Let $G:=\Bbb R$. Consider the regular representation $$R:G\to GL\big(L^2(G)\big)$$ given by $$\big(R(t)f\big)(x)=f(x+t) \ \text{for $x,t\in G$ and $f\in L^2(G)$}.$$ We show that $R$ has no irreducible subrepresentation. Arguing by a contradiction, suppose that $(R_o,V)$ be a an irreducible subrepresentation of $(R,L^2(G))$. Let $f$ be a basis element for $V$. We know that $\dim V=1$. Then for any $t,x\in G$ we have \begin{align} f(t+x)=:\big(R(t)f\big)(x)=c_tf(x) \qquad\text{for some $c_t\in\Bbb{C}\setminus\{0\}$.} \end{align} So we get $$\|f\|_2^2=\int_G |f(x)|^2dx=\frac{1}{|c_t|^2}\int_G|f(t+x)|^2dx=\frac{1}{|c_t|^2}\int_G|f(x)|^2dx=\frac{1}{|c_t|^2}\|f\|_2^2$$ which implies that \begin{align} |c_t|=1 \qquad \text{for all $t\in G$.} \end{align} Then we have $$|f(x)|=\left|\frac{1}{c_t}f(t+x)\right|=|f(t+x)|\qquad \text{for all $x,t\in G$.}$$ which implies that there is some $c>0$ such that $|f(x)|=c$ for all $x\in G$. So we get $\|f\|^2=\int_G |f(x)|^2dx=\infty$, contradiction.
My question. Let $\lambda\in\Bbb R$ and $V$ be a one-dimensional subspace of $L^2(G)$. Consider the representation $\pi_\lambda: G\to GL(V)$ given by $\pi_\lambda(t)v=e^{i\lambda t}v$ for $v\in V$, $t\in G$. But $\pi_\lambda$ is an irreducible subrepresentation of $(R,L^2(G))$. I have confusion here. Could anyone help me? Thanks.