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For $n\ge 3$ and a bounded smooth domain, a Dirichlet Laplacian eigenfunction with eigenvalue $\lambda$ satisfies the displayed estimate. A signed power test for a Laplacian eigenfunction and the zero-boundary Sobolev inequality give a Moser iteration with exponents $p_j=2[n/(n-2){]}^j$. The geometric series ${\sum }_{j}1/p_j=n/4$ gives the exact power of $\lambda$, while ${\sum }_j(\log p_j)/p_j<\infty$ controls the remaining constant. The constant depends only on dimension, because the zero-boundary Sobolev inequality follows by extension by zero to the whole space.

Use a signed power test for a Laplacian eigenfunction. For $\gamma \ge 2$, test the Dirichlet Laplacian eigenfunction equation with $|u{|}^{\gamma -2}u$. Its derivative is $(\gamma -1)|u{|}^{\gamma -2}Du$; at $\gamma =2$ the test function is just $u$. Integration by parts gives the exact energy estimateSet $v=|u{|}^{\gamma /2}$. It belongs to the zero-boundary Sobolev space and obeys the Sobolev chain rule:For $\gamma >2$, the power map is $C^1$ with bounded derivative on the bounded range of $u$. For $\gamma =2$, the absolute value map is Lipschitz, and $D|u|=\mathrm{sgn}(u)Du$ almost everywhere; on the zero set, use the fact that a gradient of a Sobolev function vanishes on a level set. These facts justify the identity even when $u$ changes sign.

Apply the given Sobolev inequality with the Sobolev conjugate exponent $2\kappa =2n/(n-2)$:Testing with a signed power raises the integrability exponent from $\gamma$ to $\gamma \kappa$. Absorbing the factor $1/4$ into $C(n)$ gives precisely the requested inequality.