34 Let
$$1^{x}+2^{x}+\cdots+(n-1)^{x}+n^{x} \cdot a=k,$$

then
$$\begin{array}{c} 1^{2}+\left(\frac{k}{n}\right)^{2} \geqslant \frac{2 k}{n}, 2^{2 x}+\left(\frac{k}{n}\right)^{2} \geqslant 2 \cdot \frac{2^{x} k}{n}, \cdots, \\ n^{2 x} a^{2}+\left(\frac{k}{n}\right)^{2} \geqslant \frac{2 n^{x} \cdot a k}{n} . \end{array}$$

Adding these inequalities, we get:
$$\begin{aligned} & 1+2^{2 x}+\cdots+(n-1)^{2 x}+n^{2 x} a^{2}+n \cdot\left(\frac{k}{n}\right)^{2} \\ \geqslant & 2 k\left[1+2^{x}+\cdots+(n-1)^{x}+n^{x} a\right] / n \\ = & \frac{2 k^{2}}{n}, \end{aligned}$$

which simplifies to
$$1+2^{2 x}+\cdots+(n-1)^{2 x}+n^{2 x} a^{2} \geqslant \frac{k^{2}}{n} .$$

Substituting the value of $k$ gives
$$1+2^{2 x}+\cdots+(n-1)^{2 x}+n^{2 x} \cdot a^{2} \geqslant \frac{\left[1+2^{x}+\cdots+(n-1)^{x}+n^{x} a\right]^{2}}{n} .$$

Since $0 < a \leqslant 1$, we have $a \geqslant a^{2}$. Therefore,
$$1+2^{2 x}+\cdots+(n-1)^{2 x}+n^{2 x} \cdot a \geqslant \frac{\left[1+2^{x}+\cdots+(n-1)^{x}+n^{x} a\right]^{2}}{n}$$

which means $f(2 x) \geqslant 2 f(x)$.