In discrete bais, we can express a vector as $$ |\psi\rangle=\sum_{i} c_i|e_i\rangle $$ with orthonormality $$ \langle e_i|e_j\rangle=\delta_{ij}.$$ $\delta_{ij}$ is usual kronecker delta. If we promote the basis to continuum basis, we use dirac delta function instead, i.e., $$ \langle x|x'\rangle=\delta (x-x').$$ My question is that, if the definition of delta is $$ \delta(x-x')= \left\{\begin{array} 1\infty, & x=x' \\ 0, & x\ne x' \end{array} \right.$$ Does orthonormality hold? since $\langle x|x\rangle=\infty$, it is not 1 as same as discrete basis.
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