Is my conjecture true or false? It seems it may be true based on the given proof.
Conjecture: Let $T:H_1\rightarrow H_2$ be a bounded linear operator between Hilbert spaces $H_1$ and $H_2$. Assume $H_1$ is separable. Suppose there exists an orthonormal basis $\{e_j\}$ so that $Te_j\rightarrow 0$ in norm as $j\rightarrow \infty$. Then $T$ is compact.
Here is my proof:
Let $h_k\rightarrow 0$ weakly in $H_1$ as $k\rightarrow \infty$. Then write $h_k=\sum_{j=1}^{\infty}\langle h_k, e_j\rangle e_j$. And so let $\varepsilon>0$. Then there exists $j_{\varepsilon}\in \mathbb{N}$ and $j_{\varepsilon}>1$ so that for all $j\geq j_{\varepsilon}$, $\|Te_j\|^2<\varepsilon $. Now we apply $T$ to the series representation for $h_k$ and split the series.
\begin{align}
\|Th_k\|^2&< \sum_{j=1}^{j_{\varepsilon}-1}|\langle h_k, e_j\rangle |^2 \|Te_j\|^2+\varepsilon\sum_{j=j_{\varepsilon}}^{\infty}|\langle h_k, e_j\rangle|^2\\
&<\sum_{j=1}^{j_{\varepsilon}-1}|\langle h_k, e_j\rangle |^2 \|T\|^2+\varepsilon\sup_{k\in \mathbb{N}}\|h_k\|^2
\end{align}
for all $k\in \mathbb{N}$. Since $h_k\rightarrow 0$ weakly as $k\rightarrow \infty$, one can show that $\|h_k\|^2$ is a bounded sequence using the uniform boundedness principle. Thus it remains to show that
$\sum_{j=1}^{j_{\varepsilon}-1}|\langle h_k, e_j\rangle|^2$ can be made arbitrarily small for $k$ sufficiently large. Because $h_k$ converges to $0$ weakly, for $\varepsilon>0$ and each $j\in \{1,2,…, j_{\varepsilon}-1\}$ there exists $k_{j,\varepsilon}\in \mathbb{N}$ so that $|\langle h_k, e_j\rangle |^2<\frac{\varepsilon}{j_{\varepsilon}-1}$ for $k\geq k_{j,\varepsilon}$. Then for
$k\geq k_{\varepsilon}:=\max_{j\in \{1,2,…,j_{\varepsilon}-1\}}\{k_{j,\varepsilon}\}+1$, we have
$\sum_{j=1}^{j_{\varepsilon}-1}|\langle h_k, e_j\rangle|^2<\varepsilon$ . This shows that $Th_k$ is strongly convergent to $0$ for any sequence $h_k$ weakly converging to $0$. Hence $T$ is compact.
Best Answer
I think this is a very clever and sensible conjecture but unfortunately it is not true.
When a reasonable statement turns out to be false the counter-examples tend to be a bit complicated but here is a recipe to produce one.
Let $H$ be a separable Hilbert space with orthonormal basis $\{e_0, e_1, e_2, \ldots \}$.
For every bounded operator $T$ on $H$, the matrix of $T$ is defined to be the matrix $A = (a_{i,j})_{i, j=0}^\infty$, given by $$ a_{i,j}=\langle T(e_j),e_i\rangle . $$
A bounded operator $T$ on $H$ is said to be a Hankel operator if there exists a function $\varphi :{\mathbb N}\to{\mathbb C}$ such that the matrix of $T$ is given by $$ a_{i,j}=\varphi (i+j), $$ for every $i$ and $j$. Equivalently, the entries ot the matrix of $T$ are constant along every diagonal perpendicular to the main diagonal.
