= Solution
Let $u_i$ denote player $i$'s expected <payoff>, so the definition also covers <mixed strategies>. If $p_i^*$ is a <dominant strategy>, then
$$
u_i(p_i^*,p_{-i})\geq u_i(q_i,p_{-i})
$$
for every opponents' profile $p_{-i}$ and every alternative $q_i$. In particular this holds at $p_{-i}=p_{-i}^*$. Thus no player can improve its <payoff> by a unilateral deviation from $(p_1^*,\ldots,p_n^*)$, which is exactly the definition of a <Nash equilibrium>. \b[A profile of <dominant strategies> is a <Nash equilibrium>.]
For the <second-price sealed-bid auction>, use the usual <private-value auction> assumptions: a bidder with value $v$ has <quasilinear utility> $v$ minus its payment if it wins, and zero if it loses. Fix the highest opposing bid $h$. If $h<v$, winning gives $v-h>0$, and bidding $v$ wins. If $h>v$, winning gives $v-h<0$, and bidding $v$ loses. If $h=v$, winning or losing gives zero, irrespective of the tie rule. Any other bid can only change whether the bidder wins, not the payment $h$ conditional on winning. It therefore cannot improve on the truthful bid in any of these cases. Hence \b[bidding the true value is a weakly <dominant strategy>]. Neither symmetry nor independence of the opposing values is needed for this argument.
For part (a), in a <first-price sealed-bid auction> with $n\geq2$, continuous nonnegative bids and $v>0$, no fixed bid is a <dominant strategy>. A positive bid $b$ can be improved against opponents all bidding zero: replace it by $b/2>0$, still win, and increase the <payoff> by $b/2$. Bid zero can be improved against an opposing highest bid $h=v/2$: bidding $3v/4$ wins and earns $v/4>0$, whereas zero loses. This proves the <no dominant positive-value bid in a first-price auction> result.
Allowing <mixed strategies> does not change that conclusion. Against all-zero opponents, a random bid has expected <payoff> strictly below $v$ unless it is always zero and the tie rule awards this bidder the item with certainty. Any positive bid incurs a strictly positive payment, and any <probability> of losing at zero also reduces the <payoff> below $v$. In the strict case choose a sufficiently small positive fixed bid $\delta$ so that $v-\delta$ exceeds the random bid's expected <payoff>. In the exceptional case the preceding deviation against $h=v/2$ still improves it. The assumptions matter: for value zero, bid zero is weakly dominant; with only one bidder it is also optimal. Consequently \b[there is no <dominant strategy> for a positive-value bidder in the standard first-price setting].
Part (b)'s printed nonexistence request is false under the stated <symmetric independent private values model>. Lack of a <dominant strategy> does not imply lack of a <Nash equilibrium>. For incomplete private information the appropriate equilibrium is a <Bayesian Nash equilibrium>, a <Nash equilibrium> of the game whose strategies are bidding rules as functions of private values. Here is an explicit counterexample with a full deviation check.
Let the $n\geq2$ <risk-neutral> bidders have independent valuations with the <uniform distribution> on $[0,1]$, and put $a=(n-1)/n$. Suppose every opponent bids $\beta(w)=aw$. A bidder with value $v$ who bids $b\in[0,a]$ wins with <probability> $(b/a)^{n-1}$, since the opposing values must all be below $b/a$. Ties have <probability> zero, including at either endpoint. Its expected <quasilinear utility> is
$$
U_v(b)=(v-b)(b/a)^{n-1}.
$$
For $b>0$,
$$
U_v'(b)=a^{-(n-1)}b^{n-2}\bigl((n-1)v-nb\bigr).
$$
Thus $U_v$ increases up to $b=av$ and decreases thereafter. For $v=0$, bid zero is optimal directly. A bid above $a$ wins with certainty and pays more than bidding $a$, so it cannot improve on the maximum already found in $[0,a]$. A randomized bid is a mixture of these <payoffs> and also cannot improve on their maximum. The <best response> for every value is therefore the proposed bidding rule itself. This proves the <uniform private-value first-price bidding equilibrium>:
$$
\boxed{\beta(v)=\frac{n-1}{n}v\quad\text{is a symmetric Bayesian Nash equilibrium}.}
$$
The value-$v$ expected equilibrium <payoff> is $v^n/n$. The bidding rule is a <pure strategy> in the Bayesian game, even though values are random. Therefore \b[the requested general nonexistence claim in (b) has a counterexample], rather than a valid proof. The defect is present in the original PDF, not introduced by the converted text.
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