If $H$ is a complex Hilbert space and $T:H\to H$ is a bounded linear operator, we say that $T$ is diagonalizable if there exists an orthonormal basis of $H$ formed by eigenvectors of $T$ ($0\neq{x}\in H$ is a eigenvector if there exists $\lambda \in \mathbb{C}$ with $T(x)=\lambda x$).
When $H$ is finite dimensional, it is known that if $T$ is normal then it is diagonalizable. However, by the spectral theorem for normal compact operator, in infinite dimensional Hilbert spaces we need to make an extra hypothesis: $T$ is compact. So, if I have understood correctly, there must exist bounded linear operator which are normal but not diagonalizable. Could someone give me an example of this?, because I have not found such operator.
Best Answer
There are many counterexamples. Before I give one I want to give some context. Recall that for any normal matrix $A$ there exists a basis of eigenvectors. Let $\sigma(A)=\{\lambda_1,\ldots,\lambda_n\}$ be the eigenvalues with eigenspaces $\{V_{\lambda_1}, \ldots, V_{\lambda_n}\}$. Let $P_k$ be the orthogonal projection onto $V_{\lambda_k}$. Rewriting the spectral theorem for finite dimensional spaces a bit shows that $$ A = \sum_{k=1}^n \lambda_k P_k. $$ Note that sums can be written as integrals over atomic measures and we can loosely write this as $$ A = \int_{\sigma(A)} \lambda dP(\lambda). $$ More precisely, we do not integrate w.r.t. a measure in the usual sense. But integrate w.r.t a "projection-valued measure". I won't go to deep however.
Why did I write the spectral theorem for finite dimensional operators in such a convoluted way? Well there is a general spectral theorem that says that for any normal bounded operator $A$ on a Hilbert space $\mathcal{H}$ there exists a so called spectral measure $P$ such that we can write $$ A = \int_{\sigma(A)} \lambda(A) dP(\lambda). $$ However, while previously, the set $\sigma(A)$ consisted of the eigenvalues of $A$ (i.e. the values $\lambda \in \mathbb{C}$ such that $A - \lambda 1$ is not injective), we now need to take the following definition: $$ \sigma(A) = \{ \lambda \in \mathbb{C} \mid A-\lambda 1 \text{ is not invertible}. \} $$ In the finite dimensional case (and even when $A$ is compact) this boils down to $\sigma(A)$ being the eigenvalues of $A$. However in general, $A-\lambda 1$ might not be invertible while there exists no non-zero vector $v \in \mathcal{H}$ such that $A v =\lambda v$. While the generalized spectrum $\sigma(A)$ may be defined for any bounded operator (and having some really nice properties popping-up from complex analysis), it can be split up (for normal operators) into a discrete part (the real eigenvalues) and a continuous part. The continuous part makes general normal operators so different than the finite dimensional ones. This also makes the domain of the integral above over a non-discrete set.
Now to give a counterexample, take the operator $$A: L^2([0,1]) \rightarrow L^2([0,1]), A(f)(x) = x f(x) \quad \text{(multiplication by }x). $$ $A$ has no eigenvalues (for every $z \in \mathbb{C}$, there is no non-zero function $f$ with $A(f) = z f$, this would mean that $x = z$ for all $x$ where $f(x)$ is non-zero, so $f$ is zero a.e.). In particular, there is no basis of eigenvectors. However the spectrum $\sigma(A)$ can be shown to be equal to $[0,1]$ (the image of the function $x$ on $[0,1])$.
Edit: To show that $\sigma(A) = [0,1]$ for the example above, assume that $\lambda \notin \sigma(A) $. This means that there exists some $T \in \mathcal{B}(L^2([0,1])$ such that $$T \circ (A-\lambda \text{Id}) = (A-\lambda \text{Id}) \circ T = \text{Id}. $$ Put $g = T(1) \in L^2([0,1])$. It follows that$(x-\lambda 1) g(x) = 1$ for all $x \in [0,1]$ a.e. or written differently that $$ g(x) = \frac{1}{x-\lambda} \quad \forall x \in [0,1] \text{ a.e.} $$ However $g$ is not square integrable if $\lambda \in [0,1]$. This shows that $[0,1] \subset \sigma(A)$. For other inclusion, it is easy to see that whenever $\lambda \notin [0,1]$, then multiplication by the function $g(x) = \frac{1}{x-\lambda}$ is an inverse to $A-\lambda 1$. Because $g$ is uniformly bounded, the multiplication operator by $g$ is a bounded operator.
As a side note, $A$ from above is not a compact operator, as you can tell from the fact that is has a continuous spectrum (compact operators are diagonisable in the classical sense (having a countable orthonormal basis of eigenvectors), hence they have a discrete spectrum).