive semi-definite, self-adjoint operator $\sqrt{S}:H\to H$
such that $S=\sqrt{S}\sqrt{S}$. Now, let $\left(e_{n}\right)_{n\in\mathbb{N}}$
be any orthonormal sequence in $H$. We have
\begin{eqnarray*}
\left\Vert \sqrt{S}e_{n}\right\Vert _{H}^{2} & = & \left\langle \sqrt{S}e_{n},\sqrt{S}e_{n}\right\rangle =\left\langle \sqrt{S}^{\ast}\sqrt{S}e_{n},e_{n}\right\rangle =\left\langle \sqrt{S}\sqrt{S}e_{n},e_{n}\right\rangle \\
 & = & \left\langle Se_{n},e_{n}\right\rangle =\left\langle T^{\ast}Te_{n},e_{n}\right\rangle =\left\langle Te_{n},Te_{n}\right\rangle =\left\Vert Te_{n}\right\Vert _{H}^{2}\xrightarrow[n\to\infty]{}0,
\end{eqnarray*}
so that $\sqrt{S}$ fulfills our assumptions. By the self-adjoint
case, we conclude that $\sqrt{S}$ is a compact operator.
Using (the proof of) the polar decomposition theorem (see http://en.wikipedia.org/wiki/Polar_decomposition#Bounded_operators_on_Hilbert_space),
there is a partial isometry (which is in particular a bounded linear
operator) $V:H\to H$ such that
$$
T=V\sqrt{T^{\ast}T}=V\sqrt{S}.
$$
As the set of compact operators forms an ideal in the algebra of bounded
operators $\mathcal{B}\left(H\right)$, we conclude that $T$ is compact.
