I would expect you to say that the functions map into R and that the we are taking the supremum in the r norm. Then, the set of bounded function from some set X into R form a metric space induced from the metric on R. There could be cases where the subtlety of induced metrics is less trivial. |f(u)-g(u)|1 and assume that the prior property holds between a,p. Consider the function v=u^q-1. Clearly the inverse function is v^p-1=u. (Recall that the product of (q-1)(p-1)=1. Then if one is the power of a single term polynomial to that power, the power of its inverse is the other. The integral of the first over [0,α], the 2nd over [0,β] yields the desired result. It is clear that equality holds, that is, the area of the rectangle is identical to the integral of the function up to a point in its argument, and the inverse function upto the output of the function at that argument, if and only if the point (α,β) lays on the graph of the function (and its inverse). (This is not a rigorous proof.) Holder’s Inequality. Σ |ξ_i*η_j|≤ (Σ |ξ_i|^p)^(1/p)(Σ |η_i|^q)^1/q where p,q are conjugate transposes. Proof.
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Sir, left ear voice not working, I checked laptop and phone. Both are same problem
I would expect you to say that the functions map into R and that the we are taking the supremum in the r norm.
Then, the set of bounded function from some set X into R form a metric space induced from the metric on R. There could be cases where the subtlety of induced metrics is less trivial.
|f(u)-g(u)|1 and assume that the prior property holds between a,p.
Consider the function v=u^q-1. Clearly the inverse function is v^p-1=u. (Recall that the product of (q-1)(p-1)=1. Then if one is the power of a single term polynomial to that power, the power of its inverse is the other.
The integral of the first over [0,α], the 2nd over [0,β] yields the desired result. It is clear that equality holds, that is, the area of the rectangle is identical to the integral of the function up to a point in its argument, and the inverse function upto the output of the function at that argument, if and only if the point (α,β) lays on the graph of the function (and its inverse).
(This is not a rigorous proof.)
Holder’s Inequality.
Σ |ξ_i*η_j|≤ (Σ |ξ_i|^p)^(1/p)(Σ |η_i|^q)^1/q
where p,q are conjugate transposes.
Proof.
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Extremely wonderful lecture sir. Thank you iit
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Mistakes are there in your lecture
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