On Peskin & Schroeder's QFT page 352, the book uses the Linear sigma model to illustrate the renormalization and symmetry.

We can write the Lagrangian of Linear sigma model with $$ \begin{aligned} & \mathcal{L}=\frac{1}{2}\left(\partial_\mu \phi^i\right)^2+\frac{1}{2} \mu^2\left(\phi^i\right)^2-\frac{\lambda}{4}\left[\left(\phi^i\right)^2\right]^2 \\ &+\frac{1}{2} \delta_Z\left(\partial_\mu \phi^i\right)^2-\frac{1}{2} \delta_\mu\left(\phi^i\right)^2-\frac{\delta_\lambda}{4}\left[\left(\phi^i\right)^2\right]^2 . \end{aligned} \tag{11.14} $$ Written in term of $\pi$ and $\sigma$ fields, the second line take the form $$ \begin{aligned} \frac{\delta_Z}{2}\left(\partial_\mu \pi^k\right)^2 & -\frac{1}{2}\left(\delta_\mu+\delta_\lambda v^2\right)\left(\pi^k\right)^2+\frac{\delta_Z}{2}\left(\partial_\mu \sigma\right)^2-\frac{1}{2}\left(\delta_\mu+3 \delta_\lambda v^2\right) \sigma^2 \\ & -\left(\delta_\mu v+\delta_\lambda v^3\right) \sigma-\delta_\lambda v \sigma\left(\pi^k\right)^2-\delta_\lambda v \sigma^3 \\ & -\frac{\delta_\lambda}{4}\left[\left(\pi^k\right)^2\right]^2-\frac{\delta_\lambda}{2} \sigma^2\left(\pi^k\right)^2-\frac{\delta_\lambda}{4} \sigma^4 \end{aligned} \tag{11.15} $$

Now, let's consider the renormalization conditions. One of the conditions is to set the tadpole diagram equal to zero.

I think this condition leads to (11.16) $$ \left\langle\phi^N\right\rangle=\frac{\mu}{\sqrt{\lambda}} \tag{11.16} $$

which is satisfied to all orders in pertubation theory.

This is where I am troubled. Why setting all the tadpole terms equal to zero can guarantee the vacuum expectation value (vev, $v$) $\frac{\mu}{\sqrt{\lambda}}$ in all orders?

I think $v=\frac{\mu}{\sqrt{\lambda}}$ only holds for the first line of (11.14). In the counter terms, there are two-points, three-points and four points interactions of the $\sigma$ field. Shouldn't all these counterterms will affect the potential shape and then affect the vev?

On Peskin & Schroeder's QFT pages 353-355, the book uses the Linear sigma model to illustrate the renormalization and symmetry.

We can write the Lagrangian of Linear sigma model with

$$ \begin{aligned} & \mathcal{L}=\frac{1}{2}\left(\partial_\mu \phi^i\right)^2+\frac{1}{2} \mu^2\left(\phi^i\right)^2-\frac{\lambda}{4}\left[\left(\phi^i\right)^2\right]^2 \\ &+\frac{1}{2} \delta_Z\left(\partial_\mu \phi^i\right)^2-\frac{1}{2} \delta_\mu\left(\phi^i\right)^2-\frac{\delta_\lambda}{4}\left[\left(\phi^i\right)^2\right]^2 . \end{aligned} \tag{11.14} $$ Written in term of $\pi$ and $\sigma$ fields, the second line take the form $$ \begin{aligned} \frac{\delta_Z}{2}\left(\partial_\mu \pi^k\right)^2 & -\frac{1}{2}\left(\delta_\mu+\delta_\lambda v^2\right)\left(\pi^k\right)^2+\frac{\delta_Z}{2}\left(\partial_\mu \sigma\right)^2-\frac{1}{2}\left(\delta_\mu+3 \delta_\lambda v^2\right) \sigma^2 \\ & -\left(\delta_\mu v+\delta_\lambda v^3\right) \sigma-\delta_\lambda v \sigma\left(\pi^k\right)^2-\delta_\lambda v \sigma^3 \\ & -\frac{\delta_\lambda}{4}\left[\left(\pi^k\right)^2\right]^2-\frac{\delta_\lambda}{2} \sigma^2\left(\pi^k\right)^2-\frac{\delta_\lambda}{4} \sigma^4. \end{aligned} \tag{11.15} $$

Now, let's consider the renormalization conditions. One of the conditions is to set the tadpole diagram equal to zero.

I think this condition leads to (11.16)

$$ \left\langle\phi^N\right\rangle=\frac{\mu}{\sqrt{\lambda}} \tag{11.16} $$ which is satisfied to all orders in pertubation theory.

This is where I am troubled. Why setting all the tadpole terms equal to zero can guarantee the vacuum expectation value (vev, $v$) $\frac{\mu}{\sqrt{\lambda}}$ in all orders?

I think $v=\frac{\mu}{\sqrt{\lambda}}$ only holds for the first line of (11.14). In the counter terms, there are two-points, three-points and four points interactions of the $\sigma$ field. Shouldn't all these counterterms will affect the potential shape and then affect the vev?