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The relation $1/q=1/p-\alpha /n>0$ implies $\alpha p<n$. Fix $r>0$ and split the Riesz potential into the regions $|x-y|<r$ and $|x-y|\ge r$.

For the near part, decompose the ball into the dyadic annuliThe definition of the Hardy-Littlewood maximal function then givesFor the far part, Hölder's inequality giveswhere convergence follows from $\alpha p<n$.

If $Mf(x)>0$, chooseThe two bounds then have the same size. If $Mf(x)=0$, the result is immediate. Consequently the Hedberg inequality isHere the dimensional constant is required for the unnormalized kernel and Lebesgue measure used in the question.

Use the centered Hardy-Littlewood maximal functionIf $x>0$, choosing $0<r<x$ makes the whole interval lie in the positive half-line, so $M\theta (x)=1$. If $x=0$, every centered interval has exactly half its length in the positive half-line, so $M\theta (0)=1/2$.

If $x<0$, the average is zero for $r\le -x$. For $r>-x$, it isand these values tend to $1/2$ as $r\to \infty$. HenceFor the uncentered maximal-function convention, intervals extending arbitrarily far to the right make the supremum equal to $1$ at every $x$.

For $1<p\le \infty$, the Hardy-Littlewood maximal function is bounded on $L^p({\mathbb{R}}^n)$. A direct proof for finite $p$ applies the weak $(1,1)$ estimate to $f{1}_{\{|f|>s/2\}}$, integrates the resulting distribution-function bound against $p{s}^{p-1}$, and uses Tonelli theorem.