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If I do the derivative of the log-likelihood when the data follows a $Po(\lambda)$ distribution, we obtain the estimator $\hat \lambda =\bar X$. However, how do we prove that indeed this estimator does maximize the likelihood?

I've tried the 2nd order conditions, $\frac{\partial^2}{\partial \lambda^2}l(\lambda|\mathbf{x})=\frac{-\sum x_i}{\lambda^2}\leq 0$. If the inequality was strict, then we would be good. However, there's the not null chance that $\sum x_i=0$. In this case, we get $l(\lambda|\mathbf{x})=-\lambda n$, and would like to minimize the value of lambda, but however $\lambda>0$. How to proceed in this case?

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    The case $\lambda=0$, actually, makes sense and means that the distribution is a constant zero. So, I'd maximize the loglikelihood over $\lambda\ge0$ instead of $\lambda>0$. – 2017-02-21
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    @SergeiGolovan but then what's $P(X=0)$? we get $0^0$ – 2017-02-21
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    Look at the case $\lambda=0$ as the limiting case when $\lambda\to0$. This way you'll get that $P(X=0)\to1$. – 2017-02-21
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    If you see nothing then it is reasonable to conclude the maximum likelihood estimate of the parameter is zero. If you have other information that the rate is in fact non-zero, you may need to move to Bayesian methods – 2017-02-21
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    @SergeiGolovan Thanks ;) – 2017-02-21
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    A Poisson distribution is non-negative. So, aside from the case where all the samples are zero, you are good (since the sum will be positive) . And if all the samples are zero, what is the likelihood function? $e^{-n \lambda}$ – 2017-02-21

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