4 votes

If $x,y,\alpha>0$ and $x>y$, then $x^\alpha>y^\alpha$

Hint: Let $\displaystyle k = \frac{x}{y} \implies k > 1.$ How does $(ky)^\alpha$ compare with $y^\alpha$?
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  • 21.6k
4 votes

Spectrum of the operator on $L^2[0,1]$

An eigenfunction $f$ corresponding to $\lambda\neq 0$ satisfies $f(0)=0.$ Moreover $$f'(x)=-{1\over \lambda}\,f(1-x)$$ Hence $f'(1) =0.$ Applying next derivative gives $$f''(x)=-{1\over \lambda^2}\,f(...
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3 votes

If $x,y,\alpha>0$ and $x>y$, then $x^\alpha>y^\alpha$

To compare $x^\alpha$ and $y^\alpha$, we can divide the two. $(\frac{x}{y})^\alpha$ is what you get when you divide the two. Since $x>y$, $(\frac{x}{y})>1$. Also, $\alpha>0$. When we raise a ...
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1 vote

How will I prove that the simultaneous limits for the following exist?

Hint: Note that $\sqrt{x^2+y^2}\ge\sqrt{x^2}=|x|$. So $$0\le\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\le|y|.$$
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1 vote

Speed of Convergence for some series (double sum)

We will show that $S_n=o(n^{-(\alpha-1)})$. We claim $$n^{(\alpha-1)}S_n \approx n^{(\alpha-1)} \sum_{i=1}^n i^{-\alpha}\left[(n-i+1)^{-(\alpha-1)}-n^{-(\alpha-1)}\right] =\sum_{i=1}^n i^{-\alpha}\...
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  • 315
1 vote

Prove the constancy of a harmonic function with $\lim_{\vert x\vert\rightarrow\infty}\frac{\vert f(x)\vert}{\ln\vert x\vert}=0$.

The assumption can be weakened to $$\limsup_{|z|\to \infty}{f(x,y)\over \ln|z|}=:r_0<\infty, \ z=x+iy\qquad(*)$$ Fix $r>r_0.$ By $(*)$ we obtain $$f(x,y)-r\ln|z|< 0,\qquad |z|> R$$ for ...
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