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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def estimate_frame_bounds(self, x=None):
 r"""Estimate lower and upper frame bounds. A filter bank forms a frame if there are positive real numbers :math:`A` and :math:`B`, :math:`0 < A \leq B < \infty`, that satisfy the *frame condition* .. math:: A \|x\|^2 \leq \| g(L) x \|^2 \leq B \|x\|^2 for all signals :math:`x \in \mathbb{R}^N`, where :math:`g(L)` is the analysis operator of the filter bank. As :math:`g(L) = U g(\Lambda) U^\top` is diagonalized by the Fourier basis :math:`U` with eigenvalues :math:`\Lambda`, :math:`\| g(L) x \|^2 = \| g(\Lambda) U^\top x \|^2`, and :math:`A = \min g^2(\Lambda)`, :math:`B = \max g^2(\Lambda)`. Parameters x : array_like Graph frequencies at which to evaluate the filter bank `g(x)`. The default is ``x = np.linspace(0, G.lmax, 1000)``. The exact bounds are given by evaluating the filter bank at the eigenvalues of the graph Laplacian, i.e., ``x = G.e``. Returns ------- A : float Lower frame bound of the filter bank. B : float Upper frame bound of the filter bank. See Also -------- compute_frame: compute the frame complement: complement a filter bank to become a tight frame Examples -------- Estimation quality (loose, precise, exact):
 A=1.883, B=2.288 A=1.708, B=2.359 A=1.723, B=2.359 The frame bounds can be seen in the plot of the filter bank as the minimum and maximum of their squared sum (the black curve):
 The heat kernel has a null-space and doesn't define a frame (the lower bound should be greater than 0 to have a frame):
 A=0.000, B=1.000 Without a null-space, the heat kernel forms a frame: A=0.135, B=2.000 A kernel and its dual form a tight frame (A=B):
 A=1.000, B=1.000 The Itersine filter bank forms a tight frame (A=B):
 A=1.000, B=1.000 """

        if x is None:
            x = np.linspace(0, self.G.lmax, 1000)
        else:
            x = np.asanyarray(x)

        sum_filters = np.sum(self.evaluate(x)**2, axis=0)

        return sum_filters.min(), sum_filters.max()