Observe that the coefficients of the vector $T(e_j)$, namely the $j^{\text{th}}$ column of its matrix, are precisely given by $\{\varphi (i+j)\}_{i=0}^\infty$. In other words, $$ T(e_j)=\sum_{i=0}^\infty \varphi (i+j)e_i. $$
Since $T(e_0)$ is supposed to lie in $H$, we see that $\{\varphi (i)\}_{i=0}^\infty$ is a square summable sequence. Consequently, $$ \Vert T(e_j)\Vert ^2 = \sum_{i=0}^\infty |\varphi (i+j)|^2 = \sum_{i=j}^\infty |\varphi (i)|^2 \quad {\buildrel j\to\infty \over \longrightarrow}\quad 0. $$ So we see that all Hankel operators satisfy the condition required by the OP!
If we are to find a counter-example, we are therefore left with the task of finding a non-compact Hankel operator, but this is where things get a bit more sophisticated, not least because Hankel operators have a strong tendency to be compact!
So here are two of the most important results about Hankel operators.
Theorem. (Nehari) Given a function $\varphi :{\mathbb N}\to{\mathbb C}$, the matrix $A = (a_{i,j})_{i, j=0}^\infty$, given by $a_{i,j}=\varphi (i+j)$, represents a bounded operator on $H$ iff there exists some bounded measurable function $f$ on $S^1$ whose nonegative Fourier coefficients satisfy $$ \hat f(n) = \varphi (n), \quad \forall n\geq 0. $$
The function $f$ referred to above is sometimes called the symbol of the operator, and the operator itself is often denoted by $H_f$.
Theorem. (Hartman) Given a function $\varphi :{\mathbb N}\to{\mathbb C}$, the corresponding Hankel operator is compact iff it admits a continuous symbol.
We must therefore find a function $\varphi $ which satisfies Nehari but not Hartman!
At first sight, one might think that it is enough to take a discontinuous symbol $f$ and produce a Hankel operator using its Fourier coefficients but, unfortunately, life is not that simple. Since the function $\varphi $ refers only to the positive Fourier coefficients of $f$, we risk being under the situation in which $f$ is discontinuous, but some other continuous function $g$ shares its positive Fourier coefficients with $f$, so $H_f$ will turn out to be compact :-(
A class of operators where one can immediately check for compactness is the class of partial isometries. Such an operator is easily seen to be compact iff its range is finite dimensional. It is therefore nice to know that partial isometric Hankel operator can easily be characterized!
Theorem. A Hankel operator is a partial isometry iff it admits a symbol of the form $\bar z\vartheta$, where $\vartheta$ is an inner function.
For those who do not yet know it, an inner function is any measurable function $\vartheta$ on $S^1$ such that $|\vartheta(z)|=1$, for almost all $z$, and also such that $\hat \vartheta(n)=0$, whenever $n< 0$.
There is an enormous supply of inner functions, such as the so called Blachke factors, namely functions of the form $$ \vartheta(z) = {|a|\over a}{a-z \over 1-\bar a z}, \quad \text { for } z\in S^1, $$ for a fixed complex number $a$ with $|a|<1$ (when $a=0$ it is custumary to replace this by $\vartheta(z)=z$). However the trouble, at least from our point of view, is that $\vartheta$ is continuous, and hence by Hartman's Theorem $H_{\bar z\vartheta}$ is compact :-( In fact $H_{\bar z\vartheta}$ actually has rank one.
If we instead take a finite product of Blachke factors, namely what is known as a $\ldots $ Blachke product $$ \vartheta(z) = \prod_{k=1}^n{|a_k|\over a_k}{a_k-z \over 1-\bar a_k z}, $$ we still get an inner function, unfortunately still continuous, but the corresponding Hankel operator turns out to have a bigger rank, namely $n$.
This suggests that if we want a partial isometric Hankel operator with infinite rank we need an infinite product of Blachke factors $$ \vartheta(z) = \prod_{k=1}^\infty{|a_k|\over a_k}{a_k-z \over 1-\bar a_k z}, $$ and, thanks to a Theorem proved by Blachke himself, such an infinite product converges provided it satisfies the so called Blachke condition, namely $$ \sum_{k=1}^\infty(1-|a_k|) <\infty . $$
Chosing any infinite sequence $\{a_k\}_k$ satisfying this condition, the corresponding Blachke product $\vartheta$ therefore leads to a non-compact Hankel operator $H_{\bar z\vartheta}$, which is then the counter example required!