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Let's consider two Banach Spaces $X, Y$ and a bounded operator $\alpha\colon X \rightarrow Y.$

Question 1: How do you prove that this operator (or certain given operators) have no representation as an integral operator?

Question 2: Are there certain necessary conditions, such that an operator has a representation as an integral operator?

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    What do you mean by integral operator (for general Banach spaces)? – 2017-02-06
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    I see. I just thought of the general representation $(\alpha(x))(t) = \int \alpha(s,t)x(s) ds.$ Actually $X,Y$ should content functions (at least) measurable function from $[0,1]$ to $\mathbb{R}.$ – 2017-02-06
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    I think this is way too broad. – 2017-02-07
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    Ok. My problem is to understand the distinction between integral operators and Hilbert Schmidt operators. Very often Hilbert Schmidt operators are integral operators and the other way arround. But when does this argument not hold? – 2017-02-07

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