filters - Is Ideal LPF BIBO unstable?

In one of other discussions : How to find frequency response, stability, and causality of a linear system?

I found a comment which was quite strong and definitely caught my attention.

An ideal low-pass filter is an example of a system that is not BIBO stable even though its frequency response is bounded for all $f$

I am following the definition of stability as per here in wiki http://en.wikipedia.org/wiki/BIBO_stability

Can anyone give me a proof that ideal LPF can indeed be BIBO unstable?

Of course, ideal LPF with infinite gain can produce unbounded output. The question is restricted to LPF when gain is finite.

Answer

This answer is a response to a comment by the OP on on yoda's answer.

Suppose that $h(t)$, the impulse response of a continuous-time linear time-invariant system, has the property that $$\int_{-\infty}^{\infty} |h(t)| \mathrm dt = M$$ for some finite number $M$. Then, for each and every bounded input $x(t)$, the output $y(t)$ is bounded also. If $|x(t)| \leq \hat{M}$ for all $t$ where $\hat{M}$ is some finite number, then $|y(t)| \leq \hat{M}M$ for all $t$ where $\hat{M}M$ is also a finite number. The proof is straightforward. $$\begin{align*} |y(t)| &= \left |\int_{-\infty}^\infty h(\tau)x(t - \tau)\mathrm d\tau\right |\\ &\leq \int_{-\infty}^\infty |h(\tau)x(t - \tau)|\mathrm d\tau\\ &\leq \int_{-\infty}^\infty |h(\tau)|\cdot|x(t - \tau)|\mathrm d\tau\\ &\leq \hat{M}\int_{-\infty}^\infty |h(\tau)|\mathrm d\tau\\ &= \hat{M}M. \end{align*}$$ In other words, $y(t)$ is bounded whenever $x(t)$ is bounded.

Thus, the condition $\displaystyle\int_{-\infty}^{\infty} |h(t)| \mathrm dt < \infty$ is sufficient for BIBO-stability.

The condition $\displaystyle\int_{-\infty}^{\infty} |h(t)| \mathrm dt < \infty$ is also necessary for BIBO-stability.

Assume that every bounded input produces a bounded output. Now consider the input $x(t) = \text{sgn}(h(-t)) ~\forall~ t$. This is clearly bounded, ($|x(t)| \leq 1$ for all $t$), and at $t=0$, it produces output $$\begin{align*} y(0) &= \int_{-\infty}^\infty h(0-\tau)x(-\tau)\mathrm d\tau\\ &= \int_{-\infty}^\infty h(-\tau)\text{sgn}(h(-\tau))\mathrm d\tau &= \int_{-\infty}^\infty |h(-\tau)|\mathrm d\tau\\ &= \int_{-\infty}^\infty |h(t)|\mathrm dt. \end{align*}$$ Our assumption that the system is BIBO stable means that $y(0)$ is necessarily finite, that is, $$\int_{-\infty}^{\infty} |h(t)| \mathrm dt < \infty$$

The proof for discrete-time systems is similar with the obvious change that all the integrals are replaced by sums.

Ideal LPFs are not BIBO-stable systems because the impulse response is not absolutely integrable, as stated in the answer by yoda. But his answer does not really answer the question

Can anyone give me a proof that ideal LPF can indeed be BIBO unstable?

A specific example of a bounded input signal that produces an unbounded output from an ideal LPF (and thus proves that the system is not BIBO-stable) can be constructed as outlined above (see also my comment on the main question).